Natural parametrization of curves

Question

I want to find curvatures and torsions for the following curves but get stuck with their natural parametrizations ($s$ is natural if $|\dot{\gamma}(s)| = 1$). Can anyone help me?

(a) $e^t(\cos t,\sin t,1)$

(b) $(t^3+t,t^3-t,\sqrt{3}t^2)$

(c) $3x^2+15y^2=1, z=xy$

Update:

Here are my attempts on solving (a):

$\dot{\gamma}(t) = (\dot{t}e^t \cos t - \dot{t} e^t \sin t, \dot{t} e^t \sin t + \dot{t}e^t \cos t, \dot{t} e^t)$ which gives $|\dot{\gamma}(t)| = \sqrt{2}\dot{t}e^t=1$ and the solution for this ODE is $t = \ln\frac{\tau}{2}$.

But if I substitute $t$ with $t=\ln\frac{\tau}{2}$ the result will be $\dot{\gamma}(\tau) = (\frac{1}{\sqrt{2}}\cos\ln\frac{\tau}{2} - \frac{1}{\sqrt{2}}\sin\ln\frac{\tau}{2},\frac{1}{\sqrt{2}}\sin\ln\frac{\tau}{2}+\frac{1}{\sqrt{2}}\cos\ln\frac{\tau}{2},\frac{1}{2})$ and $|\dot{\gamma}(\tau)| = \sqrt{\frac{3}{2}}$. So for $|\dot{\gamma}(\tau)| = 1$ we should take $t = \ln\frac{\tau}{3}$. Where is my mistake?

Any help and hints will be very appreciative.

Answer 1

$(a)$ \begin{align*} \mathbf{\dot{r}}(t) &= e^{t}(\cos t-\sin t, \sin t+\cos t,1) \\ |\mathbf{\dot{r}}(t)| &= e^{t}\sqrt{(\cos t-\sin t)^2+(\sin t+\cos t)^2+1} \\ &= e^{t}\sqrt{3} \\ s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\ &= \sqrt{3}(e^{t}-1) \\ t &= \ln \left( 1+\frac{s}{\sqrt{3}} \right) \end{align*}

$(b)$ \begin{align*} \mathbf{\dot{r}}(t) &= (3t^2+1, 3t^2-1,2t\sqrt{3}) \\ |\mathbf{\dot{r}}(t)| &= \sqrt{(3t^2+1)^2+(3t^2-1)^2+12t^2} \\ &= \sqrt{2(9t^4+6t^2+1)} \\ &= \sqrt{2}(3t^2+1) \\ s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\ &= \sqrt{2}(t^3+t) \\ t &= \sqrt[3]{\sqrt{\frac{s^2}{8}+\frac{1}{27}}+\frac{s}{2\sqrt{2}}}- \sqrt[3]{\sqrt{\frac{s^2}{8}+\frac{1}{27}}-\frac{s}{2\sqrt{2}}} \\ \end{align*}

$(c)$ \begin{align*} \mathbf{r}(t) &= \left( \frac{\cos t}{\sqrt{3}}, \frac{\sin t}{\sqrt{15}}, \frac{\cos t \sin t}{3\sqrt{5}} \right) \\ \mathbf{\dot{r}}(t) &= \left( -\frac{\sin t}{\sqrt{3}}, \frac{\cos t}{\sqrt{15}}, \frac{\cos^2 t-\sin^2 t}{3\sqrt{5}} \right) \\ |\mathbf{\dot{r}}(t)| &= \sqrt{\frac{15\sin^2 t+3\cos^2 t+(\cos^2 t-\sin^2 t)^2}{45}} \\ &= \frac{3-\cos 2t}{3\sqrt{5}} \\ s &= \int_{0}^{t} |\mathbf{\dot{r}}(t)| \, dt \\ &= \frac{6t-\sin 2t}{6\sqrt{5}} \end{align*} there is no close form for $t$ in terms of $s$.

You still can find $\kappa$ and $\tau$ by keeping $t$ parametrization \begin{align*} \kappa &= \frac{|\mathbf{\dot{r}} \times \mathbf{\ddot{r}}|} {|\mathbf{\dot{r}}|^3} \\ \tau &= \frac{\mathbf{\dot{r}} \cdot \mathbf{\ddot{r}} \times \mathbf{\dddot{r}}} {|\mathbf{\dot{r}} \times \mathbf{\ddot{r}}|^2} \end{align*}

Answer 2

You do not need an arclength or "natural" parametrization to do these computations. It's just a matter of using the chain rule. I've posted about this numerous times on here. Rather than write out another lecture, I'll just refer you to my (free) differential geometry text. Of course, if you have further questions, feel free to ask :)