Solution:
Given,
σ:R2→R3(r,θ)↦(rcosθ,rsinθ,2θ)We have to find the tangent plane of σ at P(3,1,π/3)=σ(2,π/6)
Now,
We have that
∂r∂σ=σr=(cosθ,sinθ,0)∂θ∂σ=σθ=(−rsinθ,rcosθ,2)
Thus,
σr(2,π/6)=(3/2,1/2,0)
and
σθ(2,π/6)=(−1,3,2)
Now we find the normal to the plane
n=σr(2,π/6)×σθ(2,π/6)=∣∣i3/2−1j1/23k02∣∣=(1,−3,2)
Hence,the equation of the tangent plane at point P is:
1(x−3)−3(y−1)+2(z−π/3)=0⟹x−3y+2z−2π/3=0
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