Curvature of a curve whose unit tangent creates a constant angle with z-axis 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 $\gamma$.


*

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

*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 $|h'(s)|\leq \frac{1}{\sqrt 2}$.

*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!

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 $s$.
