A dog is tied to circular pillar by a rope. Radius of this pillar is $1m$ and length of rope is $\pi m$. What is an area where dog can roam?
I tried to find the area of all semicircles and then to find its sum. It is easy to find an area at front side of the pillar. It is $\displaystyle\frac12\pi^2\pi=\frac{\pi^3}{2}$. Problem is how to find remaining area. I tried to write this area using compass and straightedge, but I couldn't. Then I wrote this in AutoCAD and it looks like this:
Is it possible to find the exact value of this area?
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Martin Sleziak
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3Although, given that this question already has more votes and a complete answer, perhaps we should close that older question as a duplicate of this one... – Sep 11 '14 at 21:10
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Is not the bottom portion an involute? It is, like the cycloid, easy to evaluate by integration. – Narasimham Sep 11 '14 at 21:00
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1Supplementary question. What's the area if the rope has arbitrary length L? – MartinG Sep 11 '14 at 22:16
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3By the way, this is described as the goat problem on MathWorld (don't be misled by the first diagram on the page; it deals with both the interior and the exterior cases). – Sep 12 '14 at 01:23
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You can find a few results searching for bull tied circular silo or some similar phrases. (You can try various animals.) – Martin Sleziak Sep 12 '14 at 13:02
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The length of the perimeter of the roaming area can also be calculated, and turns out to be equal to the circumference of the circle of radius $\pi$ - the same as if the pillar weren't there. – MartinG Sep 12 '14 at 21:49
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Got same result as MartinG – Narasimham Sep 12 '14 at 22:06
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Related: https://math.stackexchange.com/questions/1942222/ – Watson Nov 28 '18 at 09:22
1 Answers
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[Edit: redefined $\theta$ for simplicity and introduced $a$ for generality.]
Let $a$ be the radius of the pillar ($a=1$ in the example). The length of the rope is $L=\pi a.$
Let the length of the rope (assumed taut) not in contact with the pillar be $l(\theta)$, where $\theta$ is the angle between a vertical line descending from the pillar centre (O) and a line between O and the rope's point of attachment. Then $$ l = a\theta $$ $\theta$ is also the angle of the rope measured from the horizontal.
The area swept out as the rope turns anticlockwise through angle increment $d\theta$ is $$ dA = \tfrac{1}{2}l^2 d\theta $$ The total area swept out in the lower right quadrant is therefore $$ A = \int dA = \int_0^{\pi}\tfrac{1}{2}a^2\theta^2 d\theta = \tfrac{1}{6}\pi^3a^2 $$ Doubling up for the other side and adding the semicircular area at the top gives a total area of $$ 2 (\tfrac{1}{6}\pi^3a^2) + \tfrac{1}{2}\pi(\pi a)^2 = \tfrac{5}{6}\pi^3a^2 $$
MartinG
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No, but that wasn't necessary because it's not included in the calculation. – MartinG Sep 13 '14 at 16:27
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For a related physics problem, see http://physics.stackexchange.com/questions/135325/giant-leap-of-the-space-goat – MartinG Oct 14 '14 at 12:51