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I am looking at the following exercise:

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The line of striction is $\Gamma=\gamma-\frac{\dot\delta\cdot\dot\gamma}{\|\dot\delta\|^2}\delta$.

For the first part I have done the following:

We have that the line of striction is $\Gamma=\gamma-\frac{\dot\delta\cdot\dot\gamma}{\|\dot\delta\|^2}\delta$, where $\delta$ is a parallel vector to the surface. So, a parallel vector to the tangent developable of $\gamma$ is the tangent vector $\dot\gamma$. Therefore, $$\Gamma=\gamma-\frac{\ddot\gamma\cdot\dot\gamma}{\|\ddot\gamma\|^2}\dot\gamma$$ Since $\gamma$ is of unit speed, we have that $$\|\dot\gamma\|^2=\dot\gamma\cdot\dot\gamma=1 \Rightarrow (\dot\gamma\cdot\dot\gamma)'=0 \Rightarrow \ddot\gamma\cdot\dot\gamma+\dot\gamma\cdot\ddot\gamma=0 \Rightarrow 2\ddot\gamma\cdot\dot\gamma=0 \Rightarrow \ddot\gamma\cdot\dot\gamma=0$$ So, we have that $\Gamma=\gamma$.

Is this correct?

Could you give me some hints how we could show that the intersection of the surface and the plane is of the form $$\Gamma (v)=\gamma (u_0)-\frac{1}{2}\kappa (u_0)v^2\textbf{n}(u_0)+\frac{1}{3}\kappa(u_0)\tau (u_0)v^3\textbf{b}(u_0)$$ ?

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    $\begingroup$ This is problem 6.2.4 from Pressley's book. The answers to the problems are in the back of the book. Don't you like the book's answers? $\endgroup$ – bubba Jan 22 '16 at 12:57
  • $\begingroup$ I haven't really undestood how they solved it? They use Taylor expansion and the Frenet-Serret equations, right? @bubba $\endgroup$ – Mary Star Jan 22 '16 at 13:01

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