Parametric Surfaces — Question 2

Question 2 – Solution

Strategy. Use $x=u$, $y=v$, and let the defining graph supply the third component.

Step 1: Parametrize $$\mathbf r(u,v)=\langle u,v,u^2+v^2\rangle, \qquad (u,v)\in\mathbb R^2.$$ The point $P$ corresponds to $(u,v)=(1,-1)$.

Step 2: Normal vector $$\mathbf r_u=\langle 1,0,2u\rangle, \qquad \mathbf r_v=\langle 0,1,2v\rangle,$$ so $$\mathbf r_u\times\mathbf r_v=\langle-2u,-2v,1\rangle.$$ At $P$, an upward normal is $\boxed{\langle-2,2,1\rangle}$.

Step 3: Tangent plane $$-2(x-1)+2(y+1)+(z-2)=0,$$ or $$\boxed{z=2x-2y-2}.$$

Verification Substitution of $(1,-1,2)$ gives $2=2+2-2$. The normal from the graph equation $x^2+y^2-z=0$ is $\langle 2,-2,-1\rangle$, the negative of the chosen normal.