Question #138195
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).
1
Expert's answer
2020-10-15T18:44:30-0400

(a) σ(u,v)=(u,v,u2+v2)\sigma (u,v)=(u,v,u^2+v^2) , P=(1,2,3)=σ(1,2)P=(1,\sqrt{2}, 3)=\sigma(1,\sqrt{2})

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

So, σu(1,2)=(1,0,2)\sigma_u(1,\sqrt{2})=(1,0,2) and σv(1,2)=(0,1,22)\sigma _v(1,\sqrt{2})=(0,1,2\sqrt{2})

Now we find the normal to the plane

n=σu(1,2)×σv(1,2)=ijk1020122=(2,22,1)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 PP is:

2(x1)22(y2)+1(z3)=0-2(x-1)-2\sqrt{2}(y-\sqrt{2})+1(z-3)=0

or 2x+22yz3=02x+2\sqrt{2}y-z-3=0


(b) σ(r,θ)=(rcosθ,rsinθ,2θ)\sigma(r,\theta)=(r\cos \theta,r\sin \theta, 2\theta) , P(3,1,π/3)=σ(2,π/6)P(\sqrt{3}, 1,\pi/3)=\sigma(2,\pi/6)

We have that σr=(cosθ,sinθ,0)\sigma_r=(\cos\theta,\sin\theta, 0) and σθ=(rsinθ,rcosθ,2)\sigma_\theta=(-r\sin\theta, r\cos\theta,2)

So, 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)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 PP is:

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

or x3y+2z2π/3=0x-\sqrt{3}y+2z-2\pi/3=0


Answer: (a) 2x+22yz3=02x+2\sqrt{2}y-z-3=0

(b) x3y+2z2π/3=0x-\sqrt{3}y+2z-2\pi/3=0


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