Question

Find the directional derivative of the function f(x,y)=2cos(y)exf(x,y)=2cos⁡(y)ex at the point (0,π6)(0,π6) in the direction of the vector v=⟨−8,6⟩v=⟨−8,6⟩.

Answer 1

Answer:

$D_u f(0,\pi/6)=\frac{4\sqrt{3}-3}{5}$

Step-by-step explanation:

We are given that

$f(x,y)=2cos(y)e^{x}$

Point=($0,\pi/6$)

Vector v=<-8,6>

We have to find the directional derivative of the function.

$f_x(x,y)=2cos(y)e^{x}$

$f_x(0,\pi/6)=\sqrt{3}$

$f_y(x,y)=-2sin(y)e^{x}$

$f_y(0,\pi/6)=-1$

$|v|=\sqrt{(-8)^2+6^2}=\sqrt{100}=10$

$\hat{u}=\frac{v}{|v|}=\frac{-8i+6j}{10}=\frac{1}{10}(-8i+6j)$

$\nabla f(0,\pi/6)=<f_x(0,\pi/6),f_y(0,\pi/6)>=<\sqrt{3},-1>$

Now, the directional derivative in the direction of v at point (0,$\pi/6$) is given by

$D_u f(0,\pi/6)=\nabla f(0,\pi/6)\cdot \hat{u}$

By using the formula

$D_u f(x,y)=\nabla f(x,y)\cdot \hat{u}$

$D_u f(0,\pi/6)=<\sqrt{3},-1>\cdot <-\frac{8}{10},\frac{6}{10}>$

$D_u f(0,\pi/6)=-\frac{4\sqrt{3}}{5}-\frac{3}{5}=-\frac{1}{5}(4\sqrt{3}+3)$