VC3.41

From Exampleproblems

We shall use the identity $\mathrm{grad}(uv)=u\mathrm{grad}v+v\mathrm{grad}u\,$ --(1)

Now, $\mathrm{grad}u=i\frac{\partial}{\partial x}(e^{2x}+x^2z)+j\frac{\partial}{\partial y}(e^{2x}+x^2z)+k\frac{\partial}{\partial z}(e^{2x}+x^2z)=i(2e^{2x}+2xz)+j(0)+k(x^2)\,$ --(2)

and $\mathrm{grad}v=i\frac{\partial}{\partial x}(2z^2y-xy^2)+j\frac{\partial}{\partial y}(2z^2y-xy^2)+k\frac{\partial}{\partial z}(2z^2y-xy^2)=i(-y^2)+j(2z^2-2xy)+k(4zy)\,$ --(3)

Using (2) and (3),(1) becomes

$\mathrm{grad}(uv)=(e^{2x}+x^2z)[-y^2i+(2z^2-2xy)j+4zyk]+(2z^2y-xy^2)[(2e^{2x}+2xz)i+x^2k]\,$

Putting x=1,y=0,z=2 in the above equation,we get

$\mathrm{grad}(uv)\,$ at(1,0,2)= $(e^2+2)[8j]+0=8(2+e^2)j\,$

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VC3.41

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