Thread: Grid Curves View Single Post
P: 99
 Quote by HallsofIvy To find the tangent vector in the "u" direction, differentiate r(u,v)= with respect to u. To find the tangent vector in the "v" direction, differentiate r(u,v)= with respect to v. To find them at the given point, substitute $u= \pi/4$ and $v= \sqrt{2}$.
So, when r(u,v)=<vcos(u), vsin(u), v2>, the derivative with respect to u is <-vsin(u), vcos(u), 0>. And the derivative with respect to v is <cos(u), sin(u), 2v>. Now, I plug in $$\sqrt{2}$$ for v and $$\pi/4$$ for u. So I get, <-1, 1, 0> and <$$\sqrt{2}/2$$, $$\sqrt{2}/2$$, 2$$\sqrt{2}$$>. Is this correct so far?