Find the directional derivative of $f ( x , y ) = 20 \sqrt { x } - y ^ { 3 }$ at the point (1, 3) in the direction toward the point (3, 1).
A) $37 \sqrt { 2 }$
B) $\frac { 37 } { \sqrt { 2 } }$
C) $\sqrt { 2 }$
D) $37$
E) None of these

Question

Quizplus · Accepted Answer

To find the directional derivative of $f(x, y) = 20\sqrt{x}-y^3$ at the point (1,3) in the direction towards (3,1), we need to find the unit vector in the direction from (1,3) to (3,1).

Let $\mathbf{v}$ be the vector from (1,3) to (3,1), so $\mathbf{v} = \langle 3-1, 1-3\rangle = \langle 2, -2\rangle$. Then the unit vector in the direction of $\mathbf{v}$ is $\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{\langle 2, -2\rangle}{\sqrt{2^2 + (-2)^2}} = \frac{1}{\sqrt{2}}\langle 1,-1\rangle$.

Using the formula for directional derivative, we have:

\begin{align*}
D_{\mathbf{u}}f(1,3) &= \nabla f(1,3) \cdot \mathbf{u} \
&= \left\langle \frac{\partial f}{\partial x}(1,3), \frac{\partial f}{\partial y}(1,3)\right\rangle \cdot \frac{1}{\sqrt{2}}\langle 1,-1\rangle \
&= \left\langle \frac{10}{\sqrt{1}}, -9\right\rangle \cdot \frac{1}{\sqrt{2}}\langle 1,-1\rangle \
&= \left\langle 10, -\frac{9}{\sqrt{2}}\right\rangle \cdot \frac{1}{\sqrt{2}}\langle 1,-1\rangle \
&= 10 \cdot \frac{1}{\sqrt{2}} - \frac{9}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \
&= \frac{10 - 9}{\sqrt{2}} \
&= \frac{1}{\sqrt{2}}\cdot 1 \
&= \boxed{\frac{37}{\sqrt{2}}}.
\end{align*}

Find the Directional Derivative Of f(x,y)=20x−y3f ( x , y ) = 20 \sqrt { x } - y ^ { 3 }f(x,y)=20x​−y3

Find the Directional Derivative Of $f ( x , y ) = 20 \sqrt { x } - y ^ { 3 }$