Question

Find the directional derivative of f(x,y,z)=z3−x2yf(x,y,z)=z3−x2y at the point (−5,5,2)(−5,5,2) in the direction of the vector v=⟨−3,2,−4⟩v=⟨−3,2,−4⟩.

Answer 1

We are given

$f=z^3 -x^2y$

Firstly, we can find gradient

so, we will find partial derivatives

$f_x=0 -2xy$

$f_x=-2xy$

$f_y=0 -x^2$

$f_y=-x^2$

$f_z=3z^2$

now, we can plug point (-5,5,2)

$f_x=-2*-5*5=50$

$f_y=-(-5)^2=-25$

$f_z=3(2)^2=12$

so, gradient will be

$gradf=(50,-25,12)$

now, we are given that

it is in direction of v=⟨−3,2,−4⟩

so, we will find it's unit vector

$|v|=\sqrt{(-3)^2+(2)^2+(-4)^2}$

$|v|=\sqrt{29}$

now, we can find unit vector

$v'=(\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

now, we can find dot product to find direction of the vector

$dir=(gradf) \cdot (v')$

now, we can plug values

$dir=(50,-25,12) \cdot (\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })$

$dir=(-\frac{150}{\sqrt{29} } - \frac{50}{\sqrt{29} } - \frac{48}{\sqrt{29} })$

$dir=-\frac{248\sqrt{29}}{29}$.............Answer

Answer 2

Answer:

what he/she said

Step-by-step explanation: