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Difference between revisions of "Calc3.2"

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As we are about to see, these conveniences can make the calculation of the cross product relatively simple.  Hopefully, the student will remember the definition of the '''determinant''' of a matrix, because this will make calculating cross products much, much easier.  For those who are unfamiliar with the idea, we will review the concept.
 
As we are about to see, these conveniences can make the calculation of the cross product relatively simple.  Hopefully, the student will remember the definition of the '''determinant''' of a matrix, because this will make calculating cross products much, much easier.  For those who are unfamiliar with the idea, we will review the concept.
  
−
For a matrix <math>\begin{pmatrix}a&b\\c&d\end{pmatrix} \ </math>, we define the determinant <math>\begin{vmatrix}a&b\\c&d\end{vmatrix}</math> to be <math>ad - bc \ </math>.  To compute the determinant of a larger square matrix, then, we choose any row of the matrix and take the left entry, and multiply that number by the determinant of the matrix formed by eliminating the first row and column of the original matrix.  From this, we subtract the next entry in our row times the matrix formed by eliminating THAT row and column of the orginal matrix.  Then we add the next term, and so on and so forth until we have a complete expansion.   
+
For a matrix <math>\begin{pmatrix}a&b\\c&d\end{pmatrix} \ </math>, we define the determinant <math>\begin{vmatrix}a&b\\c&d\end{vmatrix}</math> to be <math>ad - bc \ </math>.  To compute the determinant of a larger square matrix, then, we choose any row of the matrix and take the left entry, and multiply that number by the determinant of the matrix formed by eliminating the first row and column of the original matrix.  From this, we subtract the next entry in our row times the matrix formed by eliminating THAT row and column of the original matrix.  Then we add the next term, and so on and so forth until we have a complete expansion.   
  
 
Below is an example of this technique, expanding along the first row.  
 
Below is an example of this technique, expanding along the first row.  
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Examine this equation, and make sure you understand how each of the matrices on the right hand side were formed.  Once we have reached this point, we can use our definition of a 2x2 determinant to arrive at a value.
 
Examine this equation, and make sure you understand how each of the matrices on the right hand side were formed.  Once we have reached this point, we can use our definition of a 2x2 determinant to arrive at a value.
  
−
We now arrive at a means of calculating the cross product of two vectors.  It turns out that the definition we gave of cross product amounts to
+
We now arrive at a means of calculating the cross product of two vectors.  It [[Calc3.2/Cross Product Proof|turns out]] that the definition we gave of cross product amounts to
  
 
<math>\vec{u}\times\vec{w}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\u_1&u_2&u_3\\w_1&w_2&w_3\end{vmatrix} \ </math>
 
<math>\vec{u}\times\vec{w}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\u_1&u_2&u_3\\w_1&w_2&w_3\end{vmatrix} \ </math>
  
 
Normally, we don't allow vectors (like <math>\vec{i} \ </math>) to be elements of a matrix or a determinant, but in this case, we make an exception because the formula is so useful and easy to remember.  We will encounter a similar exception to the rules when we give an easy way to remember the formula for '''curl.'''
 
Normally, we don't allow vectors (like <math>\vec{i} \ </math>) to be elements of a matrix or a determinant, but in this case, we make an exception because the formula is so useful and easy to remember.  We will encounter a similar exception to the rules when we give an easy way to remember the formula for '''curl.'''
 +
 +
'''In this course, we will frequently switch between the notations <math>(x,y,z), x\vec{i}+y\vec{j}+z\vec{k},</math> and <math>\begin{pmatrix}x\\y\\z\end{pmatrix} \ </math> without discussion of the switch: constantly keep in mind that these are all ways of saying the same thing.'''
  
 
==Vector Functions and Vector Fields==
 
==Vector Functions and Vector Fields==
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Nevertheless, the kinds of phenomenon we see in the multivariate case is very different.  Note that while it is still possible to define a limit from a direction, we don't bother to do so - unlike the one-variable case when a one-sided deriative told us half of the functions behavior at a point, in higher dimensions such a limit tells us almost nothing.  Indeed, the only kind of phenomenon that these kinds of limits are capable of describing are "point discontinuties," which can be "fixed" just be defining the function to be the limit at that point.
 
Nevertheless, the kinds of phenomenon we see in the multivariate case is very different.  Note that while it is still possible to define a limit from a direction, we don't bother to do so - unlike the one-variable case when a one-sided deriative told us half of the functions behavior at a point, in higher dimensions such a limit tells us almost nothing.  Indeed, the only kind of phenomenon that these kinds of limits are capable of describing are "point discontinuties," which can be "fixed" just be defining the function to be the limit at that point.
  
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=[[Derivative#Partial_derivatives|Partial Derivatives]]=
+
=[[Derivative_(calculus)#Partial derivatives|Partial Derivatives]]=
 
:''See [[Calc3.2/Partial derivatives proof|the proof page]] for details.''
 
:''See [[Calc3.2/Partial derivatives proof|the proof page]] for details.''
 
When dealing with functions of more than one variable, one has to be careful about what a "derivative" is.  The operative concept is the ''partial derivative'', which is the derivative of the function with respect to a given argument ''while holding all the other arguments constant''.  It is written like a derivative, but with a different "d" character.
 
When dealing with functions of more than one variable, one has to be careful about what a "derivative" is.  The operative concept is the ''partial derivative'', which is the derivative of the function with respect to a given argument ''while holding all the other arguments constant''.  It is written like a derivative, but with a different "d" character.
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−
Here are some alternative notations for partial derivatives:
+
Here are some other notations for partial derivatives that one might encounter.  We won't use them.
 
:<math>\frac{\partial}{\partial x} f(x,y,z) = \partial_x f(x,y,z) = f_x (x,y,z)</math>
 
:<math>\frac{\partial}{\partial x} f(x,y,z) = \partial_x f(x,y,z) = f_x (x,y,z)</math>
  
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The answer is yes, ''if the derivatives are continuous''.  In problems that arise in practice, the derivatives are always continuous, and switching derivative order is a staple of work in multivariate calculus.
 
The answer is yes, ''if the derivatives are continuous''.  In problems that arise in practice, the derivatives are always continuous, and switching derivative order is a staple of work in multivariate calculus.
  
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=The Jacobian=
+
=More About Coordinate Systems=
  
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Suppose you have a series of functions <math>f_1(x_1,...,x_n), f_2(x_1,...,x_n),..., f_n(x_1,...,x_n) \ </math>.  We define the '''Jacobian''' of these functions to be
+
The study of various coordinate systems (sometimes called "curvilinear coordinate systems") is central to the subject of multivariable calculus. You are already familiar with some of the common coordinate systems such as polar coordinates in 2 dimensions, and spherical or cylindrical coordinates in 3 dimensions. In this section we will examine the general topic of coordinate systems, and how one can perform calculations&mdash;involving vectors, surfaces, integrals, and so on&mdash;directly in specialized coordinate systems.  Physical problems can often be solved more easily in a coordinate system that matches the symmetries of the problem.
  
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<math>\frac{\partial(f_1,f_2,...,f_n)}{\partial(x_1,x_2,...,x_n)} = \begin{vmatrix} \frac{\partial f_1}{\partial x_1} & ... & \frac{\partial f_1}{\partial x_n} \\ ... & & ... \\ \frac{\partial f_3}{\partial x_1} &...&\frac{\partial f_n}{\partial x_n}\end{vmatrix} </math>
+
An extremely important consideration to be aware of is that geometrical or physical phenomena (points, vectors, regions, surfaces, etc.) have a fundamental existence and meaning that is independent of any coordinate system. Coordinate systems simply give us ways to attach numbers to them. Different coordinate systems will attach different numbers to a given vector field, for example, but the vector field has an underlying geometric meaning that does not change.  The various vector field operations have an underlying geometrical or physical meaning, and coordinate systems simply let us perform mathematical calculations on them.
  
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We'll be using Jacobians soon, so it was important to introduce them now.  They are particuarly useful for describing changes in coordinate systems.
+
You have already seen the notion of "geometrical universality" in the definitions of the dot product and the cross product.  They were given purely geometrical definitions that can be visualized without regard to any coordinate system.  Than, for any Cartesian coordinate system, these operations were defined in terms of the components of the vectors in that system, and it was shown that those results matched the purely geometrical definitions.  What we are going to do next is develop the tools to define vectors, and the dot product and cross product, in ''any coordinate system at all''.  Later, we will introduce new operations&mdash;the divergence, the curl, and various integrals, and show how to calculate them in any coordinate system.
  
 +
For example, Stokes' theorem makes a very powerful geometric statement about integrals of various vector fields over very general surfaces and lines.  The concepts appearing in the theorem are purely geometric, and therefore independent of the coordinate system.  Once we know how to manipulate those concepts (vector fields, integrals, and the "curl" operator) in any coordinate system, we will be able to choose a coordinate system that makes the surface simple.  Specifically, we will change from the usual x/y/z Cartesian system to a u/v/w system for which the surface is defined by holding w=0.  (Another way of looking at this is that the surface has been "parameterized" in terms of parameters u and v.)  Once this is done, we will be able to calculate the curl operator, perform the integration, and prove Stokes' theorem in the u/v/w system.
  
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:::::::::::::::''expand''
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As another example, Maxwell's equations make statements about the curl and divergence of the electric and magnetic field.  While the formulas for curl and divergence are simpler in Cartesian coordinates than in spherical, some problems, such as the electric field in the vicinity of a point charge, have symmetry that makes spherical coordinates more natural.  When we work out the divergence theorem in spherical coordinates, we will be able to solve problems of this sort.
  
−
=Introduction to Integrals in Multivariable Calculus=
+
We will use "u" and "v" in many of our general examples of alternative coordinate systems in 2 dimensions, and u, v, and w in 3 dimensions.  For the very common cases of polar coordinates in 2 dimensions and spherical in 3 dimensions, we will often use the more familiar r/<math>\theta</math> or r/<math>\theta</math>/<math>\phi</math>.
−
In multivariable calculus, and in the related areas of physics, we will be extending the notion of the integral well beyond the simple case of the [[Riemann Integral]] from elementary calculus.  These extensions will consist of
 
−
*Integration over higher-dimensional regions, such as surfaces and volumes, with sophisticated ways of specifying these regions.
 
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*Integration of functions, typically representing physical quantities, such as integration of density to get total mass, or integration of a gravitational or electric field along a path to get potential energy or electric potential.
 
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*Integration of vector fields, for example, a magnetic field, across a surface to get the total flux.
 
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*Integration in coordinate systems other than the familiar Cartesian coordinates, such that the physically correct result is always obtained, independently of the coordinate system used.
 
−
In all cases, the additional sophistication in the integrals will just involve things that are done to set up the mathematical problem.  The actual integration will always be the same&mdash;one-dimensional Riemann integration of some function.  That is, the integration that we will do will always reduce to the familiar definite integration for elementary calculus:
 
−
::<math>\int_a^bf(x) dx</math>
 
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Recall that the solution to such a problem is easy to state, though maybe not easy to solve:  Find a function <math>g(x)</math> such that <math>f</math> is the derivative of <math>g</math>:
 
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::<math>f(x) = \frac{dg}{dx}</math>
 
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Then the integral is the difference  between the values of <math>g</math> at <math>a</math> and <math>b</math>:
 
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::<math>\int_a^bf(x) dx = g(b) - g(a)</math>
 
  
−
=Double and Triple Integrals=
+
We always have the coordinates of each system representable as functions of the other.  For example, the transformation between polar coordinates and Cartesian coordinate is this:
 +
::<math>x = r \cos \theta\,</math>
 +
::<math>y = r \sin \theta\,</math>
  
−
We will not give rigorous proofs of the various theorems here.  Such proofs are taught in college-level analysis and differential geometry courses, and often dwell upon esoteric issues of singularities and infinities.
+
::<math>r = \sqrt{x^2 + y^2}\,</math>
 +
::<math>\theta = \tan^{-1} (y/x)\,</math>
  
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The first thing we do is develop more sophisticated ways of measuring areas of plane figures.  Ordinary integration can give us the area of a figure with a straight horizontal bottom edge, straight vertical left and right edges, and an arbitrary function giving the top edge.  Suppose we have a circle of radius R centered at (A, B), as shown in the figure at the right.
+
In spherical coordinates:
−
[[File:Dblintegralarea.JPG|thumb|The blue area is the first integral, and the blue and yellow areas combined are the second integral.]]
+
::<math>x = r \sin \theta \cos \phi\,</math>
−
(We have placed the circle completely in the upper-right quadrant so that we won't have to deal with visualizing negative areas&mdash;this is just for illustrative purposes.  The technique works in all cases.)
+
::<math>y = r \sin \theta \sin \phi\,</math>
 +
::<math>z = r \cos \theta\,</math>
  
−
The blue area is given by:
+
::<math>r = \sqrt{x^2 + y^2 + z^2}\,</math>
−
::<math>\int_{A-R}^{A+R}\ B-\sqrt{R^2 - (x-A)^2}\ dx</math>
+
::<math>\theta = \tan^{-1} (\sqrt{x^2 + y^2} / z)\,</math>
 +
::<math>\phi = \tan^{-1} (y/x)\,</math>
  
−
The blue and yellow areas combined is:
+
==The Jacobian==
−
::<math>\int_{A-R}^{A+R}\ B+\sqrt{R^2 - (x-A)^2}\ dx</math>
 
  
−
The area of the circle is the yellow area alone, which is the difference between the two:
+
Much of what happens in multivariable calculus involves the ''[[partial derivative]]s'' of the coordinates in one system with respect to the coordinates in the other.  It is very useful to represent these as a matrix:
−
::<math>\int_{A-R}^{A+R}\ 2\ \sqrt{R^2 - (x-A)^2}\ dx</math>
+
::<math>\begin{bmatrix} \displaystyle\frac{\partial x}{\partial r} & \displaystyle\frac{\partial x}{\partial \theta} \\ \\ \displaystyle\frac{\partial y}{\partial r} & \displaystyle\frac{\partial y}{\partial \theta}\end{bmatrix}\,</math>
 +
Or, going in the other direction:
 +
::<math>\begin{bmatrix} \displaystyle\frac{\partial r}{\partial x} & \displaystyle\frac{\partial r}{\partial y} \\ \\ \displaystyle\frac{\partial \theta}{\partial x} & \displaystyle\frac{\partial \theta}{\partial y}\end{bmatrix}\,</math>
  
−
Before going any further, we use the change-of-variable theorem to shift x (which also shows that A and B don't matter; the area is the same anywhere):
+
This is the ''derivative matrix'' of the transformation from one coordinate system to the other.  It is often called the '''Jacobian matrix''', though some authors use the term "Jacobian" specifically to refer to the matrix's determinant. To avoid any confusion on this point, we will always put square brackets around matrices, and use absolute value signs to denote determinants.
−
::<math>\int_{-R}^{R}\ 2\ \sqrt{R^2 - x^2}\ dx</math>
 
−
(We'll have a lot more to say about the change-of-variable theorem later, but now we are just using the version from elementary calculus, that helps us calculate ordinary integrals.)
 
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::Of course we could evaluate this integral, perhaps with a table of integrals, or a computer program, or a web site, getting:
 
−
::::<math>x \sqrt{R^2 - x^2} + R^2 \sin^{-1} (x/R)\left|\right.^{R}_{-R}</math>
 
  
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Let's look at the integrand, and its geometrical interpretation, more carefully.  <math>2\ \sqrt{R^2 - x^2}</math> is the height of a thin vertical strip of the circle, at a given x.  We could write that as an integral in its own right:
+
The notation for the Jacobian matrix of u/v/w in terms of x/y/z is written:
−
::<math>\int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}}\ 1\ dy</math>
+
::<math>J(u, v, w / x, y, z) = \frac{\partial(u, v, w)}{\partial(x, y, z)} = \begin{bmatrix} \displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y} & \displaystyle\frac{\partial u}{\partial z} \\ \\ \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y} & \displaystyle\frac{\partial v}{\partial z} \\ \\ \displaystyle\frac{\partial w}{\partial x} & \displaystyle\frac{\partial w}{\partial y} & \displaystyle\frac{\partial w}{\partial z}\end{bmatrix} </math>
−
This means that the area of the circle is:
 
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::<math>\int_{-R}^{R}\left(\int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}}\ 1\ dy\right) dx</math>
 
  
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The geometrical interpretation is that we have divided the region (circle, in this case) into thin vertical strips abutting each other from left to right, and then divided each strip into tiny squares, abutting each other from bottom to top, and added everything up.
+
More generally, it is:
 +
::<math>J(f_1,f_2,...,f_n / x_1,x_2,...,x_n) = \frac{\partial(f_1,f_2,...,f_n)}{\partial(x_1,x_2,...,x_n)} = \begin{bmatrix} \displaystyle\frac{\partial f_1}{\partial x_1} & ... & \displaystyle\frac{\partial f_1}{\partial x_n} \\ ... &  & ... \\ \displaystyle\frac{\partial f_3}{\partial x_1} &...&\displaystyle\frac{\partial f_n}{\partial x_n}\end{bmatrix} </math>
 +
and the determinant is:
 +
::<math>\begin{vmatrix} \frac{\partial f_1}{\partial x_1} & ... & \frac{\partial f_1}{\partial x_n} \\ ... &  & ... \\ \frac{\partial f_3}{\partial x_1} &...&\frac{\partial f_n}{\partial x_n}\end{vmatrix} </math>
  
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In general, the area of a region is:
+
For polar coordinates, the Jacobian matrices are:
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::<math>\int_{leftmost point of the region}^{righttmost point of the region}\left(\int_{bottom of the region at a given x}^{top of the region at a given x}\ 1\ dy\right) dx</math>
+
::<math>J(r, \theta / x, y) = \frac{\partial(r, \theta)}{\partial(x, y)} = \begin{bmatrix} \displaystyle\frac{\partial r}{\partial x} & \displaystyle\frac{\partial r}{\partial y} \\ \\ \displaystyle\frac{\partial \theta}{\partial x} & \displaystyle\frac{\partial \theta}{\partial y} \end{bmatrix}</math>
 +
and:
 +
::<math>J(x, y / r, \theta) = \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{bmatrix} \displaystyle\frac{\partial x}{\partial r} & \displaystyle\frac{\partial x}{\partial \theta} \\ \\ \displaystyle\frac{\partial y}{\partial r} & \displaystyle\frac{\partial y}{\partial \theta} \end{bmatrix}</math>
  
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This is two nested integrals, or a ''double integral''.  The quantity in parentheses is the integrand of the outer integral.  As such, it is a function of x, so its limits are permitted to depend on x.  In more complicated problems, such as finding the total electric charge, we might replace the inner integrand ("1" in the present example) with the density of electric charge, which could be a function of x and y.
+
Working these out, we have
 +
::<math>J(x, y / r, \theta) = \begin{bmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix}</math>
 +
with determinant r, and:
 +
::<math>J(r, \theta / x, y) = \begin{bmatrix} \displaystyle\frac{x}{\sqrt{x^2 + y^2}} & \displaystyle\frac{y}{\sqrt{x^2 + y^2}} \\ \\ \displaystyle\frac{- y}{x^2 + y^2} & \displaystyle\frac{x}{x^2 + y^2} \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ \\ \displaystyle\frac{- \sin \theta}{r} & \displaystyle\frac{\cos \theta}{r} \end{bmatrix}</math>
 +
with determinant 1/r.
  
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This is the essence of how double (and later, triple) integrals are used to calculate areas, volumes, and other things over 2 or 3-dimensional regions.  The big parentheses are usually omitted, but their implict meaning must be followed.  The innermost integral, and its limits of integration, relate to the innermost "d" symbol, and so on.
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An interesting theorem is that these matrices are inverses of each other.  (It follows that their determinants are reciprocals of each other.)
  
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A triple integral could be written:
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Exercise: Prove that the matrix product of the J matrices of two sequential coordinate changes is the J matrix of the combined change.  That is, if
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::<math>\int_{A}^{B}\int_{C(x)}^{D(x)}\int_{E(x, y)}^{F(x, y)}\ G(x, y, z)\ dz dy dx</math>
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::<math>f_1,f_2,...,f_n \to g_1,g_2,...,g_n</math>
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Each successive integral may use the values of all the outer integration variables in specifying its limits.
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and
 +
::<math>g_1,g_2,...,g_n \to h_1,h_2,...,h_n</math>
 +
are two coordinate changes, so that
 +
::<math>f_1,f_2,...,f_n \to h_1,h_2,...,h_n</math>
 +
is the result of doing both, then
 +
::<math>\frac{\partial(h_1,h_2,...,h_n)}{\partial(f_1,f_2,...,f_n)} = \frac{\partial(h_1,h_2,...,h_n)}{\partial(g_1,g_2,...,g_n)}\ \ \frac{\partial(g_1,g_2,...,g_n)}{\partial(f_1,f_2,...,f_n)}</math>
  
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Exercise [belongs in a separate section, no doubt]: Set up the triple integral giving the volume of a sphere of radius R, and evaluate same.
+
Hint: This is really just the chain rule for partial derivatives.  The adding up of terms in the chain rule is the same as the adding up of products in a matrix multiplication.
  
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This use of double integrals to calculate areas may seem like excessive make-work, and using "1" as the integrand may seem boring, but this sort of analysis is the basis for everything we will do.
 
  
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We could have performed the double integration in the other order, dividing the circle into horizontal strips first, and then subdividing those, so that dx is the "inner" integral and dy the "outer" one.  This would have gotten the same result.  The theorem that says that this is so is ''Fubini's theorem''.  Proving it is outside the scope of this course, or of conservapedia.
+
It follows from this that, if the second change is just the inverse of the first, bringing us back to the original coordinate system, the two matrices are inverses.
  
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Now, given that Fubini's theorem says that the order of the integrations doesn't matter, we can "abstract away" that order, say that we are really just "integrating over a region", and use a more abstract notation.  Once we have defined a 2-dimensional region R, we can just write two successive integral signs with the subscript R instead of the written-out limits of integration, and use the symbol "dA" to mean "infinitesimal piece of area".  (Recall that "dx" means, in an informal way, "infinitesimal bit of length".)  The integral would look like:
+
The coordinates should always be chosen such that the Jacobian determinant is positive.  If it isn't, exchange the order of two of the coordinates.  This ensures that the coordinate system is "right handed".  (Or more precisely, that it has the same handedness as the Cartesian system.)  As an example, the coordinate order (r/<math>\theta</math>) is correct for polar coordinates; (<math>\theta</math>/r) is not.  The order (r/<math>\theta</math>/<math>\phi</math>) is correct for spherical coordinates; (r/<math>\phi</math>/<math>\theta</math>) is not.
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::<math>\iint_{R}\ 1\ dA</math>
 
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or, more generally:
 
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::<math>\iint_{R}\ f(r)\ dA</math>
 
−
where r is some way of indicating a point in the region.  We will always use a coordinate system, so, for example, if we are using polar coordinates, the integral might look like:
 
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::<math>\iint_{R}\ f(r, \theta)\ dA</math>
 
  
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For triple integrals over a volume, we do something similar:
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The Jacobian determinant must not be zero.  (Equivalently, the Jacobian matrix must not be singular.)  The astute reader will notice that this means that polar coordinates don't work at the origin, and spherical coordinates don't work at the north and south poles.
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::<math>\iiint_{R}\ f(r, \theta, \phi)\ dV</math>
 
  
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The symbols dA and dV are often called the "area element" and "volume element", respectively.
+
==Vectors in Arbitrary Coordinate Systems==
  
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==Change of Variable==
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Having gotten the preliminary concepts out of the way, we next examine what vectors, and vector operations such as the dot product, look like in arbitrary coordinate systems.
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The change-of-variable theorem is an extremely important tool of integral calculus.  It is even more important in multivariate calculus, since it is central to the concepts of coordinate systems and coordinate system change operations.  It is implicit in nearly all integrals over curves, surfaces, and volumes.
 
  
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We would like to be able to perform the integrations of the previous section in coordinate systems other than plain Cartesian coordinates.  For example, the integral to find the area of a circle would be easier to set up if we were using polar coordinates, because a circle is trivial to describe in polar coordinates.  To do this, we need to revisit the change-of-variable theorem from elementary calculus, and extend it to higher dimensions.
+
A vector has different components in different coordinate systems.  (It's still the same vector, of course.)  If the components of the vector <math>\vec{A}</math>, in the u/v/w coordinate system, are:
 +
::<math>\vec{A} = \begin{bmatrix} A_u \\ A_v \\ A_w \end{bmatrix}</math>
 +
and, in the standard Cartesian system the components are:
 +
::<math>\vec{A} = \begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix}</math>
 +
then the transformation rule is:
 +
::<math>\begin{bmatrix} A_u \\ A_v \\ A_w \end{bmatrix} = \begin{bmatrix} \displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y} & \displaystyle\frac{\partial u}{\partial z} \\ \\ \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y} & \displaystyle\frac{\partial v}{\partial z} \\ \\ \displaystyle\frac{\partial w}{\partial x} & \displaystyle\frac{\partial w}{\partial y} & \displaystyle\frac{\partial w}{\partial z}\end{bmatrix} \begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix}</math>
  
−
Basically, the change-of-variable theorem of elementary calculus says that algebraic substitution and manipulations actually work, even when the manipulations involve the "d" symbol.  (Remember that things like "dx" are ''infinitesimals''; their actual value would have to be considered to be zero.  By a miracle of calculus notation, we can manipulate them anyway.)
+
In other words, the transformation of vector components from x/y/z to u/v/w is just a matrix multiplication by the Jacobian matrix
 +
::<math>\begin{bmatrix}J(u, v, w / x, y, z)\end{bmatrix}</math>
 +
To go the other way, multiply by the matrix
 +
::<math>\begin{bmatrix}J(x, y, z / u, v, w)\end{bmatrix}</math>
  
−
Here is a simple example, not (yet) involving integrals.  Suppose u is a function of y, and y is a function of x, as follows:
+
We make a simplification to avoid having our notation get completely out of control.  In keeping with our goal of expressing everything just in terms of the new coordinate system:
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::<math>y = \log u\,</math>
+
*The matrix that we have been calling [J(x, y, z / u, v, w)] we will call just [J].  That is, [J] for a coordinate system means [J(Cartesian/the coordinate system under discussion)].
−
::<math>u = \sin x\,</math>
+
*The matrix that goes the other way, formerly called [J(u, v, w / x, y, z)] we will call <math>[J]^{-1}</math>.  It's just the inverse of the first matrix.
−
We know that
+
We will call [J] the "Jacobian of the coordinate system".
−
::<math>\frac{dy}{du} = \frac{1}{u}</math>
 
−
::<math>\frac{du}{dx} = \cos x</math>
 
−
Can we get <math>\frac{dy}{dx}</math> from this?  The change-of-variable theorem says that we can multiply the derivatives and "cancel" the du:
 
−
::<math>\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{1}{u}\ \cos x</math>
 
−
We can remove u from this, getting the answer in terms of x:
 
−
::<math>\frac{dy}{dx} = \frac{\cos x}{\sin x} = \cot x</math>
 
−
This is the same answer as the [[chain rule]] would have given:
 
−
::<math>y = \log (\sin x)\,</math>
 
−
::<math>\frac{dy}{dx} = \frac{1}{\sin x} \cos x</math>
 
−
It's easy to see why this is true&mdash;the multiplication in the chain rule is just the multiplication from which we canceled the dy.
 
  
−
We can do the same ting for integrals, observing why the "dx" symbol in an integral is such an important part of the notation that integrals use.  Given:
+
::The astute reader will notice that this is not a good definition.  [J] depends on the choice of the "reference" Cartesian system, so we're not justified in calling it just "the Jacobian".  But it turns out that it won't make any difference.  We will always get the same answers, no matter what reference system we used.  See the exercises.
−
::<math>\int \cot x\ dx</math>
 
−
we introduce a new variable u:
 
−
::<math>u = \sin x\,</math>
 
−
so the integral is
 
−
::<math>\int \frac{\cos x}{u} dx</math>
 
−
We have
 
−
::<math>\frac{du}{dx} = \cos x</math>
 
−
So, taking the usual liberties with the notation:
 
−
::<math>dx = \frac{du}{\cos x}</math>
 
−
so the integral is:
 
−
::<math>\int \frac{\cos x}{u} \frac{du}{\cos x} = \int \frac{du}{u} = \log u = \log \sin x</math>
 
−
Tricks like this are staples of integration technique.
 
  
−
What we have really done is a change of coordinate system.  Whenever we change coordinate systems while calculating integrals, we are actually using the change-of-variable theorem.
+
Therefore:
 +
::<math>\begin{bmatrix} A_u \\ A_v \\ A_w \end{bmatrix} = [J]^{-1} \begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix}</math>
 +
and:
 +
::<math>\begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix} = [J] \begin{bmatrix} A_u \\ A_v \\ A_w \end{bmatrix}</math>
  
−
Now consider what happens when we try to change the previous integral over a circle in Cartesian coordinates to the equivalent integral in polar coordinates.  The earlier integral was:
+
For polar coordinates in two dimensions, we have:
−
::<math>\int_{-R}^{R}\int_{-\sqrt{R^2 - x^2}}^{\sqrt{R^2 - x^2}}\ 1\ dy dx</math>
+
::<math>\begin{bmatrix} A_r \\ A_\theta \end{bmatrix} = \begin{bmatrix}J(r, \theta / x, y)\end{bmatrix} \begin{bmatrix} A_x \\ A_y \end{bmatrix} = \begin{bmatrix} \displaystyle\frac{\partial r}{\partial x} & \displaystyle\frac{\partial r}{\partial y} \\ \\ \displaystyle\frac{\partial \theta}{\partial x} & \displaystyle\frac{\partial \theta}{\partial y}\end{bmatrix} \begin{bmatrix} A_x \\ A_y \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ \\ \displaystyle\frac{- \sin \theta}{r} & \displaystyle\frac{\cos \theta}{r} \end{bmatrix} \begin{bmatrix} A_x \\ A_y \end{bmatrix}</math>
−
That would change to:
+
And, going the other way:
−
::<math>\int_{0}^{2\pi}\int_{0}^{R}\ 1\ ??\ dr d\theta</math>
+
::<math>\begin{bmatrix} A_x \\ A_y \end{bmatrix} = [J] \begin{bmatrix} A_r \\ A_\theta \end{bmatrix} = \begin{bmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix} \begin{bmatrix} A_r \\ A_\theta \end{bmatrix}</math>
−
The limits of integration are, as expected, easy to set up.  But we need something that is the equivalent of
 
−
::<math>\frac{d\ x, y}{d\ r, \theta}</math>
 
−
to use as the coordinate-change multiplier.
 
  
−
The multiplier to use is called the ''[[Jacobian]]'' of the coordinate change.  It is the [[determinant]] of the matrix of partial derivatives. That matrix is:
+
Let's look at some vectors whose components are given in general coordinates&mdash;there are some important and striking aspects of this.
  
−
::<math>\begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{pmatrix}</math>
+
In polar coordinates, a "radial basis vector" at any point has components:
 +
::<math>(A_r=1, A_\theta=0)\,</math>
 +
Its Cartesian components (from the formulas above) are:
 +
::<math>(A_x=\cos \theta, A_y=\sin \theta)\,</math>
 +
The actual (Cartesian) direction in which the vector points depends on its location.  It always points radially outward.  Its length is <math>\sqrt{A_x^2 + A_y^2} = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1\,</math>.  So the radial basis vector is a unit vector.
  
−
The Jacobian is written <math>J(x, y / r, \theta)\,</math>.  Working out the derivatives and the determinant, its value in thsi case is r.
+
The "angular basis vector" has components:
 +
::<math>(A_r=0, A_\theta=1)\,</math>
 +
Its Cartesian components are:
 +
::<math>(A_x=- r \sin \theta, A_y=r \cos \theta)\,</math>
 +
It always points "laterally", at right angles to the radial direction.  More surprisingly, its length is <math>\sqrt{A_x^2 + A_y^2} = r\,</math>.  The angular basis vector is ''not'' a unit vector!  It is bigger when farther away from the origin, even though the sum of the squares of its components is 1.<ref>Some textbooks use a definition of vector components such that basis vectors always have length 1.  Such a treatment imposes the requirement that the coordinate lines be perpendicular everywhere.  The formulas for vector operations are more complicated and less elegant when this is done.</ref>  In general curvilinear coordinates, the "square root of the sum of the squares of the components" rule is not correct for the length of a vector.  The correct rule will be given [[Calc3.2#The Length of a Vector in Arbitrary Coordinates|below]].
  
−
So the area of a circle is:
+
===The Dot Product in Arbitrary Coordinates===
−
::<math>\int_{0}^{2\pi}\int_{0}^{R}\ 1\ J(x, y / r, \theta) dr d\theta = \int_{0}^{2\pi}\int_{0}^{R}\ r dr d\theta = \int_{0}^{2\pi}\frac{R^2}{2} d\theta = \pi R^2</math>
 
  
−
==Integration Over Parametrically Defined Regions==
+
We need a better formula for the dot product than the rule that works for Cartesian coordinates:
−
A variation of this method is used when integrating over curves in 2 or 3-dimensional spaces, or surface in 3-dimensional spaces.  Recall that a parametric description of such a thing is closely related to a change of coordinate system, but with different numbers of coordinates in the two systems.  This means that the matrix of partial derivatives is not square, so it has no determinant.
+
::<math>\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z\,</math>
  
−
The general problem, in arbitrary dimensions, goes deeply into advanced topics of differential geometry.  We will state the methods, without proof, for the cases of parametrically defined curves and surfaces.
+
Given two vectors <math>\vec{A}</math> and <math>\vec{B}</math>, described in a u/v/w coordinate system, we can find their dot product by converting to Cartesian coordinates and using the simple rule that we know, which is the sum of the pairwise products of the vector components.  This means that, at least momentarily, we will have to keep track of the vectors' components in both the Cartesian system, which we have been calling <math>A_x</math>, <math>A_y</math>, and <math>A_z</math>, and in the new coodinate system, which we have been calling <math>A_u</math>, <math>A_v</math>, and <math>A_w</math>.  To do the required matrix manipulations, we have to give the components numbers instead of names.  Things like <math>A_{u_1}</math>, <math>A_{u_2}</math>, and <math>A_{u_3}</math> are just too unwieldy.<ref>This is a well known notational problem in multivariable calculus.  It is one of the reasons why tensor algebra textbooks are so hard to read.</ref>  So we will make this simplification:
 +
*The components of a vector in the new coordinate system (formerly <math>A_u</math>, <math>A_v</math>, and <math>A_w</math>) will just be called <math>A_1</math>, <math>A_2</math>, and <math>A_3</math>.
 +
*The components of a vector in the "reference" Cartesian system will be written with a tilde: <math>\tilde{A_1}</math>, <math>\tilde{A_2}</math>, and <math>\tilde{A_3}</math>.
  
−
==Curve Integrals==
+
Once we reformulate the rules for vectors in terms of the coordinate system of interest, we will only use <math>A_1</math>, <math>A_2</math>, and <math>A_3</math> (or <math>A_u</math>, <math>A_v</math>, and <math>A_w</math>), and we will never have to look at another tilde again.
−
For a curve in a 2-dimensional plane or 3-dimensional volume, let the parameter (the single coordinate in the 1-dimensional curve) be denoted by t.  Work out the partial derivatives, and form the vector
 
−
::<math>\left(\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}\right)</math>
 
−
for a curve in a plane, or
 
−
::<math>\left(\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}, \frac{\partial z}{\partial t}\right)</math>
 
−
for a curve in a volume.
 
−
Let J (not really the Jacobian, but we treat it that way) be the [[norm]] of that vector, that is, the square root of the sum of the squares of its 2 or 3 components.
 
−
::<math>J = \Bigg\Vert\left(\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}, \frac{\partial z}{\partial t}\right)\Bigg\Vert = \sqrt{\left(\frac{\partial x}{\partial t}\right)^2 + \left(\frac{\partial y}{\partial t}\right)^2 + \left(\frac{\partial z}{\partial t}\right)^2}</math>
 
−
Then
 
−
::<math>\int f(t)\ J dt</math>
 
−
is the integral over the curve, in terms of the parameter t.
 
  
−
Example:
+
We can now calculate the dot product of <math>\vec{A}</math> and <math>\vec{B}</math>, from their components and the Jacobian.  We know that
 +
::<math>\vec{A} \cdot \vec{B} = \sum_i \tilde{A_i} \tilde{B_i}</math>
 +
in terms of the components in the reference system.  We also have the transformation rule in terms of the Jacobian matrix.
 +
::<math>\tilde{A_i} = \sum_j [J(x, y, z / u, v, w)]_{ij} A_j</math>
 +
This is just matrix multiplication.  The summation is from 1 to 3, or whatever the dimension is.
  
−
Suppose a circle of radius R is described in terms of a parameter t, as
+
By our new convention, this is just the Jacobian matrix [J], so
−
::<math>x = R \cos t\,</math>
+
::<math>\tilde{A_i} = \sum_j [J]_{ij} A_j</math>
−
::<math>y = R \sin t\,</math>
+
::<math>\tilde{B_i} = \sum_k [J]_{ik} B_k</math>
−
We have
+
::<math>\vec{A} \cdot \vec{B} = \sum_i \tilde{A_i} \tilde{B_i} = \sum_{ijk} [J]_{ij} [J]_{ik} A_j B_k</math>
−
::<math>\frac{\partial x}{\partial t} = - R \sin t</math>
+
As promised, no more tildes!
−
::<math>\frac{\partial y}{\partial t} = R \cos t</math>
 
−
::<math>J = \sqrt{R^2 \sin^2 t + R^2 \cos^2 t} = \sqrt{R^2} = R</math>
 
−
So, to integrate any function over the full circle (with t running from 0 to <math>2 \pi</math>), we have:
 
−
::<math>\int_0^{2\pi} f(t)\ R dt</math>
 
−
The length of the curve (that is, the circumference of the circle) is:
 
−
::<math>\int_0^{2\pi} 1\ R\ dt = 2 \pi R</math>
 
  
−
==Surface Integrals==
+
We can combine the two Jacobians into one matrix, and then use that whenever we need a dot product.  Let:
−
For a surface in a 3-dimensional volume, let the parameters be u and v.  Work out the six partial derivatives, and form these two vectors:
+
::<math>[g]_{jk} = \sum_i [J]_{ij} [J]_{ik} = \sum_i [J]^t_{ji} [J]_{ik}</math>
−
::<math>\stackrel{\textstyle{\rightarrow}}{U} = \left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u}\right)</math>
+
That is, [g] is just the matrix product of the transpose of [J] and [J] itself.
−
::<math>\stackrel{\textstyle{\rightarrow}}{V} = \left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}, \frac{\partial z}{\partial v}\right)</math>
+
::<math>[g]_{jk} = ([J]^t [J])_{jk}\ \ \ or\ \ \ [g] = [J]^t [J]\,</math>
−
Calculate their cross product, and let J be the norm of that.  Use the [[cross product]] formula.
+
Then:
−
::<math>J = ||\stackrel{\textstyle{\rightarrow}}{U} \times\stackrel{\textstyle{\rightarrow}}{V}||</math>
+
::<math>\vec{A} \cdot \vec{B} = \sum_{jk} [g]_{jk} A_j B_k</math>
−
Then
+
This is the rule for the dot product in general coordinates.  The matrix [g] is called the ''metric tensor'' or just the ''metric''.  We will have occasion to use its determinant, which we will call just g.  Actually, its square root is more commonly used.
−
::<math>\int\int f(u, v)\ J\ dv\ du</math>
 
−
is the integral over the surface, in terms of the parameters u and v.
 
  
−
Unless one is careful to choose the order of the parameters so the "outward" direction of the surface follows a right-hand rule, and use that order in forming the cross product, there may be ambiguity in the sign of J.  In practice, one just figures out what to do.
+
For those who really like equations that express things in terms of matrix multiplication, this could be written as:
 +
::<math>\vec{A} \cdot \vec{B} = \begin{bmatrix} A_1 ... A_N \end{bmatrix} [g] \begin{bmatrix} B_1 \\ ... \\ B_N \end{bmatrix}</math>
 +
The first vector is written in the form of a row instead of the more traditional column.  This gives a 1xN matrix times am NxN matrix times an Nx1 matrix, which is a 1x1 matrix.  The single value in that matrix is the dot product.
  
−
Example:
+
To review what we have for two popular systems, in 2-dimensional polar coordinates:
  
−
Parameterize a sphere of radius R in terms of <math>\theta</math> and <math>\phi</math>.  (These are two of the same coordinates that are used in 3-dimensional spherical coordinates&mdash;<math>r</math>, <math>\theta</math> and <math>\phi</math>&mdash;but <math>r</math> is held constant, so it is no longer a coordinate.)
+
::<math>[J] = \begin{bmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix}</math>
−
::<math>x = R \sin \theta \cos \phi\,</math>
 
−
::<math>y = R \sin \theta \sin \phi\,</math>
 
−
::<math>z = R \cos \theta\,</math>
 
  
−
::<math>\frac{\partial x}{\partial \theta} = R \cos \theta \cos \phi</math>
+
::<math>[J]^{-1} = \begin{bmatrix} \cos \theta & \sin \theta \\ \\ \displaystyle\frac{- \sin \theta}{r} & \displaystyle\frac{\cos \theta}{r} \end{bmatrix}</math>
−
etc.
 
−
::<math>\stackrel{\textstyle{\rightarrow}}{U} = \left(R \cos \theta \cos \phi, R \cos \theta \sin \phi, - R \sin \theta\right)</math>
 
−
::<math>\stackrel{\textstyle{\rightarrow}}{V} = \left(- R \sin \theta \sin \phi, R \sin \theta \cos \phi, 0\right)</math>
 
−
::<math>\stackrel{\textstyle{\rightarrow}}{U} \times\stackrel{\textstyle{\rightarrow}}{V} = R^2 \left(\sin^2 \theta \cos \phi, \sin^2 \theta \sin \phi, \sin \theta \cos \phi\right)</math>
 
−
::<math>J = ||\stackrel{\textstyle{\rightarrow}}{U} \times\stackrel{\textstyle{\rightarrow}}{V}|| = R^2 \sin \theta</math>
 
  
−
So an integral over the entire surface would be:
+
::<math>[g] = \begin{bmatrix} 1 & 0 \\ 0 & r^2 \end{bmatrix}</math>
−
::<math>\int_0^{\pi}\int_0^{2\pi} f(\theta, \phi)\ R^2 \sin \theta\ d\phi\ d\theta</math>
 
−
The area of a sphere is:
 
−
::<math>\int_0^{\pi}\int_0^{2\pi} R^2 \sin \theta\ d\phi\ d\theta = \int_0^{\pi} R^2 \sin \theta \left(\int_0^{2\pi} d\phi\right) d\theta = 2 \pi R^2 \int_0^{\pi} \sin \theta\ d\theta = 4 \pi R^2</math>
 
  
−
To find, for example, the area of a sphere only from <math>\theta = 0\,</math> to <math>\theta = T\,</math> (that is, the area of the Earth north of latitude <math>\pi/2 - T\,</math>), we have
+
::<math>\sqrt{g} = r</math>
−
::<math>\int_0^{T}\int_0^{2\pi} R^2 \sin \theta\ d\phi\ d\theta = 2 \pi R^2 \int_0^{T} \sin \theta\ d\theta = 2 \pi R^2\ (1 - \cos T)</math>
 
  
−
==Definition of the Double Integral==
+
For spherical coordinates in 3 dimensions (we will need these soon):
−
The definition of the double integral follows much that same pattern as the definition in the single variable case.  As in the single variable case, we divide the part of the domain we're interested in into pieces, and multiply the size of those pieces by a value of the function on that piece.  Adding up all our products gives us a Riemann sum, and taking the limit as the pieces' size goes to zero gives us the integral.
 
  
−
In the case of a function of two variables, say, ''x'' and ''y'', the domain will naturally be the ''xy''-plane, or some region in it ''R''.  A region could be a disc (a circle plus its interior), a square and its interior, or any other shape you can think of.
+
::<math>[J] = \begin{bmatrix} \sin \theta \cos \phi & r \cos \theta \cos \phi & - r \sin \theta \sin \phi \\ \sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\ \cos \theta & - r \sin \theta & 0 \end{bmatrix}</math>
  
−
While technically it doesn't matter too much what shapes we divide the region into, or that they be the same size, etc., we're going to use as simple and straightforward a definition as we can, to make our lives easier when we try to actually compute these things. 
+
::<math>[J]^{-1} = \begin{bmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \\ \displaystyle\frac{\cos \theta \cos \phi}{r} & \displaystyle\frac{\cos \theta \sin \phi}{r} & - \displaystyle\frac{\sin \theta}{r} \\ \\ - \displaystyle\frac{\sin \phi}{r \sin \theta} & \displaystyle\frac{\cos \phi}{r \sin \theta} & 0 \end{bmatrix}</math>
−
[[Image:Approximation.jpg|thumb|200px|right|As the size of the squares decrease, they approximate the region better and better.]]
 
−
So we divide ''R'' into ''n'' square subregions <math>R_k, 1\leq k\leq n</math>, and in each region pick a point <math>(x_k,y_k)\in R_k</math>.  Note that all the <math>R_k</math> might not cover all of our original region ''R''; but as the size of our squares decreases, they approximate the region better and better - in the limit, they will cover the whole thing.
 
  
−
The Riemann sum for a function ''f'' is then simple enough to construct, using <math>A</math> to note the area of one of our squares: <math>S= \sum_{k=1}^n f(x_k,y_k)A</math>.  If we use <math>dA=dxdy</math> to note we are differentiating with respect to area, we can define our double integral:
+
::<math>[g] = \begin{bmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2 \theta \end{bmatrix}</math>
  
−
<math>\iint_R f(x,y)dA = \lim_{n\to\infty,A\to 0} \sum_{k=1}^n f(x_k,y_k)A</math>
+
::<math>\sqrt{g} = r^2 \sin \theta</math>
  
−
==Computation of the Double Integral==
+
So, in polar coordinates:
 +
::<math>\vec{A} \cdot \vec{B} = A_1 B_1 + r^2 A_2 B_2</math>
 +
or, going back to giving the coordinates names:
 +
::<math>\vec{A} \cdot \vec{B} = A_r B_r + r^2 A_\theta B_\theta</math>
  
−
It would be difficult to actually use this definition to compute a double integral.  Instead, we find it convenient to use our familiar techniques from single variable to evaluate double integrals.
+
In spherical coordinates:
 +
::<math>\vec{A} \cdot \vec{B} = A_r B_r + r^2 A_\theta B_\theta + r^2 \sin^2 \theta A_\phi B_\phi</math>
  
−
Let's examine a rectangular region first, whose sides are aligned with the ''x'' and ''y'' axes, with the lower ''x'' bound at ''x=a'', the upper ''x'' bound at ''x=b'', the lower ''y'' bound at ''y=c'', and the upper ''y'' bound at ''y=d''.  See illustration.
+
The metric is always a symmetric matrix.  (Exercise:  Prove this.  Use the definition in terms of <math>[J]</math> and <math>[J]^{-1}</math>.)  For all coordinate systems that have the coordinate lines crossing at right angles everywhere (which includes all the common systems), the metric is a ''diagonal matrix''&mdash;the only nonzero items are on the main diagonal.
  
−
The double integral of a function couldn't be simpler:  we just integrate the function twice, each time treating one variable like a constant (just like we do when we take partial derivatives. 
+
===The Length of a Vector in Arbitrary Coordinates===
  
−
<math>\iint_{Rectangle} f(x,y)dA = \int_c^d \left({  \int_a^b f(x,y)dx }\right) dy</math>.
+
The length of a vector (sometimes called the "norm") is the square root of the dot product of the matrix with itself. Note that this is defined in terms of something that we have defined in a purely geometrical way, independent of any coordinate system. Note also that it is consistent with our earlier definition of the dot product of two vectors as the product of their lengths, times the cosine of the angle between them.
  
−
Let's do an example problem on the square with corners at ''(1,1), (1,2), (2,1), (2,2)'':
+
The length of a vector is usually written with a sort of double absolute value sign.
  
−
<math>\iint_{Square} \frac{1}{x+y} dA = \int_1^2 \left({\int_1^2 \frac{1}{x+y} dx}\right)dy = \int_1^2 \left({ \log(2+y) - \log(1+y) }\right) dy = \int_1^2 \log \left({ \frac{2+y}{1+y} }\right)dy </math>
+
In polar coordinates, we have:
 +
::<math>\Vert\vec{A}\Vert = \sqrt{A_r^2 + r^2 A_\theta^2}</math>
 +
In spherical coordinate:
 +
::<math>\Vert\vec{A}\Vert = \sqrt{A_r^2 + r^2 A_\theta^2 + r^2 \sin^2 \theta A_\phi^2}</math>
  
−
which is a problem of single variable calculus.
+
===The Cross Product in Arbitrary Coordinates===
  
−
But how useful is such a technique, since it is so easy to think of regions which are NOT rectangles?  Let's move on to consider a more general kind of region.
+
The rule for the cross product in the general u/v/w coordinate system makes use of all the things we have been developing. First, make a vector from the determinant with basis vectors (in the u/v/w system!) across the top, the way we did previously:
−
[[Image:Dub int 1.jpg|right|thumb|250px|The kind of region under discussion.]]
+
::<math>\vec{P}=\begin{vmatrix}\hat{u}&\hat{v}&\hat{w}\\ A_u & A_v & A_v \\ B_u & B_v & B_w \end{vmatrix}</math>
−
Suppose a region ''R'' happens to be such that we can identify its lowest extreme ''x'' value, which we'll call ''a'', and its highest ''x'' value, which we'll call ''b'', and that a vertical line drawn through the region intersects the boundary in exactly two points everywhere except those extremes, where a vertical line only intersects that boundary once.  Then there is some function of ''x'' which will give the ''y'' value of the lower of all of those intersections of the boundary with vertical lines, and some some function which will give the ''y'' value of the upper of those intersections.  Let's call these functions <math>y_1, y_2 </math> respectively.  See the illustration to the right.
+
If the coordinate system is Cartesian, we are done.  Otherwise, apply the inverse of [g] to it, and then multiply by <math>\sqrt{g}</math>.  (Remember that just "g", without brackets, is the determinant of the matrix.)
 +
::<math>\vec{A} \times \vec{B} = \sqrt{g}\ [g]^{-1}\ \vec{P} = \sqrt{g}\ [g]^{-1}\ \begin{vmatrix}\hat{u}&\hat{v}&\hat{w}\\ A_u & A_v & A_v \\ B_u & B_v & B_w \end{vmatrix}</math>
  
−
A region like this is very easily integrated, just like the rectangle.  We simply use these boundary functions in place of the ''y''-extremes:
+
In the case that [g] is a diagonal matrix, this can be simplified.  [g] is a diagonal matrix for any "orthogonal" coordinate system, that is, one in which the coordinate lines always cross at right angles.  This is true for all common coordinate systems, including polar, spherical, and cylindrical, so we will frequently make use of this simplification.
  
−
<math>\iint_{R} f(x,y)dA = \int_a^b \left({ \int_{y_1(x)}^{y_2(x)} f(x,y) dy    }\right) dx </math>
+
If <math>[g]</math> is diagonal, the only nonzero elements are <math>g_{11}</math>, <math>g_{22}</math>, and <math>g_{33}</math>.  The only nonzero elements of <math>[g]^{-1}</math> are <math>\frac{1}{g_{11}}</math>, <math>\frac{1}{g_{22}}</math>, and <math>\frac{1}{g_{33}} </math>.  In this case, the cross product formula is:
 +
::<math>\vec{A} \times \vec{B} = \sqrt{g_{11}g_{22}g_{33}}\ \begin{vmatrix}\frac{1}{g_{11}}\hat{u}&\frac{1}{g_{22}}\hat{v}&\frac{1}{g_{33}}\hat{w}\\ A_u & A_v & A_v \\ B_u & B_v & B_w \end{vmatrix} = \begin{vmatrix}\sqrt{\frac{g_{22}g_{33}}{g_{11}}}\hat{u}&\sqrt{\frac{g_{11}g_{33}}{g_{22}}}\hat{v}&\sqrt{\frac{g_{11}g_{22}}{g_{33}}}\hat{w}\\ A_u & A_v & A_v \\ B_u & B_v & B_w \end{vmatrix}</math>
  
−
Let's do an example of this.  The region we'll be integrating over is the unit disc, so the ''x'' minimum and maximum values will be <math>-1,1</math> and the ''y'' functions will be <math>y_2 = \sqrt{1-x^2}, y_1=-\sqrt{1-x^2}</math>.  Let's integrate the function <math>f(x,y)=(x+y)^2 </math>.
+
In spherical coordinates, this becomes:
 +
::<math>\vec{A} \times \vec{B} = \begin{vmatrix}r^2 \sin \theta\ \hat{r}&\sin \theta\ \hat{\theta}&\frac{1}{\sin \theta}\hat{\phi}\\ A_u & A_v & A_w \\ B_u & B_v & B_w \end{vmatrix}</math>
  
−
<math>\iint_{R}(x+y)^2dA=\int_{-1}^1\left({\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}(x+y)^2dy}\right)dx=\int_{-1}^1\left({\frac{(x+\sqrt{1-x^2})^3}{3}-\frac{(x-\sqrt{1-x^2})^3}{3}}\right)dx</math>
+
Example:  Suppose <math>\vec{A}</math> is a radial basis vector:
 +
::<math>\vec{A} = \hat{r} = \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}</math>
 +
(The length of <math>\vec{A}</math> is 1.)
 +
Let <math>\vec{B}</math> be a <math>\theta</math>-pointing basis vector:
 +
::<math>\vec{B} = \hat{\theta} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}</math>
 +
(The length of <math>\vec{B}</math> is r.)
 +
<math>\vec{B}</math> is a vector at the surface of a sphere of radius r that points "south", tangential to the surface.  <math>\vec{A}</math> points outward from the surface, that is "up".  <math>\vec{A}</math> and <math>\vec{B}</math> are orthogonal.  (Prove it; calculate <math>\vec{A} \cdot \vec{B}</math>.)
  
−
We have turned a double integral into a single-variable integral that is solved more with bookkeeping than intelligence:
+
By the formula for the cross product, we have:
 +
::<math>\vec{A} \times \vec{B} = \begin{vmatrix}r^2 \sin \theta\ \hat{r}&\sin \theta\ \hat{\theta}&\frac{1}{\sin \theta}\hat{\phi}\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix} = \frac{1}{\sin \theta}\ \hat{\phi} = \begin{bmatrix} 0 \\ 0 \\ \frac{1}{\sin \theta} \end{bmatrix}</math>
  
−
<math>\frac{1}{3}\int_{-1}^1\left({  (x+\sqrt{1-x^2})^3  -    (x-\sqrt{1-x^2})^3    }\right)dx
+
This points "east" at the surface of the sphere of radius r, is perpendicular to the other two vectors, and has length r, which is the product of the lengths of the other two vectors.  The geometrical properties of the cross product (perpendicular to each of the given vectors, and of length equal to the product of their lengths times the sine of the angle between them) are obeyed.  The cross product, like all other vector operations, is a ''geometric invariant''.  It doesn't matter what coordinate system was used to calculate the components&mdash;the answer is the same.
  
−
</math>
+
===The Divergence in Arbitrary Coordinates===
  
−
<math>= \frac{2}{3}\int_{-1}^1 \left({    x^2\sqrt{1-x^2}  + (1-x^2)^{3/2} }\right) dx    </math>
+
The divergence and curl will be discussed in detail in lecture 6. But, to give a flavor of how the [g] metric matrix is used, we give the definitions here.
  
−
<math>= \frac{2}{3} \frac{1}{8}\left({  0 + \frac{\pi}{2} + 0 + 3\frac{\pi}{2} -0 - \frac{-\pi}{2}    -0    -3\frac{\pi}{2}                        }\right) = \frac{\pi}{3}</math>
+
Here's the definition of the divergence operator in arbitrary curvilinear coordinates:
  
−
==Definition of the Triple Integral==
+
::<math>\nabla \cdot \vec{A} = \frac{1}{\sqrt{g}} \sum_i \frac{\partial}{\partial x_i}\ (\sqrt{g}\ A_i) = \sum_i \frac{\partial A_i}{\partial x_i} + \frac{1}{\sqrt{g}}\ \sum_i \frac{\partial \sqrt{g}}{\partial x_i}\  A_i</math>
 +
(The second form follows from the first due to the product rule for derivatives.)
 +
In practice, it's not as forbidding as it looks, for vector fields that are well suited to the particular coordinate system.
  
−
By now, defining this should be a simple matter for the student.  Since for a function of three variables, the domain is three dimensional space, we'll be considering regions in space and dividing it into cubes, just as we divided two dimensional regions into squares before.
+
::::***** Need to put in some examples, etc.  I have them worked out, but typing them in is tedious.
  
−
So we divide ''R'' into ''n'' cubic subregions <math>R_k, 1\leq k\leq n</math>, and in each region pick a point <math>(x_k,y_k,z_k)\in R_k</math>.  Just like before, it may be that all the <math>R_k</math> might not cover all of our original region ''R''; but as the size of our cubes decreases, they approximate the region better and better - in the limit, they will cover the whole thing.
+
===The Curl in Arbitrary Coordinates===
  
−
The Riemann sum for a function ''f'' is then simple enough to construct, using <math>V</math> to note the volume of one of our squares: <math>S= \sum_{k=1}^n f(x_k,y_k,z_k)V</math>.  If we use <math>dV=dxdydz</math> to note we are differentiating with respect to volume, we can define our triple integral:
+
Here's the definition of the curl operator in arbitrary curvilinear coordinates:
  
−
<math>\iiint_R f(x,y,z)dV = \lim_{n\to\infty,A\to 0} \sum_{k=1}^n f(x_k,y_k,z_k)V</math>
+
First, to find the curl of a vector field <math>\vec{A}</math>, let the matrix <math>[g]</math> operate on <math>\vec{A}</math>, yielding <math>\vec{B}</math>:
 +
::<math>\vec{B} = [g] \vec{A}</math>
 +
::Special case: If the coordinate system is orthogonal, so that <math>[g]</math> is diagonal, this is very easy:
 +
::::<math>B_u = g_{11} A_u\ \ \ B_v = g_{22} A_v\ \ \ B_w = g_{33} A_w</math>
 +
Then the curl is:
 +
::<math>\nabla \times \vec{A} = \frac{1}{\sqrt{g}} \begin{vmatrix} \hat{u} & \hat{v} & \hat{w} \\ \\ \displaystyle\frac{\partial}{\partial u} & \displaystyle\frac{\partial}{\partial v} & \displaystyle\frac{\partial}{\partial v} \\ \\ B_u & B_v & B_w \end{vmatrix}</math>
  
−
==Computation of the Triple Integral==
 
  
−
In our computation of the double integral, we considered only regions in the plan such that the "top" and "bottom" of the boundary could be described as functions of ''x''.  We then pointed out that all regions could be broken up into such shapes, and we could just perform each evaluation separately.  It should come as no surprise, then, that we consider 3D regions where the "top" and "bottom" of the surface enclosing the region can be described with functions of the form <math>z=g(x,y)</math>; in other words, every line which is parallel to the ''z''-axis and intersects our region ''R'' only intersects its surface boundary in at most two points. 
+
::Or, if the coordinate system is orthogonal:
  
−
Suppose ''R'' is such a region in space.  Now, suppose that the "top" of the boundary is described by <math>z=g_1(x,y)</math>, and the "bottom", by <math>z=g_2(x,y)</math>.  This is where it will get a bit tricky: for a given value of ''x'', let <math>h_1(x)</math> be the greatest value of ''y'' such that ''(x,y,z)'' is in ''R'' for any z, and <math>h_2(x)</math>, the least value of ''y''.  Finally, let the minimum ''x'' value for a point in ''R'' be ''a'', and the maximum, ''b''.
 
  
−
Then
+
::::<math>\nabla \times \vec{A} = \frac{1}{\sqrt{g}} \begin{vmatrix} \hat{u} & \hat{v} & \hat{w} \\ \\ \displaystyle\frac{\partial}{\partial u} & \displaystyle\frac{\partial}{\partial v} & \displaystyle\frac{\partial}{\partial v} \\ \\ g_{11}A_u & g_{22}A_v & g_{33}A_w \end{vmatrix}</math>
  
−
<math>\iiint_R f(x,y,z)dV = \int_a^b \int_{h_2(x)}^{h_1(x)} \int_{g_2(x,y)}^{g_1(x,y)} f(x,y,z) dz dy dx  </math>
+
::::***** Need to put in some examples, etc. I have them worked out, but typing them in is tedious.
  
−
==Polar Coordinates==
 
−
 
−
The definition of a double integral is always the same, we need only change <math>(x,y)</math> for <math>(r,/theta)</math>.  But the area element <math>dA</math>, which before was <math>dxdy</math>, is going to be slightly different.
 
−
 
−
In Cartesian coordinates, it was a simple matter to describe the area of a small rectangle - it was simply the length times the width, hence the <math>dxdy</math>.  But what is the area of a little bit of a region in polar coordinates?  The ''r'' sides of such a rectangle will be <math>dr</math>, since ''r'' is measured in length.  But <math>\theta</math> is measured in angles, which is just a number!
 
−
 
−
Remember the formula for the circumfrence of a circle of radius ''r'': <math>2\pi r</math>.  Therefore, the length of an arc of angle <math>d\theta</math> at radius ''r'' is going to be <math>rd\theta</math>.  Multiplying the length and width together gives the area element in polar coordinates: <math>dA=rdrd\theta</math>.  Hence, the double integral in polar coordinates is
 
−
 
−
<math>\iint_R f(r,\theta)dA = \iint_R f(r,\theta)rdrd\theta</math>.
 
−
 
−
For spherical coordinates in three dimensions, we can rely heavily on what we've just learned.  Since the circumference of a circle in three dimensions is the same as in two, the same reasoning tells us <math>dV = (dr)(rd\theta)(rd\phi) = r^2 d\theta d\phi dr  \ </math> and hence
 
−
 
−
 
−
<math>\iiint_R f(r,\theta,\phi)dV = \iint_R f(r,\theta,\phi)r^2drd\theta d\phi</math>.
 
  
 
=Problems=
 
=Problems=
Line 476: Line 450:
  
 
'''3.''' Suppose <math>f(x,y)</math> has continuous derivatives of all orders.  Prove <math>\partial_{xy}f = \partial_{yx}f</math>.
 
'''3.''' Suppose <math>f(x,y)</math> has continuous derivatives of all orders.  Prove <math>\partial_{xy}f = \partial_{yx}f</math>.
 +
 +
==References==
 +
<References/>
 
{{MVCalc}}
 
{{MVCalc}}

Latest revision as of 21:23, September 8, 2020

Vectors, and Vector Functions

Vectors

 
A vector in the plane.

A vector is a mathematical object which has magnitude like a number, but also direction. For example, if I wanted to describe how a person is driving at any particular time, I would need two numbers: their speed, and some measure of their direction (the angle it makes with true north, for example). These two quantities together constitute the motion vector of that person.

We also use vectors to describe position: for example, to locate an object, it suffices to know how far it is from you, and in what direction.

 
In three dimensions, three component numbers are required to describe a vector.

Vectors are the bread and butter of multivariable calculus because they pack multiple variables into a single, easily manipulated quantity. Let's look at some more specific examples.

In our first descriptions, we used a coordinate system you should be familiar with call "polar coordinates." If you aren't familiar with this system, we will review them shortly. When using Cartesian coordinates, vectors are very, very familiar. The point (x,y) has a position vector, or put more simply, is the vector written Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix}x\\y\end{pmatrix} \ } . See illustration at right. Unlike variables which may be numbers, like Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a, x, \ } etc., vectors are usually written Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{w} \ } or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{w} \ } . While the latter is common in print and online, we will use the former since it is more easily duplicated in handwriting.

In three dimensional space we use three variables to write down a vector. In Cartesian coordinates, this would look like Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{w} = \begin{pmatrix}x\\y\\z\end{pmatrix} \ } .

We can, if we wish, talk about just the length of the vector - this is its absolute value, or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left\| \vec{w} \right\| \ } . For a vector written in Cartesian coordinates, the Pythagorean theorem tells us this is just Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{x^2 + y^2} \ } .

Vectors can be added and subtracted, just like numbers. See the illustration below.

 
Vectors are added by connecting them head to tail. Vectors are subtracted by reversing heads and tails for the subtracted vector, and then connecting them head to tail.


In Cartesian coordinates, this is particularly easy: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix}x\\y\end{pmatrix} + \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}x+a\\y+b\end{pmatrix} \ } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix}x\\y\end{pmatrix} - \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}x-a\\y-b\end{pmatrix} \ } .

Multiplication by a number is also simple: the vectors direction is unchanged, but its length is multiplied by that number. Again, in Cartesian coordinates, this becomes simple: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}ca\\cb\end{pmatrix} \ } .

 
The dot product is a measure of the length of one vector in the direction of another.

There are two ways to multiply a vector by another vector. In two dimensions, we can perform only one, which is called the dot product or in more advanced terminology, inner product. Given two vectors, this dot product is how far one vector goes in the direction of the other, using the other as the "unit" of length. It is defined Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u} \cdot \vec{w} = \left\|\vec{u}\right\| \left\|\vec{w}\right\| \cos(\theta) \ } , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta \ } is the angle between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u} \ } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{w} \ } . As before, this takes a particularly simple form in Cartesian coordinates: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix}x\\y\end{pmatrix} \cdot \begin{pmatrix}a\\b\end{pmatrix} =xa+by \ } . This expression is similar for the three dimensional case: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix}x\\y\\z\end{pmatrix} \cdot \begin{pmatrix}a\\b\\c\end{pmatrix} = xa+yb+zc \ } .

 
The cross product of two vectors.

While the dot product can be used in any number of dimensions, there is a special way of multiplying vectors which only exists in three dimensions (although it is a specific case of a more general concept). It is called the cross product and written Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u} \times \vec{w} \ } . The cross product is defined as having length equal to the area of the parallelogram borderd on two sides by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u} \ } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{w} \ } , and direction perpendicular to either vector, with a right handed orientation. This means that if we take our right hand, point the palm in the direction of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u} \ } , bend our knuckles so that they point in the direction of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{w} \ } , then our thumb is already pointing in the direction of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u} \times \vec{w} \ } . This definition is not easy to calculate with, and shortly we will introduce a more convenient way of calculating these kinds of products.

While we often write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{w} = \begin{pmatrix}x\\y\\z\end{pmatrix} \ } to describe a vector, we often find it more convenient to define three vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{i}=\begin{pmatrix}1\\0\\0\end{pmatrix}, \vec{j}=\begin{pmatrix}0\\1\\0\end{pmatrix}, \ } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{k}=\begin{pmatrix}0\\0\\1\end{pmatrix} \ } , unit vectors which point along the positive Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x, y, \ } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z \ } axes, respectively.

With these vectors, we can write our original vector Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{w}=x\vec{i} + y\vec{j} + z\vec{k} \ } . It means exactly the same thing as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix}x\\y\\z\end{pmatrix} \ } , but is a little friendlier to the printed page. As an additional notational convenience, we often write Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{w} = w_1 \vec{i} + w_2 \vec{j} + w_3 \vec{k} \ } , since keeping track of all the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \ } 's and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \ } 's would get unwieldy very fast if we are working with several vectors.

As we are about to see, these conveniences can make the calculation of the cross product relatively simple. Hopefully, the student will remember the definition of the determinant of a matrix, because this will make calculating cross products much, much easier. For those who are unfamiliar with the idea, we will review the concept.

For a matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix}a&b\\c&d\end{pmatrix} \ } , we define the determinant Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{vmatrix}a&b\\c&d\end{vmatrix}} to be Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ad - bc \ } . To compute the determinant of a larger square matrix, then, we choose any row of the matrix and take the left entry, and multiply that number by the determinant of the matrix formed by eliminating the first row and column of the original matrix. From this, we subtract the next entry in our row times the matrix formed by eliminating THAT row and column of the original matrix. Then we add the next term, and so on and so forth until we have a complete expansion.

Below is an example of this technique, expanding along the first row.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{vmatrix}a&b&c\\r&s&t\\x&y&z\end{vmatrix} = a\begin{vmatrix}s&t\\y&z\end{vmatrix} - b \begin{vmatrix}r&t\\x&z\end{vmatrix} + c \begin{vmatrix}r&s\\x&y\end{vmatrix}}

Examine this equation, and make sure you understand how each of the matrices on the right hand side were formed. Once we have reached this point, we can use our definition of a 2x2 determinant to arrive at a value.

We now arrive at a means of calculating the cross product of two vectors. It turns out that the definition we gave of cross product amounts to

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}\times\vec{w}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\u_1&u_2&u_3\\w_1&w_2&w_3\end{vmatrix} \ }

Normally, we don't allow vectors (like Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{i} \ } ) to be elements of a matrix or a determinant, but in this case, we make an exception because the formula is so useful and easy to remember. We will encounter a similar exception to the rules when we give an easy way to remember the formula for curl.

In this course, we will frequently switch between the notations Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x,y,z), x\vec{i}+y\vec{j}+z\vec{k},} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{pmatrix}x\\y\\z\end{pmatrix} \ } without discussion of the switch: constantly keep in mind that these are all ways of saying the same thing.

Vector Functions and Vector Fields

 
An illustration of a vector field. The arrows show the magnitude and direction of the vector at the location of the arrow's base.

Before, we described scalar functions as taking a point in space and ascribing a number to it - for example, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x,y,z) = x^2 + y^2 +z^2 \ } . With our new notation, we can describe this as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\vec{r}) = \vec{r} \cdot \vec{r} \ } , which is a scalar function written with vectors.

There is another class of functions, though, which return not numbers but vectors. Mathematicians define a function as anything that maps some set (the "domain") to some set (the "range".) For the functions we have seen so far, the range is the real numbers (scalars.)

When the range is a vector space, it can be called a vector function. But the most common types of vector functions are those for which the domain is physical space. In that case, the function is called a vector field.

Vector fields are useful for describing many phenomena, most notably flows, winds, heat dispersion, and electric and magnetic forces. For example, at every point in the ocean, the water is moving with both a speed and a direction. If we have some vector Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{r} \ } which can be used to describe points in the ocean, we can write a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\vec{r}) = \vec{v} \ } , which will give us the velocity vector of the water at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{r} \ } . Very often, we use Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{r} \ } as a variable vector describing a position or location, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v} \ } for velocity, and later we'll encounter Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{a} \ } for acceleration.

Coordinate Systems

 
The spherical coordinate system.

We're all very familiar with the Cartesian coordinates in two dimensions, and once the student has learned the "right hand rule" and committed it to memory, they have learned all there is about the Cartesian system in three dimensions.

Hopefully, the students will be familiar with polar coordinates as well, but we shall review.

In the plane, it is frequently useful to describe shapes, not by their distance along two axes, but by their straight-line distance to a point, and their angle to a fixed line. We use Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (r,\theta) \ } instead of the usual Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x,y \ } , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r \ } is the distance and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta \ } is the angle.

We can easily move between Cartesian and polar coordinates with these equations:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = \sqrt{x^2+y^2} \ }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta = \arctan{\frac{y}{x}} \ }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=r \arccos{(\theta)} \ }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=r \arcsin{(\theta)} \ }

Cylindrical coordinates are just polar coordinates, with a z-axis. They are frequently useful for describing phenomenon which have circular motion in a plane, but are moving in some direction (like a spiral).

More common than cylindrical coordinates are spherical coordinates, which use distance from a center point together with "latitude and longitude" to describe points in three dimensional space.

In the spherical coordinate system, a point is described with its distance to a central point, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho \ } , the angle made in the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle xy \ } plane between the points projection into that plane and the x-axis Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta \ } , and the angle made between the point and the z-axis Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi \ } .

As with polar coordinates, there are equations which allow us to transfer between spherical and Cartesian coordinates.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho = \sqrt{x^2 + y^2 + z^2} \ }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta = \arctan{(\frac{y}{x})} \ }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi = \arctan{ \left({ \frac{\sqrt{x^2+y^2}}{z} }\right) } \ }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = r \sin(\phi) \cos(\theta) \ }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = r \sin(\phi) \sin(\theta) \ }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z = r \cos(\phi) \ }

Limits of Multivariate Functions

See the limit article.

The limit of a multivariate function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{(x,y)\rightarrow(x_0,y_0)} f(x,y) \ } is almost identical to the single-variable limit in concept, but complicated by the realization that (x,y) can approach the target point from any direction, not just the left or right hand sides.

The definition is not dissimilar from the single-variable case. The limit of a function is defined so that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{(x,y) \rightarrow (x_0,y_0)} f(x,y) = L \ } if, for any tiny little number Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon>0 \ } you can think of, there is some distance Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta \ } such that every point within Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta \ } of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_0,y_0) \ } is taken by the function to a number within Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon \ } of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L \ } .

Nevertheless, the kinds of phenomenon we see in the multivariate case is very different. Note that while it is still possible to define a limit from a direction, we don't bother to do so - unlike the one-variable case when a one-sided deriative told us half of the functions behavior at a point, in higher dimensions such a limit tells us almost nothing. Indeed, the only kind of phenomenon that these kinds of limits are capable of describing are "point discontinuties," which can be "fixed" just be defining the function to be the limit at that point.

Partial Derivatives

See the proof page for details.

When dealing with functions of more than one variable, one has to be careful about what a "derivative" is. The operative concept is the partial derivative, which is the derivative of the function with respect to a given argument while holding all the other arguments constant. It is written like a derivative, but with a different "d" character.

If

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x,y,z) = \frac{x y^2 \sin{x}}{z}\,}

we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} f(x,y,z) = \frac{y^2 (x \cos{x} + \sin{x})}{z}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial y} f(x,y,z) = \frac{2 x y \sin{x}}{z}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial z} f(x,y,z) = \frac{- x y^2 \sin{x}}{z^2}}


Here are some other notations for partial derivatives that one might encounter. We won't use them.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x} f(x,y,z) = \partial_x f(x,y,z) = f_x (x,y,z)}


One can take higher-order partial derivatives as well, such as

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial^2 f}{\partial x^2}}

or

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial^2 f}{\partial x \partial y}}

which means

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)}


A question that arises is: do mixed partial derivatives commute? That is, do we have:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}}

The answer is yes, if the derivatives are continuous. In problems that arise in practice, the derivatives are always continuous, and switching derivative order is a staple of work in multivariate calculus.

More About Coordinate Systems

The study of various coordinate systems (sometimes called "curvilinear coordinate systems") is central to the subject of multivariable calculus. You are already familiar with some of the common coordinate systems such as polar coordinates in 2 dimensions, and spherical or cylindrical coordinates in 3 dimensions. In this section we will examine the general topic of coordinate systems, and how one can perform calculations—involving vectors, surfaces, integrals, and so on—directly in specialized coordinate systems. Physical problems can often be solved more easily in a coordinate system that matches the symmetries of the problem.

An extremely important consideration to be aware of is that geometrical or physical phenomena (points, vectors, regions, surfaces, etc.) have a fundamental existence and meaning that is independent of any coordinate system. Coordinate systems simply give us ways to attach numbers to them. Different coordinate systems will attach different numbers to a given vector field, for example, but the vector field has an underlying geometric meaning that does not change. The various vector field operations have an underlying geometrical or physical meaning, and coordinate systems simply let us perform mathematical calculations on them.

You have already seen the notion of "geometrical universality" in the definitions of the dot product and the cross product. They were given purely geometrical definitions that can be visualized without regard to any coordinate system. Than, for any Cartesian coordinate system, these operations were defined in terms of the components of the vectors in that system, and it was shown that those results matched the purely geometrical definitions. What we are going to do next is develop the tools to define vectors, and the dot product and cross product, in any coordinate system at all. Later, we will introduce new operations—the divergence, the curl, and various integrals, and show how to calculate them in any coordinate system.

For example, Stokes' theorem makes a very powerful geometric statement about integrals of various vector fields over very general surfaces and lines. The concepts appearing in the theorem are purely geometric, and therefore independent of the coordinate system. Once we know how to manipulate those concepts (vector fields, integrals, and the "curl" operator) in any coordinate system, we will be able to choose a coordinate system that makes the surface simple. Specifically, we will change from the usual x/y/z Cartesian system to a u/v/w system for which the surface is defined by holding w=0. (Another way of looking at this is that the surface has been "parameterized" in terms of parameters u and v.) Once this is done, we will be able to calculate the curl operator, perform the integration, and prove Stokes' theorem in the u/v/w system.

As another example, Maxwell's equations make statements about the curl and divergence of the electric and magnetic field. While the formulas for curl and divergence are simpler in Cartesian coordinates than in spherical, some problems, such as the electric field in the vicinity of a point charge, have symmetry that makes spherical coordinates more natural. When we work out the divergence theorem in spherical coordinates, we will be able to solve problems of this sort.

We will use "u" and "v" in many of our general examples of alternative coordinate systems in 2 dimensions, and u, v, and w in 3 dimensions. For the very common cases of polar coordinates in 2 dimensions and spherical in 3 dimensions, we will often use the more familiar r/Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} or r/Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} /Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} .

We always have the coordinates of each system representable as functions of the other. For example, the transformation between polar coordinates and Cartesian coordinate is this:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = r \cos \theta\,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = r \sin \theta\,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = \sqrt{x^2 + y^2}\,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta = \tan^{-1} (y/x)\,}

In spherical coordinates:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = r \sin \theta \cos \phi\,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = r \sin \theta \sin \phi\,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z = r \cos \theta\,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = \sqrt{x^2 + y^2 + z^2}\,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta = \tan^{-1} (\sqrt{x^2 + y^2} / z)\,}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi = \tan^{-1} (y/x)\,}

The Jacobian

Much of what happens in multivariable calculus involves the partial derivatives of the coordinates in one system with respect to the coordinates in the other. It is very useful to represent these as a matrix:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} \displaystyle\frac{\partial x}{\partial r} & \displaystyle\frac{\partial x}{\partial \theta} \\ \\ \displaystyle\frac{\partial y}{\partial r} & \displaystyle\frac{\partial y}{\partial \theta}\end{bmatrix}\,}

Or, going in the other direction:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} \displaystyle\frac{\partial r}{\partial x} & \displaystyle\frac{\partial r}{\partial y} \\ \\ \displaystyle\frac{\partial \theta}{\partial x} & \displaystyle\frac{\partial \theta}{\partial y}\end{bmatrix}\,}

This is the derivative matrix of the transformation from one coordinate system to the other. It is often called the Jacobian matrix, though some authors use the term "Jacobian" specifically to refer to the matrix's determinant. To avoid any confusion on this point, we will always put square brackets around matrices, and use absolute value signs to denote determinants.

The notation for the Jacobian matrix of u/v/w in terms of x/y/z is written:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J(u, v, w / x, y, z) = \frac{\partial(u, v, w)}{\partial(x, y, z)} = \begin{bmatrix} \displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y} & \displaystyle\frac{\partial u}{\partial z} \\ \\ \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y} & \displaystyle\frac{\partial v}{\partial z} \\ \\ \displaystyle\frac{\partial w}{\partial x} & \displaystyle\frac{\partial w}{\partial y} & \displaystyle\frac{\partial w}{\partial z}\end{bmatrix} }

More generally, it is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J(f_1,f_2,...,f_n / x_1,x_2,...,x_n) = \frac{\partial(f_1,f_2,...,f_n)}{\partial(x_1,x_2,...,x_n)} = \begin{bmatrix} \displaystyle\frac{\partial f_1}{\partial x_1} & ... & \displaystyle\frac{\partial f_1}{\partial x_n} \\ ... & & ... \\ \displaystyle\frac{\partial f_3}{\partial x_1} &...&\displaystyle\frac{\partial f_n}{\partial x_n}\end{bmatrix} }

and the determinant is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{vmatrix} \frac{\partial f_1}{\partial x_1} & ... & \frac{\partial f_1}{\partial x_n} \\ ... & & ... \\ \frac{\partial f_3}{\partial x_1} &...&\frac{\partial f_n}{\partial x_n}\end{vmatrix} }

For polar coordinates, the Jacobian matrices are:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J(r, \theta / x, y) = \frac{\partial(r, \theta)}{\partial(x, y)} = \begin{bmatrix} \displaystyle\frac{\partial r}{\partial x} & \displaystyle\frac{\partial r}{\partial y} \\ \\ \displaystyle\frac{\partial \theta}{\partial x} & \displaystyle\frac{\partial \theta}{\partial y} \end{bmatrix}}

and:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J(x, y / r, \theta) = \frac{\partial(x, y)}{\partial(r, \theta)} = \begin{bmatrix} \displaystyle\frac{\partial x}{\partial r} & \displaystyle\frac{\partial x}{\partial \theta} \\ \\ \displaystyle\frac{\partial y}{\partial r} & \displaystyle\frac{\partial y}{\partial \theta} \end{bmatrix}}

Working these out, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J(x, y / r, \theta) = \begin{bmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix}}

with determinant r, and:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J(r, \theta / x, y) = \begin{bmatrix} \displaystyle\frac{x}{\sqrt{x^2 + y^2}} & \displaystyle\frac{y}{\sqrt{x^2 + y^2}} \\ \\ \displaystyle\frac{- y}{x^2 + y^2} & \displaystyle\frac{x}{x^2 + y^2} \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ \\ \displaystyle\frac{- \sin \theta}{r} & \displaystyle\frac{\cos \theta}{r} \end{bmatrix}}

with determinant 1/r.

An interesting theorem is that these matrices are inverses of each other. (It follows that their determinants are reciprocals of each other.)

Exercise: Prove that the matrix product of the J matrices of two sequential coordinate changes is the J matrix of the combined change. That is, if

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_1,f_2,...,f_n \to g_1,g_2,...,g_n}

and

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_1,g_2,...,g_n \to h_1,h_2,...,h_n}

are two coordinate changes, so that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_1,f_2,...,f_n \to h_1,h_2,...,h_n}

is the result of doing both, then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial(h_1,h_2,...,h_n)}{\partial(f_1,f_2,...,f_n)} = \frac{\partial(h_1,h_2,...,h_n)}{\partial(g_1,g_2,...,g_n)}\ \ \frac{\partial(g_1,g_2,...,g_n)}{\partial(f_1,f_2,...,f_n)}}

Hint: This is really just the chain rule for partial derivatives. The adding up of terms in the chain rule is the same as the adding up of products in a matrix multiplication.


It follows from this that, if the second change is just the inverse of the first, bringing us back to the original coordinate system, the two matrices are inverses.

The coordinates should always be chosen such that the Jacobian determinant is positive. If it isn't, exchange the order of two of the coordinates. This ensures that the coordinate system is "right handed". (Or more precisely, that it has the same handedness as the Cartesian system.) As an example, the coordinate order (r/Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} ) is correct for polar coordinates; (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} /r) is not. The order (r/Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} /Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} ) is correct for spherical coordinates; (r/Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi} /Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} ) is not.

The Jacobian determinant must not be zero. (Equivalently, the Jacobian matrix must not be singular.) The astute reader will notice that this means that polar coordinates don't work at the origin, and spherical coordinates don't work at the north and south poles.

Vectors in Arbitrary Coordinate Systems

Having gotten the preliminary concepts out of the way, we next examine what vectors, and vector operations such as the dot product, look like in arbitrary coordinate systems.

A vector has different components in different coordinate systems. (It's still the same vector, of course.) If the components of the vector Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} , in the u/v/w coordinate system, are:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} = \begin{bmatrix} A_u \\ A_v \\ A_w \end{bmatrix}}

and, in the standard Cartesian system the components are:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} = \begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix}}

then the transformation rule is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} A_u \\ A_v \\ A_w \end{bmatrix} = \begin{bmatrix} \displaystyle\frac{\partial u}{\partial x} & \displaystyle\frac{\partial u}{\partial y} & \displaystyle\frac{\partial u}{\partial z} \\ \\ \displaystyle\frac{\partial v}{\partial x} & \displaystyle\frac{\partial v}{\partial y} & \displaystyle\frac{\partial v}{\partial z} \\ \\ \displaystyle\frac{\partial w}{\partial x} & \displaystyle\frac{\partial w}{\partial y} & \displaystyle\frac{\partial w}{\partial z}\end{bmatrix} \begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix}}

In other words, the transformation of vector components from x/y/z to u/v/w is just a matrix multiplication by the Jacobian matrix

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix}J(u, v, w / x, y, z)\end{bmatrix}}

To go the other way, multiply by the matrix

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix}J(x, y, z / u, v, w)\end{bmatrix}}

We make a simplification to avoid having our notation get completely out of control. In keeping with our goal of expressing everything just in terms of the new coordinate system:

  • The matrix that we have been calling [J(x, y, z / u, v, w)] we will call just [J]. That is, [J] for a coordinate system means [J(Cartesian/the coordinate system under discussion)].
  • The matrix that goes the other way, formerly called [J(u, v, w / x, y, z)] we will call Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [J]^{-1}} . It's just the inverse of the first matrix.

We will call [J] the "Jacobian of the coordinate system".

The astute reader will notice that this is not a good definition. [J] depends on the choice of the "reference" Cartesian system, so we're not justified in calling it just "the Jacobian". But it turns out that it won't make any difference. We will always get the same answers, no matter what reference system we used. See the exercises.

Therefore:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} A_u \\ A_v \\ A_w \end{bmatrix} = [J]^{-1} \begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix}}

and:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix} = [J] \begin{bmatrix} A_u \\ A_v \\ A_w \end{bmatrix}}

For polar coordinates in two dimensions, we have:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} A_r \\ A_\theta \end{bmatrix} = \begin{bmatrix}J(r, \theta / x, y)\end{bmatrix} \begin{bmatrix} A_x \\ A_y \end{bmatrix} = \begin{bmatrix} \displaystyle\frac{\partial r}{\partial x} & \displaystyle\frac{\partial r}{\partial y} \\ \\ \displaystyle\frac{\partial \theta}{\partial x} & \displaystyle\frac{\partial \theta}{\partial y}\end{bmatrix} \begin{bmatrix} A_x \\ A_y \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \\ \\ \displaystyle\frac{- \sin \theta}{r} & \displaystyle\frac{\cos \theta}{r} \end{bmatrix} \begin{bmatrix} A_x \\ A_y \end{bmatrix}}

And, going the other way:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} A_x \\ A_y \end{bmatrix} = [J] \begin{bmatrix} A_r \\ A_\theta \end{bmatrix} = \begin{bmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix} \begin{bmatrix} A_r \\ A_\theta \end{bmatrix}}

Let's look at some vectors whose components are given in general coordinates—there are some important and striking aspects of this.

In polar coordinates, a "radial basis vector" at any point has components:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (A_r=1, A_\theta=0)\,}

Its Cartesian components (from the formulas above) are:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (A_x=\cos \theta, A_y=\sin \theta)\,}

The actual (Cartesian) direction in which the vector points depends on its location. It always points radially outward. Its length is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{A_x^2 + A_y^2} = \sqrt{\cos^2 \theta + \sin^2 \theta} = 1\,} . So the radial basis vector is a unit vector.

The "angular basis vector" has components:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (A_r=0, A_\theta=1)\,}

Its Cartesian components are:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (A_x=- r \sin \theta, A_y=r \cos \theta)\,}

It always points "laterally", at right angles to the radial direction. More surprisingly, its length is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{A_x^2 + A_y^2} = r\,} . The angular basis vector is not a unit vector! It is bigger when farther away from the origin, even though the sum of the squares of its components is 1.[1] In general curvilinear coordinates, the "square root of the sum of the squares of the components" rule is not correct for the length of a vector. The correct rule will be given below.

The Dot Product in Arbitrary Coordinates

We need a better formula for the dot product than the rule that works for Cartesian coordinates:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z\,}

Given two vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} , described in a u/v/w coordinate system, we can find their dot product by converting to Cartesian coordinates and using the simple rule that we know, which is the sum of the pairwise products of the vector components. This means that, at least momentarily, we will have to keep track of the vectors' components in both the Cartesian system, which we have been calling Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_x} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_y} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_z} , and in the new coodinate system, which we have been calling Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_u} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_v} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_w} . To do the required matrix manipulations, we have to give the components numbers instead of names. Things like Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{u_1}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{u_2}} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_{u_3}} are just too unwieldy.[2] So we will make this simplification:

  • The components of a vector in the new coordinate system (formerly Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_u} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_v} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_w} ) will just be called Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_2} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_3} .
  • The components of a vector in the "reference" Cartesian system will be written with a tilde: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A_1}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A_2}} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A_3}} .

Once we reformulate the rules for vectors in terms of the coordinate system of interest, we will only use Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_2} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_3} (or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_u} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_v} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_w} ), and we will never have to look at another tilde again.

We can now calculate the dot product of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} , from their components and the Jacobian. We know that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \cdot \vec{B} = \sum_i \tilde{A_i} \tilde{B_i}}

in terms of the components in the reference system. We also have the transformation rule in terms of the Jacobian matrix.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A_i} = \sum_j [J(x, y, z / u, v, w)]_{ij} A_j}

This is just matrix multiplication. The summation is from 1 to 3, or whatever the dimension is.

By our new convention, this is just the Jacobian matrix [J], so

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{A_i} = \sum_j [J]_{ij} A_j}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tilde{B_i} = \sum_k [J]_{ik} B_k}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \cdot \vec{B} = \sum_i \tilde{A_i} \tilde{B_i} = \sum_{ijk} [J]_{ij} [J]_{ik} A_j B_k}

As promised, no more tildes!

We can combine the two Jacobians into one matrix, and then use that whenever we need a dot product. Let:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [g]_{jk} = \sum_i [J]_{ij} [J]_{ik} = \sum_i [J]^t_{ji} [J]_{ik}}

That is, [g] is just the matrix product of the transpose of [J] and [J] itself.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [g]_{jk} = ([J]^t [J])_{jk}\ \ \ or\ \ \ [g] = [J]^t [J]\,}

Then:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \cdot \vec{B} = \sum_{jk} [g]_{jk} A_j B_k}

This is the rule for the dot product in general coordinates. The matrix [g] is called the metric tensor or just the metric. We will have occasion to use its determinant, which we will call just g. Actually, its square root is more commonly used.

For those who really like equations that express things in terms of matrix multiplication, this could be written as:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \cdot \vec{B} = \begin{bmatrix} A_1 ... A_N \end{bmatrix} [g] \begin{bmatrix} B_1 \\ ... \\ B_N \end{bmatrix}}

The first vector is written in the form of a row instead of the more traditional column. This gives a 1xN matrix times am NxN matrix times an Nx1 matrix, which is a 1x1 matrix. The single value in that matrix is the dot product.

To review what we have for two popular systems, in 2-dimensional polar coordinates:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [J] = \begin{bmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [J]^{-1} = \begin{bmatrix} \cos \theta & \sin \theta \\ \\ \displaystyle\frac{- \sin \theta}{r} & \displaystyle\frac{\cos \theta}{r} \end{bmatrix}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [g] = \begin{bmatrix} 1 & 0 \\ 0 & r^2 \end{bmatrix}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{g} = r}

For spherical coordinates in 3 dimensions (we will need these soon):

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [J] = \begin{bmatrix} \sin \theta \cos \phi & r \cos \theta \cos \phi & - r \sin \theta \sin \phi \\ \sin \theta \sin \phi & r \cos \theta \sin \phi & r \sin \theta \cos \phi \\ \cos \theta & - r \sin \theta & 0 \end{bmatrix}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [J]^{-1} = \begin{bmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \\ \displaystyle\frac{\cos \theta \cos \phi}{r} & \displaystyle\frac{\cos \theta \sin \phi}{r} & - \displaystyle\frac{\sin \theta}{r} \\ \\ - \displaystyle\frac{\sin \phi}{r \sin \theta} & \displaystyle\frac{\cos \phi}{r \sin \theta} & 0 \end{bmatrix}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [g] = \begin{bmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2 \theta \end{bmatrix}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{g} = r^2 \sin \theta}

So, in polar coordinates:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \cdot \vec{B} = A_1 B_1 + r^2 A_2 B_2}

or, going back to giving the coordinates names:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \cdot \vec{B} = A_r B_r + r^2 A_\theta B_\theta}

In spherical coordinates:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \cdot \vec{B} = A_r B_r + r^2 A_\theta B_\theta + r^2 \sin^2 \theta A_\phi B_\phi}

The metric is always a symmetric matrix. (Exercise: Prove this. Use the definition in terms of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [J]} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [J]^{-1}} .) For all coordinate systems that have the coordinate lines crossing at right angles everywhere (which includes all the common systems), the metric is a diagonal matrix—the only nonzero items are on the main diagonal.

The Length of a Vector in Arbitrary Coordinates

The length of a vector (sometimes called the "norm") is the square root of the dot product of the matrix with itself. Note that this is defined in terms of something that we have defined in a purely geometrical way, independent of any coordinate system. Note also that it is consistent with our earlier definition of the dot product of two vectors as the product of their lengths, times the cosine of the angle between them.

The length of a vector is usually written with a sort of double absolute value sign.

In polar coordinates, we have:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Vert\vec{A}\Vert = \sqrt{A_r^2 + r^2 A_\theta^2}}

In spherical coordinate:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Vert\vec{A}\Vert = \sqrt{A_r^2 + r^2 A_\theta^2 + r^2 \sin^2 \theta A_\phi^2}}

The Cross Product in Arbitrary Coordinates

The rule for the cross product in the general u/v/w coordinate system makes use of all the things we have been developing. First, make a vector from the determinant with basis vectors (in the u/v/w system!) across the top, the way we did previously:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{P}=\begin{vmatrix}\hat{u}&\hat{v}&\hat{w}\\ A_u & A_v & A_v \\ B_u & B_v & B_w \end{vmatrix}}

If the coordinate system is Cartesian, we are done. Otherwise, apply the inverse of [g] to it, and then multiply by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{g}} . (Remember that just "g", without brackets, is the determinant of the matrix.)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \times \vec{B} = \sqrt{g}\ [g]^{-1}\ \vec{P} = \sqrt{g}\ [g]^{-1}\ \begin{vmatrix}\hat{u}&\hat{v}&\hat{w}\\ A_u & A_v & A_v \\ B_u & B_v & B_w \end{vmatrix}}

In the case that [g] is a diagonal matrix, this can be simplified. [g] is a diagonal matrix for any "orthogonal" coordinate system, that is, one in which the coordinate lines always cross at right angles. This is true for all common coordinate systems, including polar, spherical, and cylindrical, so we will frequently make use of this simplification.

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [g]} is diagonal, the only nonzero elements are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_{11}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_{22}} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_{33}} . The only nonzero elements of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [g]^{-1}} are Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{g_{11}}} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{g_{22}}} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{g_{33}} } . In this case, the cross product formula is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \times \vec{B} = \sqrt{g_{11}g_{22}g_{33}}\ \begin{vmatrix}\frac{1}{g_{11}}\hat{u}&\frac{1}{g_{22}}\hat{v}&\frac{1}{g_{33}}\hat{w}\\ A_u & A_v & A_v \\ B_u & B_v & B_w \end{vmatrix} = \begin{vmatrix}\sqrt{\frac{g_{22}g_{33}}{g_{11}}}\hat{u}&\sqrt{\frac{g_{11}g_{33}}{g_{22}}}\hat{v}&\sqrt{\frac{g_{11}g_{22}}{g_{33}}}\hat{w}\\ A_u & A_v & A_v \\ B_u & B_v & B_w \end{vmatrix}}

In spherical coordinates, this becomes:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \times \vec{B} = \begin{vmatrix}r^2 \sin \theta\ \hat{r}&\sin \theta\ \hat{\theta}&\frac{1}{\sin \theta}\hat{\phi}\\ A_u & A_v & A_w \\ B_u & B_v & B_w \end{vmatrix}}

Example: Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} is a radial basis vector:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} = \hat{r} = \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}}

(The length of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} is 1.) Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} be a Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} -pointing basis vector:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B} = \hat{\theta} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}}

(The length of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} is r.) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} is a vector at the surface of a sphere of radius r that points "south", tangential to the surface. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} points outward from the surface, that is "up". Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} are orthogonal. (Prove it; calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \cdot \vec{B}} .)

By the formula for the cross product, we have:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A} \times \vec{B} = \begin{vmatrix}r^2 \sin \theta\ \hat{r}&\sin \theta\ \hat{\theta}&\frac{1}{\sin \theta}\hat{\phi}\\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix} = \frac{1}{\sin \theta}\ \hat{\phi} = \begin{bmatrix} 0 \\ 0 \\ \frac{1}{\sin \theta} \end{bmatrix}}

This points "east" at the surface of the sphere of radius r, is perpendicular to the other two vectors, and has length r, which is the product of the lengths of the other two vectors. The geometrical properties of the cross product (perpendicular to each of the given vectors, and of length equal to the product of their lengths times the sine of the angle between them) are obeyed. The cross product, like all other vector operations, is a geometric invariant. It doesn't matter what coordinate system was used to calculate the components—the answer is the same.

The Divergence in Arbitrary Coordinates

The divergence and curl will be discussed in detail in lecture 6. But, to give a flavor of how the [g] metric matrix is used, we give the definitions here.

Here's the definition of the divergence operator in arbitrary curvilinear coordinates:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \cdot \vec{A} = \frac{1}{\sqrt{g}} \sum_i \frac{\partial}{\partial x_i}\ (\sqrt{g}\ A_i) = \sum_i \frac{\partial A_i}{\partial x_i} + \frac{1}{\sqrt{g}}\ \sum_i \frac{\partial \sqrt{g}}{\partial x_i}\ A_i}

(The second form follows from the first due to the product rule for derivatives.) In practice, it's not as forbidding as it looks, for vector fields that are well suited to the particular coordinate system.

          • Need to put in some examples, etc. I have them worked out, but typing them in is tedious.

The Curl in Arbitrary Coordinates

Here's the definition of the curl operator in arbitrary curvilinear coordinates:

First, to find the curl of a vector field Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} , let the matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [g]} operate on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{A}} , yielding Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B}} :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{B} = [g] \vec{A}}
Special case: If the coordinate system is orthogonal, so that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [g]} is diagonal, this is very easy:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_u = g_{11} A_u\ \ \ B_v = g_{22} A_v\ \ \ B_w = g_{33} A_w}

Then the curl is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \times \vec{A} = \frac{1}{\sqrt{g}} \begin{vmatrix} \hat{u} & \hat{v} & \hat{w} \\ \\ \displaystyle\frac{\partial}{\partial u} & \displaystyle\frac{\partial}{\partial v} & \displaystyle\frac{\partial}{\partial v} \\ \\ B_u & B_v & B_w \end{vmatrix}}


Or, if the coordinate system is orthogonal:


Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \times \vec{A} = \frac{1}{\sqrt{g}} \begin{vmatrix} \hat{u} & \hat{v} & \hat{w} \\ \\ \displaystyle\frac{\partial}{\partial u} & \displaystyle\frac{\partial}{\partial v} & \displaystyle\frac{\partial}{\partial v} \\ \\ g_{11}A_u & g_{22}A_v & g_{33}A_w \end{vmatrix}}
          • Need to put in some examples, etc. I have them worked out, but typing them in is tedious.


Problems

Review Problems

1. Find an antiderivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(x)e^{2x-1}}

2. Find the Taylor series of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = e^{-x^2}, f(0)=0} at 0. What do you notice about this sum?

3. Sketch level curves of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z=\cos(x) + \sin(y)} in the xy plane.

4. Describe the level surface of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x,y,z) = x^2 + y^2 + z^2/4} .

Main Problems

1. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v} = 2\vec{i} - (1/2)\vec{j}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{w}=-\vec{i}+ (1/3)\vec{j}} , find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}+\vec{w}, \vec{u}-\vec{w}, \vec{u}\cdot\vec{w},} . Moving to three dimensional space now, what can you say about Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}\times\vec{w}} before you even compute it? Carry out the computation. Was your prediction correct?

2. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}=4\vec{i}-2\vec{j}+\vec{k}} . Find a spherical coordinate expression for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}} .

3. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{F}(r,\theta,\phi) = (-r^{-1})(\vec{i}+\vec{j}+\vec{k})} . Describe Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{F}} in words. Find a formula for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{F}} in Cartesian coordinates.

4. Find the partial derivatives Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial x}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial}{\partial y}} of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3e^{x-y}\sin(y)+2x}

5. Find the partial derivatives Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \partial_x, \partial_y} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \partial_z} of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{x}{y+z} + \frac{y}{x+z}} .

Challenging Problems

1. Prove that the geometric definition of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{a}\cdot\vec{b}} matches our coordinate expression.

2. Prove that the geometric definition of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{a}\times\vec{b}} matches our coordinate expression.

3. Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x,y)} has continuous derivatives of all orders. Prove Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \partial_{xy}f = \partial_{yx}f} .

References

  1. ↑ Some textbooks use a definition of vector components such that basis vectors always have length 1. Such a treatment imposes the requirement that the coordinate lines be perpendicular everywhere. The formulas for vector operations are more complicated and less elegant when this is done.
  2. ↑ This is a well known notational problem in multivariable calculus. It is one of the reasons why tensor algebra textbooks are so hard to read.
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