(a) σ ( u , v ) = ( u , v , u 2 + v 2 ) \sigma (u,v)=(u,v,u^2+v^2) σ ( u , v ) = ( u , v , u 2 + v 2 ) , P = ( 1 , 2 , 3 ) = σ ( 1 , 2 ) P=(1,\sqrt{2}, 3)=\sigma(1,\sqrt{2}) P = ( 1 , 2 , 3 ) = σ ( 1 , 2 )
We have that σ u = ( 1 , 0 , 2 u ) \sigma_u=(1,0,2u) σ u = ( 1 , 0 , 2 u ) and σ v = ( 0 , 1 , 2 v ) \sigma_v=(0,1,2v) σ v = ( 0 , 1 , 2 v ) .
So, σ u ( 1 , 2 ) = ( 1 , 0 , 2 ) \sigma_u(1,\sqrt{2})=(1,0,2) σ u ( 1 , 2 ) = ( 1 , 0 , 2 ) and σ v ( 1 , 2 ) = ( 0 , 1 , 2 2 ) \sigma _v(1,\sqrt{2})=(0,1,2\sqrt{2}) σ v ( 1 , 2 ) = ( 0 , 1 , 2 2 )
Now we find the normal to the plane
n = σ u ( 1 , 2 ) × σ v ( 1 , 2 ) = ∣ i j k 1 0 2 0 1 2 2 ∣ = ( − 2 , − 2 2 , 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) n = σ u ( 1 , 2 ) × σ v ( 1 , 2 ) = ∣ ∣ i 1 0 j 0 1 k 2 2 2 ∣ ∣ = ( − 2 , − 2 2 , 1 )
The equation of the tangent plane at point P P P is:
− 2 ( x − 1 ) − 2 2 ( y − 2 ) + 1 ( z − 3 ) = 0 -2(x-1)-2\sqrt{2}(y-\sqrt{2})+1(z-3)=0 − 2 ( x − 1 ) − 2 2 ( y − 2 ) + 1 ( z − 3 ) = 0
or 2 x + 2 2 y − z − 3 = 0 2x+2\sqrt{2}y-z-3=0 2 x + 2 2 y − z − 3 = 0
(b) σ ( r , θ ) = ( r cos θ , r sin θ , 2 θ ) \sigma(r,\theta)=(r\cos \theta,r\sin \theta, 2\theta) σ ( r , θ ) = ( r cos θ , r sin θ , 2 θ ) , 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 )
We have that σ r = ( cos θ , sin θ , 0 ) \sigma_r=(\cos\theta,\sin\theta, 0) σ r = ( cos θ , sin θ , 0 ) and σ θ = ( − r sin θ , r cos θ , 2 ) \sigma_\theta=(-r\sin\theta, r\cos\theta,2) σ θ = ( − r sin θ , r cos θ , 2 )
So, 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 ) 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 )
The equation of the tangent plane at point P P P is:
1 ( x − 3 ) − 3 ( y − 1 ) + 2 ( z − π / 3 ) = 0 1(x-\sqrt{3})-\sqrt{3}(y-1)+2(z-\pi/3)=0 1 ( x − 3 ) − 3 ( y − 1 ) + 2 ( z − π /3 ) = 0
or x − 3 y + 2 z − 2 π / 3 = 0 x-\sqrt{3}y+2z-2\pi/3=0 x − 3 y + 2 z − 2 π /3 = 0
Answer: (a) 2 x + 2 2 y − z − 3 = 0 2x+2\sqrt{2}y-z-3=0 2 x + 2 2 y − z − 3 = 0
(b) x − 3 y + 2 z − 2 π / 3 = 0 x-\sqrt{3}y+2z-2\pi/3=0 x − 3 y + 2 z − 2 π /3 = 0
Comments