I'm trying to find the equation of the tangent plane of a surface, at a given point $\vec{a}$
For a surface, say $f(x,y,z)=0$, this is how I'd normally find the tangent plane :
$\hat{n} = \frac{\vec\nabla f(x,y,z)}{|\vec\nabla f(x,y,z)|}$.
This is the unit normal to the surface. Then we can say that, the tangent plane is given by :
$(\vec{r}-\vec{a})\hat{n} = 0 $
From here, we can easily obtain the equation of the tangent plane in the cartesian form $ax+by+cz=0$.
Now, here is my first doubt.
In cartesian coordinates, doesn't the equation $ax+by+cz=0$ represent both a plane and a curve. Moreover, the equation $f(x,y,z)=0$ represents both a curve, as well as a surface. How do we know, if we are getting a tangent line or a tangent plane, or that the equation represents a curve or a surface?
My guess is, we can differentiate if we are using the parametric form, because in the case of curves, we need one parameter, and for surfaces, we need two. However, I still don't know how can we go from cartesian representation to parametric representation, for $ax+by+cz=0$ in the first place, if we don't know if it's a line or a plane.
Suppose, we have the equations giving in parametric forms.
For a curve, we have $\vec{r_c}=f(t)\hat{i} + g(t)\hat{j}+ h(t)\hat{k}$.
For a surface, it becomes $\vec{r_c}=f(u,v)\hat{i} + g(u,v)\hat{j}+ h(u,v)\hat{k}$.
However, given in this form, how do we find the tangent/tangent plane since we cannot use the gradient operator anymore?
For the curve, we can just take the partial w.r.t to the variable $t$, at the point $\vec{a}$, and that will give us the tangent vector ( not the tangent plane ). We have found the vector parallel to the plane and passing through the plane. How do I find the normal vector to this point, in order to find an expression for the tangent plane ? Basically, $(\vec{r}-\vec{a})\hat{n}=0$ and I've managed to find a vector parallel to $(\vec{r}-\vec{a})$ at that point. How do I proceed?
However, how do I do this for the parametric surface? My intuition is to take the partial with respect to the parametric variables $u$ and $v$ , and then take the cross product to get the normal vector, and hence the tangent plane.
Is this intuition correct?
Can anyone clear these doubts for me? How to know if an expression represents a curve or a surface in cartesian coordinates, and how do I find the tangent plane to them, in parametric form? How do I find the tangent plane to a curve in parametric form?
Best Answer
Yes. Your intuition is correct. For a parametrically specified surface, find $\dfrac{\partial \vec{r}}{\partial u} $ and $\dfrac{\partial \vec{r}}{\partial v} $, then find their cross product,
$ \vec{n} = \dfrac{\partial \vec{r}}{\partial u} \times \dfrac{\partial \vec{r}}{\partial v} $
then $\vec{n}$ is a normal vector to the surface, at $\vec{a}$, which you can use to find the equation of the tangent plane, as follows:
$\vec{n} \cdot (\vec{r} - \vec{a}) = 0 $