Given the following equation:


Part 1: At point $P(3,4,5)$, find the rate of change in the direction of the vector $\langle1,1,-1\rangle$

Part 2: Find the direction in which $V$ changes most rapidly at $P(3,4,5)$

Part 3: Find the maximum rate of change at $P(3,4,5)$

I think I've managed to do part 1. Here's what I've done so far - but I am clueless as to how to find the max rate of change at $P$ and the direction in which the change occurs most rapidly.




Based on the above partial derivatives,

$$\begin{align} \nabla V(3,4,5) &=\left.\langle10x+y(z-3),\ x(z-3),\ xy\rangle\right|_{(3,4,5)} \\ &=\langle38,6,12\rangle \end{align}$$

Unit vector of $\langle1,1,-1\rangle$ is $\vec u=\langle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\rangle$.

Directional derivative at $(3,4,5)$ in the direction of $\langle1,1,-1\rangle$ is: $$\begin{align} D_\vec uV(3,4,5) &= \nabla V(3,4,5) \cdot \vec u \\ &=\frac{32}{\sqrt{3}} \end{align}$$

How should I proceed?

  • 1
    $\begingroup$ There is a small mistake. The unit vector of $(1,1,-1)$ is $\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big)$. $\endgroup$ – Pragabhava Nov 14 '12 at 3:23
  • $\begingroup$ Thanks for the catch! Updated the question accordingly. $\endgroup$ – JTJM Nov 14 '12 at 3:32

In order to understand clearly what's going on, I'm going to restrict myself to two dimensions (i.e. $z=0$).

In this case $$ \tilde V(x,y) = V(x,y,0) = 5 x^2 -3 xy $$

The directional derivative in the direction $\vec{\alpha}$ is $$ \frac{\partial \tilde V}{\partial \vec{\alpha}} = \nabla \tilde V \cdot \frac{\vec{\alpha}}{\|\vec{\alpha}\|} $$ hence $$ \frac{\partial \tilde V}{\partial \vec{\alpha}} = \frac{1}{\|\vec{\alpha}\|}\big(\tilde V_x \alpha_1 + \tilde V_y \alpha_2) $$ where $\vec{\alpha} = (\alpha_1, \alpha_2)^T$.

As a function of $\alpha$, in $(3,4)$ $$ \frac{\partial}{\partial \vec{\alpha}} \tilde V(3,4) = \frac{18 \alpha_1 - 9 \alpha_2}{\sqrt{\alpha_1^2 + \alpha_2^2}} $$ we have $$ \nabla_{(\alpha_1,\alpha_2)}\left(\frac{\partial}{\partial \vec{\alpha}} \tilde V(3,4)\right) = \begin{pmatrix} \frac{9\alpha_2(\alpha_1 + 2\alpha_2)}{(\alpha_1^2 + \alpha_2^2)^{3/2}}\\ -\frac{9\alpha_1(\alpha_1 + 2\alpha_2)}{(\alpha_1^2 + \alpha_2^2)^{3/2}}\\\end{pmatrix} $$ and if $\alpha_1 = -2 \alpha_2$, $\nabla_{(\alpha_1,\alpha_2)} \left(\frac{\partial}{\partial \vec{\alpha}} \tilde V(3,4)\right) = 0$, hence, the direction in which $V$ changes critically is $$ \vec{\alpha} = \frac{1}{\sqrt{5}}\begin{pmatrix}2 \\ -1 \end{pmatrix} $$

$\hskip1.5in$In blue, the contour passing trough the point (3,4)

The maximum rate of change would be $$ \frac{\partial}{\partial \vec{\alpha}} \tilde V(3,4) = 9\sqrt{5} $$ in the direction $(2,-1)$.

Note that the contour $\tilde V(x,y) = 9$, which passes through $(3,4)$ can be expressed locally using the Implicit Function Theorem as a function $g(x)$, where $$ g'(x) = -\frac{\tilde V_x}{\tilde V_y} $$ as long as $\tilde V_y \neq 0$. In $(3,4)$ $$ g'(3) = 2 $$ The direction of min/max growth of a function is normal to the level curve passing trough it!

This is very usefull for finding local maxima/minima of a function, an there is a method called Gradient Descent Method that depends upon this property.

A fantastic excercise would be to prove this result in general ;)

  • $\begingroup$ What sort of tool did you use to draw the picture ? $\endgroup$ – guanglei Feb 26 '16 at 13:10
  • $\begingroup$ @guanglei Mathematica $\endgroup$ – Pragabhava Mar 2 '16 at 3:20

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