Function $$\mathbf{r}(u,v)=(2u+2v)\mathbf{i}+(-3+v^2)\mathbf{j}+(2u^2)\mathbf{k}$$ is a parametrization of a surface. What's the normal vector of this surface at the point $(u,v)=(1,-2)$.
What I got:
$$\begin{align}\mathbf{n}(u,v)&=\frac{d\mathbf{r}}{du}(u,v)\times\frac{d\mathbf{r}}{dv}(u,v)\\ &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & 4u \\ 2 & 2v & 0 \end{vmatrix} \\ &=(-8uv)\mathbf{i}+(8u)\mathbf{j}+(4v)\mathbf{k}\\ \mathbf{n}(1,-2)&=16\mathbf{i}+8\mathbf{j}-8\mathbf{k}\end{align}$$
Makes sense?