Question #227723

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


1
Expert's answer
2021-08-23T14:49:24-0400
f(x,y,z)=xsinyzf(x,y,z)=x\sin yz

f=(sinyz)i+(xzcosyz)j+(xycosyz)k\nabla f=(\sin yz)i+(xz\cos yz)j+(xy\cos yz)k


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

i+2jk=12+22+(1)2=6|i+2j-k|=\sqrt{1^2+2^2+(-1)^2}=\sqrt{6}

u=16i+26j16ku=\dfrac{1}{\sqrt{6}}i+\dfrac{2}{\sqrt{6}}j-\dfrac{1}{\sqrt{6}}k

Duf(1,3,0)=(0i+0j+3k)(16i+26j16k)D_uf(1,3,0)=(0i+0j+3k)(\dfrac{1}{\sqrt{6}}i+\dfrac{2}{\sqrt{6}}j-\dfrac{1}{\sqrt{6}}k)

=62=-\dfrac{\sqrt{6}}{2}


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