# Calc3.4

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# Velocity and Acceleration

When we describe a particle traveling in space (or in the plane) with a set of parametric equations $\vec{r}(t) = x(t)\vec{i} + y(t)\vec{j}+z(t)\vec{k}$, we can easily use these equations to discern two other important facts about the particle: its velocity, and its acceleration.

This shouldn't be surprising; in single variable calculus, if f(t) was the position of a particle traveling along an axis after t seconds, the derivative was its velocity, and the second derivative was its acceleration. The only difference is, now we are dealing with vectors.

Here, the particle (green dot) has position vector $\vec{r}$ (black) and velocity vector $\vec{v}$ (red) as it traces out the curve, C (blue). Note that the velocity vector is always tangent to C.

So, let's define our velocity vector as $=\vec{v}(t) = \frac{d\vec{r}}{dt} = x'(t)\vec{i}+y'(t)\vec{j}+z'(t)\vec{k}$. Unlike our position vector $\vec{r}$, which we imagine as having its base attatched to the origin, the velocity vector we picture as having its base attatched to the point that is moving. See illustration, right.

The definition of the acceleration vector $\vec{a}$ shouldn't come as any more of a surprise than the velocity vectors: $\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d^2\vec{r}}{dt^2}$.

Now we have something we can DO something with! Why? Newton's second law tells us the the force on an object is always equal to the mass times the acceleration, that is, $\vec{F}=m\vec{a}$. Now that we have a formula for $\vec{a}$, lets try our hands on a gravitational field to get some practice.

Let's consider a pair of objects, one so much larger than the other that the smaller ones presence is basically inconsequential. We may think of, for example, the planet Earth and a tiny object, like a satellite. Newton's law of gravity is that the gravitational force between two objects is proportional to the products of their mass, divided by the square of distance between their centers. So lets put the Earth at the origin of our coordinate system, and use the typical ρ in spherical coordinates. Let's also agree to write mE for the mass of the Earth, and mO for the mass of our object.

The force on an object at distance ρ is then $F=G \frac{m_E m_O}{\rho^2}$, pointing towards the origin, ie,

$\vec{F}(\vec{r}) = -G \frac{m_E m_O}{ \left\| \vec{r} \right\|^3 } \vec{r}$

By Newton's famous second law of motion, $\vec{F}=m\vec{a}$, we have

$-G\frac{m_E m_O}{ \left\| \vec{r} \right\|^3 } \vec{r} = m_O \frac{d^2\vec{r}}{dt^2}$

Therefore, if a function $\vec{r}(t)$ satisfies

$\vec{r} \ ''(t) = -Gm_E \frac{\vec{r}(t)}{\left\| \vec{r}(t) \right\|^3 }$

then it can describe the position of an object at time t. Already, not that we have demonstrated that objects of different masses do not fall at different velocities, since the mO term divided out. Finding a family of solutions to this equation would give us all the ways an object can behave in a gravitational field. A derivation of all the solutions to this equation is beyond us at the moment, but we can examine a few functions $\vec{r}(t)$ which we suspect are solutions.

### Falling Objects

First, let's consider an object which starts out at some point $\vec{r}(0)$ and has no initial velocity, ie, $\vec{r} \ ' (0)=\vec{0}$. Therefore, there is not initial motion in the angular directions, and since the gravitationial field is purely radial, by Newtons first law, there will never be any angular motion. Hence, this reduces to an equation only in the spherical coordinate ρ:

$\rho''(t) = - \frac{Gm_E}{\rho(t)^2}$

### Circular Orbits

We're familiar with objects orbiting the Earth. Here, we'll examine two kinds of orbits, circular, and elliptical. A circular orbit can be described as $\vec{r}(t) = r\cos(kt)\vec{i}+r\sin(kt)\vec{j}+0\vec{k}$, where r is the radius of the orbit, and k is the angular velocity. We wish to check that this equation satisfies the equations of gravity described above - otherwise, all the satellites will fall from the sky!

We begin by calculating a few of the terms that appear in the gravitation equations:

$\left\| \vec{r} \right\|^3 = \sqrt{r^2\cos^2(k t) + r^2\sin^2(k t)}^3 = r^3$

$\vec{r} \ ''(t) = -rk^2\cos(kt)\vec{i}-rk^2\sin(kt)\vec{j}$

So now we can check the gravitational equation:

$-rk^2\cos(kt)\vec{i}-rk^2\sin(kt)\vec{j}=-Gm_E\frac{r\cos(kt)\vec{i}+r\sin(kt)\vec{j}}{r^3}$

Dividing out, we have

$k^2=\frac{Gm_E}{r^3}$

Since k is the angular velocity, we only need to multiply k by the circumference of the orbit (2πr), to get the actual orbit velocity. Thus, we have used the equations of gravity to determine what speed is necessary for an object to orbit a planet at a certain height.

# Tangents and Normals

If a curve is described parametrically as $\vec{r}(t) = x(t)\vec{i} + y(t)\vec{j}+z(t)\vec{k}$, the vector $=\vec{v}(t) = \frac{d\vec{r}}{dt} = x'(t)\vec{i}+y'(t)\vec{j}+z'(t)\vec{k}$ will always be tangent to that curve if we imagine the basepoint at attatched to (x,y,z) instead of the origin.

However, when people speak of the tangent vector to a curve, they almost always mean the normalization of $\vec{v}$, which means a vector that points in the same direction as $\vec{v}$ but with constant length =1. We can compute this tangent vector with the formula

$\vec{T} = \vec{v}/\left\|{\vec{v}}\right\|$.

This is not the only vector which commonly arises in the analysis of curves: we often also wish to have a description of how the curve is turning at any specific point. Since this amounts to knowing how the tangent is changing, it should not surprise anybody that we examine the derivative of the tangent for this purpose. Frequently in applications, we find it doesn't matter how long this derivative is, only it's direction. Hence we define the normal vector using the same process as above:

$\vec{N} = \vec{T'}/\left\|{\vec{T'}}\right\| = \frac{d\vec{T}}{dt}\div \left\|{ \frac{d\vec{T}}{dt} }\right\|$.

Since the tangent vector $\vec{T}$ always has unit length, it only change direction, not length. This means that the derivative vector $\vec{N}$ must always be perpendicular to $\vec{T}$, since if it were not, there would be a component of $\vec{N}$ in the direction of $\vec{T}$; that is to say, the length of $\vec{T}$ would be changing.

# Binormals, Curvature, and Torsion

Geometers also define a third vector associated with a curve, called the binormal. It is defined as $\vec{T}\times\vec{N}$, and obviously it also has unit length. These three vectors, each imagined as having their base at a single point on the curve, are called the Frenet frame of the curve.

# Arc Length

If we imagine a very very small portion of a curve, that piece will be almost straight, and so we can almost use the Pythagorean theorem to know the length of that piece. See the diagram at left. By adding a number of tiny portions, we should get a good apprximation to the length of the curve. Let's formalize this by letting the length of a parametrically-defined curve from t=a to t=b be divided into n tiny pieces, denote the x length of the kth piece as Δxk (and so on for the other coordinate variables), and adding up all these Pythagorean approximations:

$L\approx\sum_{k=1}^{n} \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2 + (\Delta z_k)^2 } \$

As we go to the limit, the approximation becomes exact:

$L = \lim_{n\to\infty} \sum_{k=0}^{n-1} \sqrt{(\Delta x_k)^2 + (\Delta y_k)^2 + (\Delta z_k)^2 } = \int_a^b \sqrt{dx^2+dy^2+dz^2} = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2 }dt$.

# Surface Normals

See the proof page for details.

Given a surface S defined as z = f(x,y), we can compute a line which passes through a point of the surface and is perpendicular (normal) to it there using the direction ratios

$\frac{\partial z}{\partial x} : \frac{\partial z}{\partial y} : -1$

If the surface is described parametrically, ie, as some subset of values (u,v) and parametric equations x = x(u,v),y = y(u,v),z = z(u,v), we can describe a normal vector fairly easily. We define

$\vec{u} = \begin{pmatrix}\partial_u x\\\partial_u y\\\partial_u z\end{pmatrix}, \vec{v} = \begin{pmatrix}\partial_v x\\\partial_v y\\\partial_v z\end{pmatrix}$.

Note that since $\vec{u}$ is tangent to a curve in the surface (name, v=constant), it it tangent to the surface. Similarly, $\vec{v}$ is tangent to the surface. These two vectors both lie in the tangent plane of the surface, and, since the cross product will be perpendicular to both of them, it will be perpendicular to the tangent plane at that point. Hence, the surface normal will simply be $\vec{n}=\vec{u}\times\vec{v}$.

There is an equivalent, but more difficult, formulation of a normal which passes through a point of the surface which uses Jacobians. You'll be asked to demonstrate this in the challenging problems below.

# Problems

## Main Problems

1. Most orbits are not perfectly circular, but are instead ellipses. An ellipse has equation $\vec{r}(t)=a\cos(kt)\vec{i}+b\sin(kt)\vec{j}$. Demonstrate that the influence of gravity can cause a particle to act in this way.

## Challenging Problems

1. In the lecture, we described two trajectories possible for a body in free fall in a gravitational field: straight down and circular. In the main problems, you investigated another kind of orbit, elliptical. Based on this, what other kind of trajectories do you think there could be? Describe one of your suspected trajectories with parametric equations, and demonstrate that it does indeed satisfy the gravitational equation.