Question

Find the directional derivative of the function at the given point in the direction of vector v. g(x, y, z) = (x + 2y + 3z)3/2, (5, 4, 4), v = 4j − k dug(5, 4, 4) =

Answer 1

$g(x,y,z)=(x+2y+3z)^{3/2}$

$\implies\nabla g(x,y,z)=\dfrac32\left((x+2y+3z)^{1/2},2(x+2y+3z)^{1/2},3(x+2y+3z)^{1/2}\right)$

In the direction of $\mathbf v=4\,\mathbf j-\mathbf k=(0,4,-1)$, with norm $\sqrt{17}$, we have the directional derivative

$D_{\mathbf v}g(5,4,4)=\nabla g(5,4,4)\cdot\dfrac{(0,4,-1)}{\sqrt{17}}=\dfrac1{\sqrt{17}}\left(\dfrac{15}2,15,\dfrac{45}2\right)\cdot(0,4,-1)=\dfrac{75}{2\sqrt{17}}$