Which value(s) of k make h(t) an arc-length parametrization?

Question

Consider the helix parametrized by $\vec h(t)= \langle kt, 4 \sin \frac{t}{5},4 \cos\frac{t}{5}\rangle$ Which value(s) of $k$ make $\vec h(t)$ an arc-length parametrization?

I tried to find the speed of the $\vec h(t)$ and set it equal to 1, but I ended up getting a complex number.

Answer 1

We have the parameterization given by

$$\vec h(t)=\hat x kt+\hat y 4 \sin (t/5)+\hat z 4 \cos(t/5)$$

Then, the differential length $d\vec h(t)$ is

$$d\vec h(t)=\hat x k+\hat y \frac45 \cos (t/5)-\hat z \frac45 \sin(t/5)$$

and the magnitude of the differential length is

$$|d\vec h(t)|=\sqrt{k^2+16/25}$$

In order for the differential length to be arc length, we must have $|d\vec h(t)|=1$. Therefore,

$$\sqrt{k^2+16/25}=1 \implies k=\pm 3/5$$

and we are done!

Answer 2

$\vec h(t)$ is a unit length parameterization iff $\left|\frac{\mathrm{d} \vec h(t)}{\mathrm{d} t}\right| = 1 \forall t$, so in order to make it an arc-length parameterization by adjusting $k$ we need to solve that equation for $k$.

$$\left|\frac{\mathrm{d} \vec h(t)}{\mathrm{d} t}\right| = \sqrt{k^2+\left(\frac{4}{5}\cos\frac{t}{5}\right)^2+\left(-\frac{4}{5}\sin\frac{t}{5}\right)}\\=\sqrt{k^2+\frac{16}{25}\cos^2\frac{t}{5}+\frac{16}{25}\sin^2\frac{t}{5}}=\sqrt{k^2+\frac{16}{25}}=1$$ $$k^2+\frac{16}{25}=1$$ $$k^2=\frac{9}{25}$$ $$k=\pm\frac{3}{5}$$