Solution:
Given,
σ : R 2 → R 3 ( r , θ ) ↦ ( r cos θ , r sin θ , 2 θ ) \sigma:\R^2\rightarrow \R^3\\
(r,\theta)\mapsto(r\cos \theta,r\sin \theta, 2\theta) σ : R 2 → R 3 ( r , θ ) ↦ ( r cos θ , r sin θ , 2 θ ) We have to find the tangent plane of σ \sigma σ at P ( 3 , 1 , π / 3 ) = σ ( 2 , π / 6 ) P(\sqrt{3}, 1,\pi/3)=\sigma(2,\pi/6) P ( 3 , 1 , π /3 ) = σ ( 2 , π /6 )
Now,
We have that
∂ σ ∂ r = σ r = ( cos θ , sin θ , 0 ) \frac{\partial \sigma}{\partial r}=\sigma_r=(\cos\theta,\sin\theta, 0) ∂ r ∂ σ = σ r = ( cos θ , sin θ , 0 ) ∂ σ ∂ θ = σ θ = ( − r sin θ , r cos θ , 2 ) \frac{\partial \sigma}{\partial \theta}=\sigma_\theta=(-r\sin\theta, r\cos\theta,2) ∂ θ ∂ σ = σ θ = ( − r sin θ , r cos θ , 2 )
Thus,
σ r ( 2 , π / 6 ) = ( 3 / 2 , 1 / 2 , 0 ) \sigma_r(2,\pi/6)=(\sqrt{3}/2,1/2,0) σ r ( 2 , π /6 ) = ( 3 /2 , 1/2 , 0 )
and
σ θ ( 2 , π / 6 ) = ( − 1 , 3 , 2 ) \sigma_\theta(2,\pi/6)=(-1,\sqrt{3},2) σ θ ( 2 , π /6 ) = ( − 1 , 3 , 2 )
Now we find the normal to the plane
n ⃗ = σ r ( 2 , π / 6 ) × σ θ ( 2 , π / 6 ) = ∣ i j k 3 / 2 1 / 2 0 − 1 3 2 ∣ = ( 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) n = σ r ( 2 , π /6 ) × σ θ ( 2 , π /6 ) = ∣ ∣ i 3 /2 − 1 j 1/2 3 k 0 2 ∣ ∣ = ( 1 , − 3 , 2 )
Hence,the equation of the tangent plane at point P P P is:
1 ( x − 3 ) − 3 ( y − 1 ) + 2 ( z − π / 3 ) = 0 ⟹ x − 3 y + 2 z − 2 π / 3 = 0 1(x-\sqrt{3})-\sqrt{3}(y-1)+2(z-\pi/3)=0\\
\implies x-\sqrt{3}y+2z-2\pi/3=0 1 ( x − 3 ) − 3 ( y − 1 ) + 2 ( z − π /3 ) = 0 ⟹ x − 3 y + 2 z − 2 π /3 = 0
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