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Given the following equation:

$V(x,y,z)=5x^2-3xy+xyz$

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.

$V_x(x,y,z)=10x+y(z-3)$

$V_y(x,y,z)=x(z-3)$

$V_z(x,y,z)=xy$

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
    There is a small mistake. The unit vector of $(1,1,-1)$ is $\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big)$.2012-11-14
  • 0
    Thanks for the catch! Updated the question accordingly.2012-11-14

1 Answers 1