Thread: Grid Curves
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BoundByAxioms
#5
Jul27-08, 01:46 PM
P: 99
Quote Quote by HallsofIvy View Post
To find the tangent vector in the "u" direction, differentiate r(u,v)= <vcos u, vsin(u), v2> with respect to u. To find the tangent vector in the "v" direction, differentiate r(u,v)= <vcos u, vsin(u), v2> with respect to v. To find them at the given point, substitute [itex]u= \pi/4[/itex] and [itex]v= \sqrt{2}[/itex].
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 [tex]\sqrt{2}[/tex] for v and [tex]\pi/4[/tex] for u. So I get, <-1, 1, 0> and <[tex]\sqrt{2}/2[/tex], [tex]\sqrt{2}/2[/tex], 2[tex]\sqrt{2}[/tex]>. Is this correct so far?