Calculus question with circle, and string tracing an area

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 $A = \frac{5}{6}\pi^3 \;\approx\; 25.838564$.


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.

Tags:

Calculus

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