Question #140102
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
1
Expert's answer
2020-10-26T19:12:45-0400

Solution:

Given,

σ:R2R3(r,θ)(rcosθ,rsinθ,2θ)\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 σ\sigma at P(3,1,π/3)=σ(2,π/6)P(\sqrt{3}, 1,\pi/3)=\sigma(2,\pi/6)

Now,

We have that

σr=σr=(cosθ,sinθ,0)\frac{\partial \sigma}{\partial r}=\sigma_r=(\cos\theta,\sin\theta, 0)σθ=σθ=(rsinθ,rcosθ,2)\frac{\partial \sigma}{\partial \theta}=\sigma_\theta=(-r\sin\theta, r\cos\theta,2)


Thus,

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


and

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


Now we find the normal to the plane


n=σr(2,π/6)×σθ(2,π/6)=ijk3/21/20132=(1,3,2)\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 PP is:


1(x3)3(y1)+2(zπ/3)=0    x3y+2z2π/3=01(x-\sqrt{3})-\sqrt{3}(y-1)+2(z-\pi/3)=0\\ \implies x-\sqrt{3}y+2z-2\pi/3=0




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS