a) Describe the intersection $(C)$ of sphere $x^2 + y^2 + z^2 = 1$ and the elliptic cylinder $x^2 + 2z^2 = 1$, and find out the total arc-length of this intersection.
b) Determine the points on the curve $(C)$, where the curvature have the maximum value.
I've been tackling this problem for quite awhile now, going through my notes, researching, reading my textbook but I have not gotten anything productive. Any form of help or solution would be greatly appreciated.
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From $x^2+y^2+z^2=1$ and $x^2+2z^2=1$ it immediately follows that $z^2-y^2=0$, or $z=\pm y$. Conversely, $x^2+y^2+z^2=1$ and $z=\pm y$ imply $x^2+2z^2=1$. Therefore the intersection of $S^2$ and the elliptical cylinder coincides with the intersection of $S^2$ with the two planes $z=\pm y$, i.e. with the union of two unit circles intersecting at $(1,0,0)$. The total arc length is $4\pi$, and the curvature along these circles is of course constant.
There are two curves. They can be parametrized as follows.
$\gamma_1 :[0,2\pi]\to\mathbb{R}^3$ given by $\gamma_1(t)=(\cos t,\frac{1}{\sqrt{2}}\sin t,\frac{1}{\sqrt{2}}\sin t)$, and
$\gamma_2 :[0,2\pi]\to\mathbb{R}^3$ given by $\gamma_2(t)=(\cos t,\frac{1}{\sqrt{2}}\sin t,-\frac{1}{\sqrt{2}}\sin t)$.
So $C=\gamma_1([0,2\pi])\cup \gamma_2([0,2\pi])$. Length of $\gamma_1$ is the same of $\gamma_2$, then the length of $C$ is \begin{align*} s&=2\int_0^{2\pi}\sqrt{(-\sin t)^2+2\left(\frac{1}{\sqrt{2}}\cos t\right)^2}\mathrm dt\\ &=2\int_0^{2\pi}\mathrm dt\\ &=4\pi \end{align*}
Also, it can be proved that $\gamma_1([0,2\pi])$ and $\gamma_2([0,2\pi])$ are circles of radius $r=1$ in the planes $z=y$ and $z=-y$, respectively. And is a circle in the plane. Therefore, the curvature is constant, $\kappa=1$, for every point of $C$.