Use the gradient to find the directional derivative of the function at P in the direction of Q. $f ( x , y ) = \sin ( 7 x ) \cos y , P ( 0,0 ) , Q \left( \frac { \pi } { 7 } , \pi \right)$
A) $\pi$
B) $\pi ( \sin ( 7 x ) \cos y + \cos ( 7 x ) \sin y )$
C) $\pi ( \cos ( 7 x ) \cos y - \sin ( 7 x ) \sin y )$
D) $\frac { 1 } { \sqrt { 50 } }$
E) $\frac { 7 } { \sqrt { 50 } }$

Question

Quizplus · Accepted Answer

The directional derivative of $f$ at a point $P$ in the direction of a vector $v$ is given by the dot product of the gradient of $f$ at $P$ and the unit vector in the direction of $v$. First, calculate the gradient of $f$:$$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (7\cos(7x)\cos(y), -\sin(7x)\sin(y))$$At $P(0,0)$, this becomes $\nabla f = (7, 0)$.The direction vector $v$ from $P$ to $Q$ is $\left(\frac{\pi}{7}, \pi\right)$. To find the unit vector in the direction of $v$, we normalize $v$:$$\|v\| = \sqrt{\left(\frac{\pi}{7}\right)^2 + \pi^2} = \sqrt{\frac{\pi^2}{49} + \pi^2} = \sqrt{\frac{50\pi^2}{49}} = \frac{\pi\sqrt{50}}{7}$$The unit vector $u$ in the direction of $v$ is $\left(\frac{1}{\sqrt{50}}, \frac{7}{\sqrt{50}}\right)$.The directional derivative is then the dot product of $\nabla f$ at $P$ and $u$:$$D_u f = \nabla f \cdot u = (7, 0) \cdot \left(\frac{1}{\sqrt{50}}, \frac{7}{\sqrt{50}}\right) = 7\left(\frac{1}{\sqrt{50}}\right) + 0 = \frac{7}{\sqrt{50}}$$

Use the Gradient to Find the Directional Derivative of the Function