# Cannot Solve Tangent Plane Equation to Parametric Surface

I have a problem while finding the equation of a tangent plane to a parametric surface. The surface is given by $$\pmb r = \begin{bmatrix} s^2 + t^2 \cr 2s+2t \cr 2 \end{bmatrix}$$

The point at which plane is tangent $$\pmb r_o = \begin{bmatrix} 2 \cr 4 \cr 2 \end{bmatrix} (s=1, t=1)$$

I know that for tangent-plane equation, I need to find a normal vector $$\pmb n$$ by doing a cross-product of $$\frac{\partial \pmb r}{\partial t}$$ at $$r_o$$ and $$\frac{\partial \pmb r}{\partial s}$$ at $$r_o$$.

My problem is that both $$\frac{\partial \pmb r}{\partial t}$$ and $$\frac{\partial \pmb r}{\partial s}$$ are being computed as the same vector $$\begin{bmatrix} 2 \cr 2 \cr 0 \end{bmatrix}$$. So my normal vector will always be zero, and I can't use it to find the equation of the tangent.

How to get around this problem?

Also, in general, how do we solve tangent-plane equations to parametric surfaces in such cases?

Your "surface" $$S$$ is the subset $$x\geq y^2/8$$ inside the plane $$z=2$$. As such, it is not regular along the curve $$(\tau, \tau^2/8, 2)$$ and, according to most standard definitions, one cannot talk about tangent planes at points along that curve. Since $$\pmb r(1,2)$$ lies on this curve, we have no well-defined tangent. Of course, since $$S$$ is entirely contained in the plane $$z=2$$, at any point not on its boundary curve the tangent plane is $$z=2$$.