Curvature of a curve whose unit tangent creates a constant angle with z-axis

Question

I have this question: Consider a curve $\gamma$(s) and its projection to the plane $\beta$(s), i.e $$\gamma(t) = \begin{bmatrix} f(s)\\ g(s) \\ h(s) \end{bmatrix}, \beta(t) = \begin{bmatrix} f(s) \\ g(s) \\ 0 \end{bmatrix} $$ Where s is the arc length parameter for $\gamma$.

1. Is s also the arc length parameter for $\beta$?

2. Let $T$ be the unit tangent vector of $\gamma$. Prove that the angle between $T(s)$ and the z-axis is at least $\frac{\pi}{4}$ iff $|h'(s)|\leq \frac{1}{\sqrt 2}$.

3. Assume $|h'(s)|\leq \frac{1}{\sqrt 2}$ holds for all $s$. Assume there is a constant C so that $$\kappa_{\gamma}\leq C$$ is satisfied everywhere. Prove that the curvature $$\kappa_{\beta}\leq 100C.$$

It would be great if someone could help me with this problem because I have been working on it for quite some time and have no idea how to arrive at the conclusion for part 3. Thank you in advance!

Answer 1

Here is the answer to your third question:

As $s$ is arc length along $\gamma$ we have $$\kappa_\gamma(s)=|\dot\gamma\times\ddot\gamma|=\sqrt{(\dot g\ddot h-\ddot g\dot h)^2+(\dot h\ddot f-\ddot h\dot f)^2+(\dot f\ddot g-\ddot f\dot g)^2}\ .$$ Along $\beta$ the variable $s$ is no longer arc length; therefore we have to use the formula $$|\kappa_\beta(s)|={|\dot f\ddot g-\ddot f\dot g|\over(\dot f^2+\dot g^2)^{3/2}}\ .$$

By assumption $\dot h^2(s)\leq{1\over2}$. As $\dot f^2(s)+\dot g^2(s)+\dot h^2(s)\equiv1$ it follows that $\dot f^2(s)+\dot g^2(s)\geq{1\over2}$, so that we can immediately deduce $$|\kappa_\beta(s)|\leq\sqrt{8}\>\kappa_\gamma(s)$$ for all $s$.

Answer 2

An alternative answer: $$k\gamma=\frac{\mid g'h''-h'g''h'f''-f'h''f'g''-g'f''\mid}{((f')^{2}+(g')^{2}+(h')^{2})^{3/2}}$$ greater and equal to C #.

Thanks to Hieu Hoang who wrote this short equation.