A surface is given by $$r(u,v) = \langle u, v^2, uv\rangle$$
(a) Evaluate the unit normal vector, $\vec n$, to the surface at the point corresponding to $u=2$ and $v=1$.
I've done this by computing the tangent vectors:
$r_u = \langle 1, 0, v\rangle $ and $r_v = \langle 0, 2v, u\rangle$ and taking their cross product and normalising the result, which yielded the normal unit vector:
$$n = \frac{-1}{\sqrt{3}}\langle1,1,-1\rangle$$
(b) Find the equation of the surface in non-parametric form $$f(x,y,z) = 0$$ and determine the unit normal vector to the surface at the same point as above and thus verify that you get the same result.
I have that the surface is given by $$x^2y-z^2=0.$$ But how do I verify that the unit normal vector is the same?
in parametric form your way of finding normal is correct. we use cross product of vectors tangent to u-curve and v-curve.
in the non-parametric form unit normal vector is indeed normalized gradient $\frac{\nabla f}{|\nabla f|}$. so firstly we have to find non-parametric form of surface: $$x=u \\ y=v^2 \\ z=uv $$ for a curve with equation $z=f(x,y)$ the gradient is defined as $(-\frac{\partial f}{\partial x},-\frac{\partial f}{\partial y},1)$ because the equation of surface is $z-f(x,y)=0$. so we have to find $f(x,y)$ in this equation and finally solve the problem: $$z=uv=x(\pm \sqrt y)=\pm x \sqrt y$$ we have to consider $z=+ x \sqrt y$ because $v=1>0$. so the gradient is: $$(-\sqrt y,-\frac{x}{2\sqrt y},1)$$ substituting $x$ and $y$ and normalizing gives us the normal vector as: $$\frac{1}{\sqrt 3}(-1,-1,1)$$