VC3.44

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Here \phi=x^2yz+2xz^2\,

Now,\nabla\phi=[i\frac{\partial}{\partial}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}](x^2yz+4xz^2)\,

=(2xyz+4z^2)i+x^2zj+(x^2y+8xz)k=8i-j-10k\, at the given point.

The unit vector in the direction of 2i-j-2k\,=\hat a=\frac{2i-j-2k}{\sqrt{4+1+4}}=\frac{1}{3}(2i-j-2k)\,

So the required directional derivative at(1,-2,-1)=\nabla\phi\cdot\hat a=(8i-j-10k)\cdot\frac{2i-j-2k}{3}=\frac{16+1+20}{3}=\frac{37}{3}\,

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