A student gave me the following problem, which is of some interest. A pipe with circular cross-section is mostly buried, with only its top showing. Given the horizontal distance x and the distance over the pipe y between the points where the pipe breaks the ground, find the radius r of the pipe. (See the diagram.)
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From the diagram we have \theta=\frac{y/2}{r}, \sin\theta=\frac{x/2}{r}
so \frac{\sin\theta}{\theta}=\frac{x}{y}.
First we solve this equation. Then r=y/{2\theta}.
We have \frac{x}{y}=1-\frac{\theta^2}{3!}+\frac{\theta^4}{5!}-+ \cdots .
Write \frac{\displaystyle x}{\displaystyle y}=1-\frac{\displaystyle q}{\displaystyle 6}, q=6\left (1-\frac{\displaystyle x}{\displaystyle y}\right ).
Then q=\theta^2\left (1-\frac{\theta^2}{4\times 5}+\frac{\theta^4}{4\times5\times6\times7}-+ \cdots \right ).
Write \theta^2=u.
Then q=u\left (1-\frac{u}{4\times5}+\frac{u^2}{4\times5\times6\times7}-+ \cdots \right ).
We can invert this to find u=q\left (1+\frac{q}{20}+ \cdots \right ), \theta=\sqrt{u}=\sqrt{q}\left (1+\frac{q}{20}+ \cdots \right )^{\frac12}=\sqrt{q}\left (1+\frac{q}{40}+ \cdots \right )
and r=\frac{y}{2\theta}=\frac{y}{2\sqrt{q}}\left (1+\frac{q}{40}+ \cdots \right )^{-1}=\frac{y}{2\sqrt{q}}\left (1-\frac{q}{40}+ \cdots \right ).
Using MAPLE, we find in more detail that \align r&=\frac{y}{2\sqrt{q}}\left (1-\frac{q}{40}-\frac{13}{13440}q^2-\frac{1649}{24192000}q^3-\frac{1761877}{298045440000}q^4\right.\\ &\left. -\frac{265494577}{464950886400000}q^5-\frac{138120996553}{2343352467456000000}q^6- \cdots \right ). align
For example, if x=3.464, y=4.189 then q\approx1.0384334 r\approx2.000.
The referee points out that the problem can be formulated as \sin(y/2r)-x/2r=0,
and solved numerically. Indeed, MAPLE does this very quickly. However, when I was given the problem, a `formula' was wanted.
School of Mathematics
University of New South Wales