Answer to Question #227723 in Calculus for Mis

Question

Find the directional derivative of (x,y,z) = xsin yz at (1,3,0) in the direction of i+2j-k

Answer 1

$$f(x,y,z)=x\sin yz$$

$$\nabla f=(\sin yz)i+(xz\cos yz)j+(xy\cos yz)k$$

$$\nabla f(1,3,0)=0i+0j+3k$$

$$|i+2j-k|=\sqrt{1^2+2^2+(-1)^2}=\sqrt{6}$$

$$u=\dfrac{1}{\sqrt{6}}i+\dfrac{2}{\sqrt{6}}j-\dfrac{1}{\sqrt{6}}k$$

$$D_uf(1,3,0)=(0i+0j+3k)(\dfrac{1}{\sqrt{6}}i+\dfrac{2}{\sqrt{6}}j-\dfrac{1}{\sqrt{6}}k)$$

$$=-\dfrac{\sqrt{6}}{2}$$