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Given the following sphere and cylinder,

$$\begin{align} x^2+y^2+z^2&=4R^2,\\ (x-R)^2+y^2&=R^2, \end{align}$$

find a parametric equation of their intersection.

I know that their intersection is called a hippopede and that on the $x$-$y$ plane, its parametrization is $r(t)=R(\cos t+1)\,\hat i+R\sin t\,\hat j$. However, I have no idea how to find its $\hat k$ component.

Any hint would be appreciated!

Edit: The $\hat k$ component is supposed to be

$$ 2R\sin\left(\frac t2\right), $$

but I have no idea how that was obtained.

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1 Answer 1

up vote 2 down vote accepted

I have no idea how they chose it, but I can show you it works. First let's do some algebra on our second equation.

$$x^2-2Rx+R^2+y^2=R^2,2Rx=x^2+y^2$$

Substituting into the first equation gives

$$2Rx+z^2=4R^2$$

$z=2R\sin u$ seems a reasonable enough substitution, yielding $x=2R\cos^2u$. Now plugging that into the second equation gives

$$(2R\cos^2u-R)^2+y^2=R^2$$

$$R^2(2\cos^2u-1)^2+y^2=R^2$$

$$R^2\cos^22u+y^2=R^2$$

$$y^2=R^2\sin^22u$$

Now to show the functions of x are equivalent, use the power reduction formula.

$$2R\cos^2u=R(1+\cos2u)$$

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Thank you. Your approach led me to realize why their pick works. I am still wondering how the substitution $z=2R\sin u$ can be found, though; I would have never come up with it. –  Josué Molina Oct 10 '12 at 22:52
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