How Do You Find an Equation of the Tangent Plane for a Torus?

Question

I've parameterized the torus given by $z^2 + (r - 2)^2$ = 1 using:
x = (2 + $\cos\theta$)$\cos\alpha$, y = (2 + $\cos\theta$)$\sin\alpha$, z = $\sin\theta$. I'm really stuck on how to find the equation of the tangent plane to the torus.

Answer 1

Call $\phi:\mathbb R^2\to\mathbb R^3:(\alpha,\theta)\mapsto(x,y,z)$ your parametrization. Then the vectors $\left\{\frac{\partial\phi}{\partial\alpha},\frac{\partial\phi}{\partial\theta}\right\}$ span the tangent space $T$. If you want to find an equation, take a normal vector $n=\frac{\partial\phi}{\partial\alpha}\times\frac{\partial\phi}{\partial\theta}$, and your tangent space becomes $$ T=\{v\in\mathbb R^3:n\cdot v=0\}. $$ If you are interested in the affine tangent plane, you just translate $T$ to the corresponding point on the torus.

Alternatively, from the implicit definition $$ f(x,y,z) = z^2+\left(\sqrt{x^2+y^2}-2\right)^2-1=0, $$ you know that the tangent space is defined by $\{v:v\cdot\nabla f=0\}$. Again, you can translate it to $$ T_{(x,y,z)}=\{v:[v-(x,y,z)]\cdot\nabla f=0\} $$ if you are interested in the affine plane.

Answer 2

If you're only looking for the equation of the tangent plane of the torus $$ z^2+\left(\sqrt{x^2+y^2}-2\right)^2 = 1, $$ then you can consider the function $$ f(x,y,z) = z^2+\left(\sqrt{x^2+y^2}-2\right)^2-1, $$ so that your torus is the preimage $f^{-1}(\{0\}),$ and so that the gradient $\nabla f(x_0,y_0,z_0)$ is a normal vector to the torus at the point $(x_0, y_0, z_0)$, which will immediately give you the equation you are after. Can you fill in the specific details on your own?