Question

Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xey + yez + zex, (0, 0, 0), v = 5, 2, −2

Answer 1

The directional derivative of $f$ at the point $\vec p$ in the direction of $\vec v$ is

$D_{\vec v}f(\vec p)=\nabla f(\vec p)\cdot\dfrac{\vec v}{\|\vec v\|}$

We have

$f(x,y,z)=xe^y+ye^z+ze^x\implies\nabla f(x,y,z)=\langle e^y+ze^x,xe^y+e^z,ye^z+e^x\rangle$

With $\vec p=\langle0,0,0\rangle$,

$\nabla f(0,0,0)=\langle1,1,1\rangle$

$\vec v$ has magnitude

$\|\vec v\|=\sqrt{5^2+2^2+(-2)^2}=\sqrt{33}$

and so the directional derivative is

$\langle1,1,1\rangle\cdot\dfrac{\langle5,2,-2\rangle}{\sqrt{33}}=\boxed{\dfrac5{\sqrt{33}}}$