From 6f759cb171c27c8359f654529a2baef8894579e2 Mon Sep 17 00:00:00 2001 From: Tyler Clarke Date: Tue, 4 Mar 2025 11:44:19 -0500 Subject: [PATCH] cosine rose --- index.html | 11 +- mathin-a-cos-rose.html | 700 +++++++++++++++++++++++++++++++++++++++++ 2 files changed, 708 insertions(+), 3 deletions(-) create mode 100644 mathin-a-cos-rose.html diff --git a/index.html b/index.html index 1f72465..e3b05cd 100644 --- a/index.html +++ b/index.html @@ -15,13 +15,18 @@

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Mathin’ a Cos Rose

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by Tyler Clarke

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Got me a multivariable calculus exam coming up soon. Most of the material is pretty simple – lagrange multipliers, for instance (I’ll be writing something about those soon) – but I’ve got one particular problem regarding integrating with radial coordinates that’s hard and interesting.

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I’m asked to find the area of a single loop of a rose described in radial coordinates by + + + + + r + = + 5 + + cos + + ( + + + 3 + θ + + + ) + + + . This is difficult to visualize, but luck hath smiled upon us, and Desmos supports radial coordinates:
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Not too bad. The maximum extent would appear to be 5, which makes sense, because cosine maxes out at 1 and we multiply by 5. I was a bit curious to see if the multiplier of theta controls the number of loops – as it turns out, the relation is a bit complicated, and I won’t get into that now.

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It’s looking to me like the best solution here is to integrate for + + + + + + − + π + + 2 + + ≤ + θ + ≤ + + π + 2 + + + , to grab that petal in the first and fourth quadrants, but I’m worried that that isn’t a general enough solution: other roses that I’ve got to graph might not be so conveniently situated. The more general way I can think of is integrating for all theta and dividing by the number of petals – which isn’t necessarily easily predictable for all theta, but is a damn sight better than just hoping my hand-drawn sketches appropriately capture the useful quadrants. Regardless, let’s do the first idea first, and see how it works out. In this case, we know there are 3 petals, so the integral should be + + + + + + ∫ + + + − + π + + 2 + + + π + 2 + + + + + ∫ + 0 + + 5 + cos + + ( + + + 3 + θ + + + ) + + + + r + + + dr + d + θ + + . Why   + + + + r + dr + d + θ + +  instead of just   + + + + 1 + dr + d + θ + + ? Because of the shape of the mesh we’re integrating; I won’t elaborate, but only because Paul's Online Math Notes already did.
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Integrating this isn’t too hard. The inner integral is simply + + + + + + + ∫ + 0 + + 5 + cos + + ( + + + 3 + θ + + + ) + + + + r + + dr + → + + + + + + + + r + 2 + + 2 + + + | + + 0 + + 5 + cos + + ( + + + 3 + θ + + + ) + + + + → + + 25 + 2 + + + cos + 2 + + + ( + + + 3 + θ + + + ) + + + + , giving us + + + + + 25 + 2 + + + + ∫ + + + − + π + + 2 + + + π + 2 + + + + cos + 2 + + + + ( + + + 3 + θ + + + ) + + d + θ + + . This is kind of a nasty trig integral, one which I very much don’t want to derive myself, but that’s what Wolfram is for! Wolfram kindly tells me that my disgusting integral is + + + + + + + + 25 + 24 + + + ( + + + 6 + + θ + + + sin + + + ( + + + 6 + θ + + + ) + + + + ) + + + + | + + + + − + π + + 2 + + + π + 2 + + + . Because + + + + sin + + + ( + + + 3 + π + + + ) + + = + 0 + + + , this works out nicely to… + + + + + 25 + 24 + + 6 + + π + = + + 25 + 4 + + + π + + . This is wrong.
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I’ve kept the wrong work up because it’s subtly wrong, and (I think) does a pretty good job of illustrating one of the most annoying problems with polar coordinates. Do you see it?

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The problem lies in how cosine behaves with multipliers. + + + + cos + + ( + + θ + + ) + + +  makes a full trip from 1 to 0 in + + + + 0 + ≤ + θ + ≤ + π + + ; + + + + cos + + ( + + + 3 + θ + + + ) + + +  makes three such trips in the same span. And our rose has three peaks.
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By some horrible mathemagic, integrating what should be a half-turn integrates the whole pattern. As it turns out, this is true for every cosine rose: regardless of the multiplier, a full span of pi (not 2pi!) covers the entire shape. Because we only want one petal, we have to cut our bounds in third: + + + + + + 25 + 2 + + + + ∫ + + + − + π + + 6 + + + π + 6 + + + + cos + 2 + + + + ( + + + 3 + θ + + + ) + + d + θ + → + + + + + + + + 25 + 24 + + + ( + + + 6 + + θ + + + sin + + + ( + + + 6 + θ + + + ) + + + + ) + + + + | + + + + − + π + + 6 + + + π + 6 + + + → + + 25 + 12 + + π + + + . And that’s right!
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What if we integrated the entire shape and divided by three, as mentioned above? This gives us actually the same integral, with a different multiplier and different bounds: + + + + + + 1 + 3 + + + 25 + 2 + + + + ∫ + + + − + π + + 2 + + + π + 2 + + + + cos + 2 + + + + ( + + + 3 + θ + + + ) + + d + θ + → + + + + + + + + 1 + 3 + + + 25 + 24 + + + ( + + + 6 + + θ + + + sin + + + ( + + + 6 + θ + + + ) + + + + ) + + + + | + + + + − + π + + 2 + + + π + 2 + + + → + + 25 + 12 + + π + + + . Exactly the same! Math is weird like that.
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To avoid boundary hell in the future, I’m just going to integrate whole shapes and divide by petals. It’s much easier to wrap my head around why that actually works.

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Stay tuned for Lagrange multipliers!

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