differential-geometry
Find the equation of the tangent plane of each of the following surface
patches at the indicated points:
$$σ(r,θ)=(r\cosh(θ),r\sinh(θ),r^2), (1, 0, 1).$$
I'm not sure what to do, any hints are greatly appreciated.
What I have:
$σ_r=(\cosh\theta,\sinh\theta,2r)$
$σ_\theta=(r\sinh\theta,r\cosh\theta,0)$
If a surface is defined by a regular function $z=g(x,y)$, the equation of the tangent plane at $(x_{0},y_{0},z_{0})$ in terms of the partial derivatives of $g$ is
$$z=z_{0}+\left. \frac{\partial g}{\partial x}\right\vert _{(x_{0},y_{0})}(x-x_{0})+\left. \frac{\partial g}{\partial y}\right\vert _{(x_{0},y_{0})}(y-y_{0}).$$
If the surface is defined implicitly by $f(x,y,z)=0$, then $z=g(x,y)=f^{-1}(0)$ (i.e. $f(x,y,g(x,y))\equiv 0$). Since
$$\frac{\partial g}{\partial x}=-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}$$
and
$$\frac{\partial g}{\partial y}=-\frac{\partial f}{\partial y}/\frac{\partial f}{\partial z},$$
the equation of the tangent plane at $(x_{0},y_{0},z_{0})$ is given by
$$z=z_{0}-\left(\frac{\partial f}{\partial x}/\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}(x-x_{0})-\left(\frac{\partial f}{\partial y}/\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}(y-y_{0})$$
or
$$(x-x_{0})\left(\frac{\partial f}{\partial x}\right)_{(x_{0},y_{0},z_{0})}+(y-y_{0})\left(\frac{\partial f}{ \partial y}\right)_{(x_{0},y_{0},z_{0})}+(z-z_{0})\left(\frac{\partial f}{\partial z}\right)_{(x_{0},y_{0},z_{0})}=0.$$
In compact notation, $\mathbf{x}=\left( x,y,z\right) ,\mathbf{x}_{0}=\left( x_0,y_0,z_0\right) $, we get
$$\left( \mathbf{x}-\mathbf{x}_{0}\right) \cdot \mathbf{\nabla }f(\mathbf{x}_{0})=0.$$
After applying a rigid motion to $S$ and $P$, we may as well assume $(x_0,y_0,z_0)=(0,0,0)$ and $P$ is the $xy$-plane. Let $U$ be a small neighborhood of the origin in $S$; by taking $U$ to be small enough, we may assume that $U$ is the image of an open disk in $\mathbb R^2$ under a local parametrization, and thus $U$ is homeomorphic to a disk. The hypothesis implies that the origin is the only point of $U$ for which the $z$-coordinate is zero, so $U\smallsetminus\{(0,0,0)\}$ is contained in the set where $z\ne 0 $. Since a disk minus a point is connected, it follows that either $z>0$ on all of $U\smallsetminus\{(0,0,0)\}$ or $z<0$ on all of $U\smallsetminus\{(0,0,0)\}$. After reflecting in the $xy$-plane, we may assume it is $z>0$.
Now suppose $v$ is a tangent vector to $S$ at the origin. Then there is a smooth curve $c:(-\varepsilon,\varepsilon)\to U$ satisfying $c(0) = (0,0,0)$ and $c'(0) = v$. If we write $c(t) = (x(t),y(t),z(t))$, we see that $z(t)$ attains a minimum at $t=0$, and therefore $z'(0)=0$. This means that every tangent vector at the origin has zero $z$-coordinate, so $T_{(0,0,0)}S$ is contained in the $xy$-plane (i.e., in $P$). Since both $T_{(0,0,0)}S$ and $P$ are two-dimensional, they must be equal.
Best Answer
Compute $\sigma_r \times \sigma_\theta = (-2r\cosh\theta, 2r^2 \sinh \theta, r)$. Since $\sigma(1,0) = (1,0,1)$, the vector $\sigma_r \times \sigma_\theta(1,0) = (-2,0,1)$ is normal to the tangent plane at $(1,0,1)$. The equation of the plane is $(-2,0,1) \cdot (x - 1, y - 0, z - 1) = 0$, or $2x - z = 1$.