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The figure shows a piece of string tied to a circle with a radius of one unit. The string is just long enough to reach the opposite side of the circle. Find the area of the region, not including the circle itself that is traced out when the string is unwound counterclockwise and continues counterclockwise until it reaches the opposite side again.

enter image description here

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    $\begingroup$ Ideally the picture would show that portion of the string that is not touching the circle as a straight line rather than a curve. $\endgroup$ Jun 13, 2016 at 15:55
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    $\begingroup$ This is the case of the goat problem with a goat exterior to a silo. See the second half of this page. $\endgroup$ Jun 13, 2016 at 16:13
  • $\begingroup$ You can see an animation of the first part of the circle involute at desmos.com/calculator/c7zwxqr8mq $\endgroup$ Jun 23, 2016 at 11:27

4 Answers 4

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I'm mostly condensing what Eric Weisstein wrote in this MathWorld article, citing a 1998 paper by Hoffman, The Bull and the Silo: An Application of Curvature. You can solve this by finding the area of the semicircle the string traces (before the string begins to wrap) and then do some fancy calculus to find the area of the regions formed when the string begins to wrap around the circle.

By the calculations in the article, given a circle of radius $a$ and a string of length $L$, the area $A$ will be $$A = \frac{\pi L^2}{2} + \frac{L^3}{3a}\;.$$

So in your case where $a=1$ and $L=\pi$ we have $A = \frac{5}{6}\pi^3 \;\approx\; 25.838564$.

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There are 3 parts to the calculation. The string unwinds until it points straight up. A semi-circle. The string winds back onto the wheel. Part 1 and part 3 trace out equal area.

Part 1: Let $(\cos\theta,\sin\theta)$ be the point where the string is starting to pull away from the wheel.

let (x,y) be the endpoint of the string.

the area traced inside of(x,y).

$-\int_0^{\pi} y(\theta) \frac {dx}{d\theta} d\theta$

It is negative, because x is moving from right to left.

Now we need to find functions for x and y in terms of theta.

The length of string on the free end = $(\pi - \theta)$ and the segment from $(\cos\theta, \sin\theta) \to (x,y)$ is tangent to the curve.

$(x,y) = (\cos\theta + (\pi-\theta) \sin \theta,\sin\theta - (\pi-\theta) \cos \theta)$

$\frac {dx}{d\theta} = -\sin\theta + (\pi-\theta) \cos\theta - \sin\theta$

$\int_0^\pi (\sin\theta - (\pi-\theta) \cos \theta)(2\sin\theta - (pi-\theta)\cos\theta) d\theta$

part 2: is a semi circle of radius $\pi$

$\frac 12\pi^3$

part 3 = part 1

By the way, this calculation has included the the are of the unit circle as being inside the area traced out by the end-point of the string.

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  • $\begingroup$ Doug M, did not follow your position equations. See my answer below. $\endgroup$ Jul 2, 2016 at 19:57
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I tried graphing the equations given in the accepted answer by Doug M, but they do not seem to trace out the path of the end of the string as it is being unwound (although his method does gives the correct answer). The way I approached the problem was to copy the equations (5) and (6) from the page http://mathworld.wolfram.com/GoatProblem.html, using $a=1$ and then use polar coordinates. I translated everything to the right one unit so the area of part 1 of the calculation looks like this:

Silo Problem

... (where the orange point on the $y-$axis is located at $(0,\pi)$) so that the equations needed are $$x=\cos\theta+\theta\sin\theta+1\quad\text{and}\quad y=\sin\theta-\theta\cos\theta.$$ Now using $r^2=x^2+y^2$ eventually gives us $$r^2=\theta^2+2\theta\sin\theta+\cos\theta+2$$ and subtracting the area of the upper half of circle $r=2\cos\theta$, we have the area between the curves as $$A=\left[\frac12\int_0^\pi(\theta^2+2\theta\sin\theta+\cos\theta+2)\,d\theta\right]-2\pi=\boxed{\frac{\pi^3}6}.$$

Doubling this value and adding the semicircle from "part 2" described by Doug M gives the final result of $\boxed{\dfrac{5\pi^3}6}$.

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Well, I cheated, and solved it numerically: it comes up at about 19.55.

Unfortunately that does not match up with Mike Pierce's answer of $\frac56 \pi$...

Here is a graph of the interesting part, and here is the Matlab code.

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  • $\begingroup$ Well, know it's not exactly what OP was looking for, but at least it's a way to check the answer. Or is there a mistake in my code? $\endgroup$ Jun 13, 2016 at 17:54
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    $\begingroup$ I haven't looked at your code, but it seems your answer differs from Mike Pierce's by exactly $2\pi$. $\endgroup$ Jun 14, 2016 at 22:08

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