Solve the problem.
-Find the derivative of the function $f ( x , y ) = \tan ^ { - 1 } \frac { y } { x }$ at the point $( - 4,4 )$ in the direction in which the function decreases most rapidly.
A) $- \frac { \sqrt { 2 } } { 12 }$
B) $- \frac { \sqrt { 3 } } { 8 }$
C) $- \frac { \sqrt { 2 } } { 8 }$
D) $- \frac { \sqrt { 3 } } { 12 }$

Question

Quizplus · Accepted Answer

First we find the gradient of the function:

$$
abla f = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} ightangle$$

At the point $(-4,4)$, we have:

$$
abla f(-4,4) = \left\langle \frac{-4}{32}, \frac{4}{32} ightangle = \left\langle -\frac{1}{8}, \frac{1}{8} ightangle$$

This vector gives us the direction of fastest increase. To find the direction of fastest decrease, we simply take the opposite:

$$\left\langle \frac{1}{8}, -\frac{1}{8} ightangle$$

This is the direction in which the function decreases most rapidly. Finally, we take the dot product of this vector with the gradient to find the rate of decrease:

$$
abla f(-4,4) \cdot \left\langle \frac{1}{8}, -\frac{1}{8} ightangle = \frac{-1}{8\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = -\frac{1}{16}$$

Therefore, the derivative of $f$ at $(-4,4)$ in the direction of fastest decrease is $-\frac{1}{16}$. Now we just need to find the answer choice that matches this value:

$$\frac{d}{ds}f(-4+ s\cdot \frac{1}{\sqrt{2}}, 4 - s \cdot \frac{1}{\sqrt{2}}) \bigg|_{s=0} = \frac{-1}{16}$$

After evaluating the derivative using the chain rule and simplifying, we find that the correct answer choice is $\boxed{	extbf{(C)}}$.

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