Puzzle time - Silo dog

[jon-nyc]

This isn’t a ‘find the trick and do the answer in your head’ problem. It’s a pencil-to-paper problem.

[bachophile]

I figure the dog wanting to mark its territory simply pees all over the place.

[Klaus]

Game plan would be:

1. Find function that computes max height/distance from any point on silo, i.e., the shape of the "wrap around" part.

2. Presuming the formula is complicated, ask computer algebra system to integrate the function to find its area.

3. Multiply by two and add area of the semi-circle.

[Axtremus]

Yeah, getting the the “game plan” is easy, actually getting step 1 is the harder part.

[jon-nyc]

Actually integrating the function is straightforward.

The trick, if you want to call it that, is just figuring out the function. Totally doable by both of you.

[bachophile]

I glad he didn’t say the three of you

[jon-nyc]

lol.

[Axtremus]

||Cycloid is key.
Imagine the silo "rolls over" by 180º, the arc traced out by the point opposite the hook is that of a cycloid, and the area under the cycloid is given by the formula 3πr^2 where r is the silo's radius.
Solution is (area of cycloid traced out by the silo)-(area of silo)+(half the area of circle with radius that is the leash's length).
Area covered by the dog = 3π(10)^2 - π(10)^2 + (π(10π)^2)/2
||

[Klaus]

@Axtremus said in Puzzle time - Silo dog:

Imagine the silo "rolls over" by 180º, the arc traced out by the point opposite the hook is that of a cycloid,

But how does that arc relate to where the dog can go? Your solution - which may well be correct - is missing an argument here.

[Klaus]

Wouldn't the resulting shape be a (part of a) cardoid and not a cycloid?

[jon-nyc]

Yes. This is the shape. Note the 10pi leash length is just perfectly half the circumference of the silo.

[Axtremus]

Right, cardioid rather than cycloid.

[jon-nyc]

Maybe you can google up a formula if you can’t figure out the integral.

[Klaus]

I haven't looked at formulas yet but it looks like one needs only a part of the cardoid area since it overlaps with the semi circle.

[jon-nyc]

Yep. It’s the quarter circle plus that bit of cartioid x 2

[Klaus]

The area formula is:

but the integral would not have to be taken to pi but to the angle at which the y value is maximal, would be the case at the angle 3/4 * sqrt(3).

But this gets rather complicated hence I assume something's wrong here...

[jon-nyc]

That's ugly don't continue.

Smallish hint:

||Look at the cardioid area in isolation and write an equation for the length of the leash as a function of the angle traversed through it.||

Biggish hint:

||Imagine the dog is at the "northern" border between the half circle and the cardioid part as shown in this diagram:

His leash is now a tangent on the circle formed by the silo.

As he moves "west" the leash shortens but continues drawing a tangent against the circle. As a first step, express the length 𝓁 of the leash as a function of the angle θ.||

[jon-nyc]

buffer post because you can see through the spoiler inadvertently from certain views on certain devices.

[jon-nyc]

Any reason I shouldn't post a solution?

[jon-nyc]

Here it is, in a spoiler in case someone still wants to solve it.

||As previously noted, the total area comprises a half circle plus that little strange part which apparently is called a cardioid (a word I don't remember hearing before this thread.

The half circle is easy, 1/2(pi * r^2), where the radius is 10pi, so 50pi^3.

Now for the cardioid area. Note in the following diagram, when stretched taut the length of the remaining leash after it has already wrapped around a portion of the silo wall is the total length of the leash minus the circumference of the arc it has traversed. So if we call the angle t the length is 10pi - 10t, or 10(pi-t).

Now for a little calculus. Increase t by a tiny amount - dt. The tangent advances correspondingly, and the area of the tiny sector of the circle you've covered is 50(pi-t)^2 dt. Integrate that from 0 to pi and you get 50/3 * pi^3.

Double that to get 100/3 * pi^3 and add the half circle and the final answer is 250/3 * pi^3.||