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Compute the curvature and torsion of the curve:

$\vec{r}(t) = \langle\frac{4}{5} \cos(t), 1 - \sin(t), -\frac{3}{5} \cos (t)\rangle$

For example, I know the curvature can be found using the following formula:

$$\frac{|| \vec{r}^{'}(t) \times \vec{r}^{''}(t) ||}{ || \vec{r}^{'}(t) ||^3}$$

... but is there an easier way to solve this problem? Is there a relationship that I'm not seeing immediately that might help me answer this in a different way? Just curious.

2 Answers 2

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Let's suppose that $t\in[0,2\pi]$, then the arc lenght function is $$s=\int_{0}^{t}\|r'(t)\|dt=\int_0^t\sqrt{(16\sin^2t+25\cos^2t+9\sin^2t)/25}dt=\int_0^t1dt=t$$ It follows that $t$ is the arc length parameter, so the curvature $\kappa$ is given by $$\kappa=\left\Vert\frac{dT}{ds}\right\Vert=\left\Vert\frac{dT}{dt}\right\Vert=\|T'(t)\|$$ where $T$ is the unit tangent vector $$T=\frac{r'(t)}{\|r'(t)\|}=\frac{r'(t)}1=(-\tfrac45\sin t,-\cos t,\tfrac35\sin t)$$ Thus $$\frac{dT}{dt}=(-\tfrac45\cos t,\sin t,\tfrac35\cos t)\qquad\implies\qquad \kappa=\left\Vert\frac{dT}{dt}\right\Vert=\|T'(t)\|=\boxed{\color{blue}{1}}$$

Also, since $\|T'(t)\|=1$ it follows $$N=\frac{T'(t)}{\|T'(t)\|}=T'(t)=(-\tfrac45\cos t,\sin t,\tfrac35\cos t)$$

Where $N$ is the unit normal vector, then the binormal vector $B$ is $$B=T\times N=(-\tfrac35,0,-\tfrac45)$$ Now, the torsion $\tau$ of the curve is a scalar function given by $$\dfrac{dB}{ds}=-\tau(s)N$$ From the fact that $dB/ds=dB/dt=0$ it follows $\tau=0$.

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    Thanks! But just wondering, why is $‖\frac{dT}{dt}‖$ = 1? – 2017-02-16
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    \begin{align*} \left\Vert\frac{dT}{dt}\right\Vert&=\sqrt{\tfrac{16}{25}\cos^2t+\sin^2t+\tfrac9{25}\cos^2t}\\ &=\sqrt{\cos^2 t + \sin^2 t}\\ &=\sqrt{1}\\ &=1 \end{align*} – 2017-02-16
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    Ah, I see. Thanks! – 2017-02-16
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    Also is there also a formula for finding the torsion? How would you find it? (Sorry, I haven't seen the one about κ=$‖\frac{dT}{ds}‖$ before but I'm going to research these formulas further). – 2017-02-16
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Observe that the distance from $r(t)$ to $(0,1,0)$ is $1$ for all $t$, hence $r$ might be circle so its torsion is zero. Indeed, $r$ is a plane curve since it is orthogonal to $(3,0,4)$ for all $t$.