Question

Find the directional derivative of f(x,y,z)=xy+z^3 at the point (2,3,1) in the direction of a vector making an angle of 3π/4 with ∇f(2,3,1).

Answer 1

Firstly , we find partial derivative

$f_x=y$

$f_y=x$

$f_z=3z^2$

now, we can plug points (2,3,1)

$(f_x,f_y,f_z)=(3,2.3)$

we are given angle between gradient and unit vector as

$\theta=\frac{3\pi}{4}$

so, we can find direction derivative

$Direction_D=(grad-f).(u)$

$Direction_D=|grad-f||u|cos(\theta)$

$Direction_D=\sqrt{3^2+2^2+3^2}*1cos(\frac{3\pi}{4})$

$Direction_D=\sqrt{22}cos(\frac{3\pi}{4})$

$Direction_D=\sqrt{22}*\frac{-\sqrt{2}}{2}$

$Direction_D=\sqrt{2}*\sqrt{11}*\frac{-\sqrt{2}}{2}$

$Direction_D=-2*\sqrt{11}*\frac{1}{2}$

$Direction_D=-\sqrt{11}$................ANswer