# curved formed by the intersection of two equal cylinders.

• Feb 9th 2011, 06:52 PM
lllll
curved formed by the intersection of two equal cylinders.
I'm wondering if anybody can help me solve the following problem, the curve I'm getting at the end is either planar or is completely incorrect.

Find the curvature and the parametrized curved formed by intersection the following two cylinders:

$\displaystyle x^2+(y-1)^2=1$
$\displaystyle y^2+z^2=1$

so what I have after some calculation is:

$\displaystyle x=\pm \sqrt{1-(\sqrt{1-z^2}-1)^2}$

$\displaystyle y=\sqrt{1-z^2}$

Now if I convert let $\displaystyle z=\sin(t)$ and substitute then the result follows. But is there a way that I don't have to deal with both positive and negative parts of the equation, as to simplify the curvature calculation?
• Feb 9th 2011, 08:42 PM
FernandoRevilla
Quote:

Originally Posted by lllll
But is there a way that I don't have to deal with both positive and negative parts of the equation, as to simplify the curvature calculation?

You can use the Implicit Function Theorem if you want to find the curvature.

For example, express $\displaystyle \vec{r}(t)=(x(t),y(t),t)$ and you'll obtain $\displaystyle \vec{r}\;'(t)=((y-1)z/xy,-z/y,1)$ if $\displaystyle xy\neq 0$ etc.

Fernando Revilla
• Feb 10th 2011, 06:06 PM
lllll
thanks, that thought hasn't occured to me.