Curve length of cycloid and Helix $\gamma(t) = r \begin{pmatrix} r \cos t \\ r \sin t \\ ht \end{pmatrix}$

Question

I have to calculate the curve length of the

(a) $ \text{ cycloids } \gamma:[0,2\pi] \to \mathbb{R^2}$

$$\gamma(t) = r \begin{pmatrix} t-\sin t \\ 1 - \cos t \end{pmatrix}$$

(b) $\text{Helix } \gamma:[0,2\pi] \to \mathbb{R}^3$

$$\gamma(t) = r \begin{pmatrix} r \cos t \\ r \sin t \\ ht \end{pmatrix}$$

Regarding (a) I have that

$$||\gamma'(t)|| = \sqrt{(1-\cos (t))^2+ (\sin(t))^2} = \sqrt{1-2\cos(t)+\cos^2(t)+\sin^2(t)} =\sqrt{2-2\cos(2 \cdot \frac{t}{2})}$$ $$= \sqrt{2} \sqrt{1- \cos^2(\frac{t}{2})+ \sin^2(\frac{t}{2})} = \sqrt{2} \sqrt{2 \sin^2 \frac{t}{2}} = 2|\sin (\frac{t}{2})|$$

$$L = \int_{0}^{2\pi} ||\gamma'(t)||dt = 2 \int_0^{2\pi}|\sin(\frac{t}{2})|dt = 8 $$ Is that correct?

Regarding (b) I don't know how it's done because the $ht$ confuses me. Can someone show me how it's done?

Answer 1

$h$ is a constant, same as $r$:

\begin{eqnarray} \frac{{\rm d}x}{{\rm d}t} &=& -r\sin t \\ \frac{{\rm d}y}{{\rm d}t} &=& -r\cos t \\ \frac{{\rm d}z}{{\rm d}t} &=& h \end{eqnarray}

So that

\begin{eqnarray} L &=& \int_0^{2\pi} ||\gamma'(t)||{\rm d}t \\ &=& \int_0^{2\pi} \left[r^2\sin^2 t + r^2\cos^2 t + h^2 \right]^{1/2}{\rm d}t \\ &=& \int_0^{2\pi} \sqrt{r^2 + h^2}{\rm d}t \\ &=& 2\pi \sqrt{r^2 + h^2} \end{eqnarray}