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Need help showing important tangent plane property:

  • Thread starter ozone
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Homework Statement



Consider the tangent surface of some regular differentiable curve given as [itex]X(t,v) = \alpha(t) + v \alpha'(t) [/itex]. Show that the tangent planes along X(t,constant) are equal.

Homework Equations



[itex] N = \frac{X_{t} \wedge X_{v}}{|X_{t} \wedge X_{v}|} [/itex]

The general tangent plane equation, [itex]T_{t_0}(S) = N|_{t_0, v_0} \cdot( (t - t_0) - (v-v_0)) = 0 [/itex]

The Attempt at a Solution



Using the equation above I solved for [itex] N = \hat{v} b(s) [/itex] where b(s) is the binormal to the curve.

However I wasn't sure if this was correct since it does not seem like I am close to answering the problem. I wanted to reach the conclusion that [itex] \frac{dN}{dt} = 0 [/itex] along this curve, but this seemed to be untrue. Another idea was to show that the tangent plane equation is the same for all initial points [itex] t_0 [/itex] but this too seemed incorrect.

I am really stuck at this point and would love it if anyone could point me in the right direction
 

Answers and Replies

  • #2
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Edit: I realized v is just a constant so what we really should have is something like [itex] N = \pm b(t) [/itex]. Also I forgot to mention I had a factor of [itex] |\alpha(t)|^4 [/itex] so that [itex] N = \pm |\alpha(t)|^4 b(t) [/itex] but I assumed that the curve is parameterized by arc-length so that I could ignore this term.. I'm not sure that it should make any difference by including it.

Also I should note that I wrote the tangent plane equation down very wrong, it should be given by

[itex] T_{t_0,v_0}(S) = N|_{t_0,v_0} \cdot ((x,y,z) - X(t_0,v_0)) [/itex]

Which would simplify to

[itex] T_{t_0,v_0}(S) = \pm |\alpha(t_0)|^4 (b(t_0) \cdot (x,y,z) - b(t_0) \cdot \alpha(t_0) ) [/itex]

Which we would then want to show is not changing as [itex] t_0 [/itex] varies
 
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