Question

Find out the unit normal to the surface xy³z²= 4 at (1,12)

Answer 1

Surface xy³z² = 4

∴ Normal vector to given surface is ∇(xy³z²)

 \(=\hat i\frac{\partial}{\partial\mathrm x}\mathrm xy^3z^2+\hat j \frac{\partial}{\partial y}\mathrm x y^3z^2+\hat k\frac{\partial}{\partial z}(\mathrm xy^3z^2)\)

= y³z²\(\hat i\) + 3xy²z²\(\hat j\) + 2xy³z\(\hat k\)

Normal vector to given surface at (1, 1, 2) is 

\(\mathrm{\left[\nabla(xy^3z^2)\right]}_{(1, 1, 2)}\) = 4\(\hat i\) + 12\(\hat j\) + 8\(\hat k\)    (By putting x = 1, y = 1, z = 2)

So, the vector normal to surface xy³z² = 4 is 4\(\hat i\) + 12\(\hat j\) + 8\(\hat k\).