I want to show that the curve
$$ \gamma(t)=(1+\cos(t),\sin(t),2\sin(t/2)) $$
is the interesection between
\begin{align*}
x^2+y^2+z^2=4, \newline (x-1)^2+y^2=1.
\end{align*}
The way I wanted to show this was by parametrizing the intersection and then finding a bijective re-parametrization.
So for the intersection curve I substituted
$$ 1=(x-1)^2+y^2=x^2-2x+1+y^2 \iff x^2+y^2=2x $$
into the sphere equation, giving
$$ 2x+z^2=4 \iff x=2-\frac{z^2}{2}. $$
From the ylinder equation I get
$$y=\sqrt{2x-x^2}=\sqrt{4-z^2-4+2z^2-\frac{z^4}{4}}=\pm z\sqrt{1-\frac{z^2}{4}} $$
Now I have the intersection curve
$$ \alpha(t)=(2-\frac{z^2}{2},\pm z\sqrt{1-\frac{z^2}{4}},z). $$
I plotted them in WolframAlpha and they seem to have the same image, but I can't find a proper map $\varphi$ with $\gamma=\alpha\circ\varphi$.
From the last parameter it looks like it should be $\varphi(z)=2\sin(z/2)$, which is correct for the first (and obviously third) entry because
$$ 2-\frac{(2\sin(z/2))^2}{2}=1+\cos(z), $$
but in the second entry I have
$$\pm2\sin(z/2)\sqrt{1-\frac{(2\sin(z/2))^2}{4}}\neq\sin(z).$$
Is there something wrong with my parametrizations?
Best Answer
I would recommend starting with the parametrization of the cylinder given by the second equation:
\begin{align*} (x-1)^{2} + y^{2} = 1 \Longleftrightarrow \begin{cases} x = 1 + \cos(\theta)\\\\ y = \sin(\theta) \end{cases} \end{align*} where $\theta\in[0,2\pi]$. If we insert such relations into the first equation, we get that \begin{align*} (1 + \cos(\theta))^{2} + \sin^{2}(\theta) + z^{2} = 4 & \Longleftrightarrow 1 + 2\cos(\theta) + 1 + z^{2} = 4\\\\ & \Longleftrightarrow z^{2} = 2 - 2\cos(\theta)\\\\ & \Longleftrightarrow z = \pm\sqrt{2-2\cos(\theta)} \end{align*}
But we can go further: according to the double-angle identity, we get that \begin{align*} \cos(\theta) & = \cos^{2}(\theta/2) - \sin^{2}(\theta/2) = 1 - 2\sin^{2}(\theta/2) \Rightarrow 2 - 2\cos(\theta) = 4\sin^{2}(\theta/2) \end{align*}
Finally, we have that \begin{align*} z = \pm\sqrt{4\sin^{2}(\theta/2)} = \pm 2\sin(\theta/2) \end{align*}
and we are done.
Hopefully this helps!