Solve the problem.
-About how much will $f ( x , y ) = \tan ^ { - 1 } x y$ change if the point $( x , y , z )$ moves from $\left( 9 \sqrt { 2 } , \frac { - 6 } { \sqrt { 2 } } \right)$ a distance of $\mathrm { ds } = \frac { 1 } { 10 }$ unit in the direction of $\mathbf { i } + \mathbf { j } ?$
A) $\frac { 3 } { 14585 }$
B) $\frac { 3 } { 5834 }$
C) $\frac { 21 } { 58340 }$
D) $\frac { 3 } { 58340 }$

Question

Quizplus · Accepted Answer

The correct answer is A because the directional derivative of $f(x, y) = \tan^{-1} xy$ in the direction of $\mathbf{i} + \mathbf{j}$ is $$\frac{y}{x^2 + y^2} \mathbf{i} + \frac{x}{x^2 + y^2} \mathbf{j}$$and at the point $(9\sqrt{2}, -\frac{6}{\sqrt{2}})$, this directional derivative is $$\frac{-\frac{6}{\sqrt{2}}}{(9\sqrt{2})^2 + \left(-\frac{6}{\sqrt{2}}\right)^2} \mathbf{i} + \frac{9\sqrt{2}}{(9\sqrt{2})^2 + \left(-\frac{6}{\sqrt{2}}\right)^2} \mathbf{j} = \frac{3}{14585} \mathbf{i} + \frac{3}{14585} \mathbf{j}$$Therefore, if the point $(x, y, z)$ moves from $(9\sqrt{2}, -\frac{6}{\sqrt{2}})$ a distance of $\frac{1}{10}$ unit in the direction of $\mathbf{i} + \mathbf{j}$, then $f(x, y)$ will change by approximately $$\frac{3}{14585} \cdot \frac{1}{10} = \frac{3}{145850}$$

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