Question

Please show all your work. I will give an upvote if all work is correct.

Use the gradient to find the directional derivative of the function at P in the direction of v. h(x, y) = (-9sin(y), P(1,5), v=-i

Answer 1

we are given

$h=e^{-9x}sin(y)$

now, we can find partial derivatives

$h_x=-9e^{-9x}sin(y)$

$h_y=e^{-9x}cos(y)$

now, we can plug

x=1 and y=pi/2

$h_x=-9e^{-9(1)}sin(\frac{\pi}{2})$

$h_x=-9e^{-9}$

$h_y=e^{-9(1)}cos(\frac{\pi}{2})$

$h_y=0$

now, we can find gradient

grad(h) = (-9e-9,0)

now, we can find directional derivative

$Dir(h)=(grad(h))\cdot v$

$Dir(h)=(-9e^{-9},0)\cdot (-1,0)$

$Dir(h)=-9e^{-9}\times -1+0\times 0$

so, we get

$Dir(h)=9e^{-9}$............Answer