Parametric Surfaces — Question 5

Question 5 – Solution

Strategy. On this cone the radius equals the height, so one parameter can serve both roles.

Step 1: Parametrize $$\mathbf r(u,v)=\langle u\cos v,u\sin v,u\rangle, \qquad 1\le u\le 2,\qquad 0\le v\le 2\pi.$$

Step 2: Cross the tangents $$\mathbf r_u=\langle\cos v,\sin v,1\rangle, \qquad \mathbf r_v=\langle-u\sin v,u\cos v,0\rangle.$$ Thus $$\mathbf r_u\times\mathbf r_v =\langle-u\cos v,-u\sin v,u\rangle.$$ This points inward horizontally. Reverse the order to obtain $$\boxed{\mathbf r_v\times\mathbf r_u =\langle u\cos v,u\sin v,-u\rangle}.$$

Step 3: Verify Its horizontal component is $\langle x,y,0\rangle$, pointing away from the axis. It also agrees with the outward direction of $x^2+y^2-z^2=0$, whose gradient is $2\langle x,y,-z\rangle$.