Answer to Question #140102 in Differential Geometry | Topology for shweta

Question

Find the equation of the tangent plane of the following surface patches
at the indicated points:
(b) σ(r, θ) = (r cos θ, r sin θ, 2θ), P =
√
3, 1,
π
3

Answer 1

Solution:

Given,

$$\sigma:\R^2\rightarrow \R^3\\ (r,\theta)\mapsto(r\cos \theta,r\sin \theta, 2\theta)$$

We have to find the tangent plane of $\sigma$ at $P(\sqrt{3}, 1,\pi/3)=\sigma(2,\pi/6)$

Now,

We have that

$$\frac{\partial \sigma}{\partial r}=\sigma_r=(\cos\theta,\sin\theta, 0)$$ $$\frac{\partial \sigma}{\partial \theta}=\sigma_\theta=(-r\sin\theta, r\cos\theta,2)$$

Thus,

$$\sigma_r(2,\pi/6)=(\sqrt{3}/2,1/2,0)$$

and

$$\sigma_\theta(2,\pi/6)=(-1,\sqrt{3},2)$$

Now we find the normal to the plane

$$\vec n=\sigma_r(2,\pi/6)\times \sigma_\theta(2,\pi/6)=\begin{vmatrix}i &j&k \\ \sqrt{3}/2&1/2&0\\ -1&\sqrt{3}&2 \end{vmatrix}=(1,-\sqrt{3},2)$$

Hence,the equation of the tangent plane at point $P$ is:

$$1(x-\sqrt{3})-\sqrt{3}(y-1)+2(z-\pi/3)=0\\ \implies x-\sqrt{3}y+2z-2\pi/3=0$$