Difference between revisions of "Gradient"
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Evidently by the Cauchy-Schwartz inequality, the directional derivative in the direction <math>u</math> is maximal in the direction of the gradient, and equal to <math>||\nabla f||</math> for <math>u</math> a unit vector in the direction of the gradient. | Evidently by the Cauchy-Schwartz inequality, the directional derivative in the direction <math>u</math> is maximal in the direction of the gradient, and equal to <math>||\nabla f||</math> for <math>u</math> a unit vector in the direction of the gradient. | ||
| + | |||
| + | ==Properties of the Gradient== | ||
| + | |||
| + | If <math>f</math> is a differentiable function with smooth level sets <math>f^{-1}(c)</math>, then the gradient vector field <math>\nabla f</math> is perpendicular to the level sets of <math>f</math>. For fix a level set <math>S = f^{-1}(c)</math>, and let <math>v</math> be a vector tangent to <math>S</math> at <math>p</math>. Then we can find a curve <math>\gamma(t)</math> on <math>S</math> with <math>\gamma'(0) = v</math>. Now | ||
| + | |||
| + | <math> | ||
| + | f\circ\gamma(t) = c | ||
| + | </math> | ||
| + | |||
| + | since <math>S</math> is a level set. Taking derivatives of both sides and applying the chain rule, we get that | ||
| + | |||
| + | <math> | ||
| + | \nabla f\cdot \gamma'(0) = \nabla f\cdot v = 0 | ||
| + | </math> | ||
| + | |||
| + | Thus, <math>\nabla f</math> is perpendicular to <math>v</math> at <math>p</math>, i.e., the gradient of <math>f</math> is perpendicular to the level sets of <math>f</math>. | ||
[[Category:mathematics]] | [[Category:mathematics]] | ||
Revision as of 15:34, July 2, 2008
In mathematics, the gradient of a real-valued differentiable function <math>f(x_1,...,x_n)</math> at a point <math>p</math> is a vector in <math>R^n</math> which points in the direction in which <math>f</math> increases most rapidly at <math>p</math>. The magnitude of the gradient at <math>p</math> is equal to the maximum directional derivative of <math>f</math> at <math>p</math>.
More precisely, we define the gradient, <math>\nabla f</math> of <math>f</math> to be the vector-field:
<math> \nabla f = (\frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n}) </math>
If <math>u</math> is a unit vector in <math>R^n</math>, then, by the chain rule, the directional derivative of <math>f</math> in the direction of <math>u</math> is simply the dot product:
<math> \nabla f \cdot u </math>
Evidently by the Cauchy-Schwartz inequality, the directional derivative in the direction <math>u</math> is maximal in the direction of the gradient, and equal to <math>||\nabla f||</math> for <math>u</math> a unit vector in the direction of the gradient.
Properties of the Gradient
If <math>f</math> is a differentiable function with smooth level sets <math>f^{-1}(c)</math>, then the gradient vector field <math>\nabla f</math> is perpendicular to the level sets of <math>f</math>. For fix a level set <math>S = f^{-1}(c)</math>, and let <math>v</math> be a vector tangent to <math>S</math> at <math>p</math>. Then we can find a curve <math>\gamma(t)</math> on <math>S</math> with <math>\gamma'(0) = v</math>. Now
<math> f\circ\gamma(t) = c </math>
since <math>S</math> is a level set. Taking derivatives of both sides and applying the chain rule, we get that
<math> \nabla f\cdot \gamma'(0) = \nabla f\cdot v = 0 </math>
Thus, <math>\nabla f</math> is perpendicular to <math>v</math> at <math>p</math>, i.e., the gradient of <math>f</math> is perpendicular to the level sets of <math>f</math>.