Question #227723

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


Expert's answer

f(x,y,z)=xsin⁡yzf(x,y,z)=x\sin yz

∇f=(sin⁡yz)i+(xzcos⁡yz)j+(xycos⁡yz)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+2j−k∣=12+22+(−1)2=6|i+2j-k|=\sqrt{1^2+2^2+(-1)^2}=\sqrt{6}

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

Duf(1,3,0)=(0i+0j+3k)(16i+26j−16k)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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