curved formed by the intersection of two equal cylinders.

**lllll** · February 9th 2011, 06:52 PM

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:

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

so what I have after some calculation is:

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

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

Now if I convert let $z=\sin(t)$ 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?

**FernandoRevilla** · February 9th 2011, 08:42 PM

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

Fernando Revilla

**lllll** · February 10th 2011, 06:06 PM

thanks, that thought hasn't occured to me.

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