Answer to Question #138195 in Differential Geometry | Topology for anjali g

Question

Find the equation of the tangent plane of the following surface patches at the indicated points: (a) σ(u,v) = (u,v,u2 + v2), P = (1,√2,3).
(b) σ(r,θ) = (rcosθ,rsinθ,2θ), P =(√3,1, π 3).

Answer 1

(a) $\sigma (u,v)=(u,v,u^2+v^2)$ , $P=(1,\sqrt{2}, 3)=\sigma(1,\sqrt{2})$

We have that $\sigma_u=(1,0,2u)$ and $\sigma_v=(0,1,2v)$ .

So, $\sigma_u(1,\sqrt{2})=(1,0,2)$ and $\sigma _v(1,\sqrt{2})=(0,1,2\sqrt{2})$

Now we find the normal to the plane

$n=\sigma_u(1,\sqrt{2})\times \sigma_v(1,\sqrt{2})=\begin{vmatrix}i &j&k \\ 1&0&2\\ 0&1&2\sqrt{2} \end{vmatrix}=(-2,-2\sqrt{2},1)$

The equation of the tangent plane at point $P$ is:

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

or $2x+2\sqrt{2}y-z-3=0$

(b) $\sigma(r,\theta)=(r\cos \theta,r\sin \theta, 2\theta)$ , $P(\sqrt{3}, 1,\pi/3)=\sigma(2,\pi/6)$

We have that $\sigma_r=(\cos\theta,\sin\theta, 0)$ and $\sigma_\theta=(-r\sin\theta, r\cos\theta,2)$

So, $\sigma_r(2,\pi/6)=(\sqrt{3}/2,1/2,0)$ and $\sigma_\theta(2,\pi/6)=(-1,\sqrt{3},2)$

Now we find the normal to the plane

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

The equation of the tangent plane at point $P$ is:

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

or $x-\sqrt{3}y+2z-2\pi/3=0$

Answer: (a) $2x+2\sqrt{2}y-z-3=0$

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