Question

find the unit normal to the surface xy^3z^2=4 at (1, 1,2)​

Answer 1

Given:

$x {y}^{3} {z}^{2} = 4$

at (1,1,2)

To find:

$f(x ,y,z,) = x {y}^{3} {z}^{2} - 4$

∴ Normal vector to given surface is

$∇(x {y}^{3} {z}^{2} )$

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

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

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

$4\hat i \: + 12\hat j + 8\hat k$

unit normal n =

$\frac{∇ f(1,1,2)}{ |∇ f(1,1,2)| }$

$= (4, 12, 8) /√( 4^2 + 12^2 + 8^2)$

$= (4, 12, 8) /√224$