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My midterms are approaching, and I was going through some of our past Calculus midterms when I stumbled upon this question from 1996:

Show that these two curves,

$$(\Gamma) : \frac {x^2}{4a^2} + \frac{y^2}{a^2} -1 = 0 $$ $$(\Omega): r=a\sin2\theta$$

have the same length.

Now, what I've done so far is find the corresponding parametric equations and use the following method to compute the lengths:

$$ \int_a^{b} \sqrt{x'(t)^2 +y'(t)^2}dt $$

And the results are:

Length of $(\Gamma)=\int_0^{2\pi} |a|\sqrt{4\sin^2\theta + \cos^2\theta}d\theta$

Length of $(\Omega)=\int_0^{2\pi} 2|a| \sqrt{ \sin^6\theta + \cos^6\theta}d\theta $

But this is where I reached an impass. I tried to compute the integrals myself, to no avail. Then I tried to compute them using an online integrator, and it turns out that they have the same value, so I think I'm on the right track. Can anyone suggest a way to prove they are equal that doesn't involve a calculator at the end (as they were banned during exams back in 1996, and I don't think that many students had access to them anyway) or a whole different method to get the lengths?

I really appreciate any help you can provide.

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i dont think its done by considering the integral. try to find some relation between the curves. – Suraj M S Apr 13 '14 at 9:31
up vote 1 down vote accepted

The length element for a polar curve $\theta\mapsto r(\theta)$ is $\mathrm{d}\ell=\sqrt{r(\theta)^2+r'(\theta)^2}\mathrm{d}\theta$. This fact is easy to prove as follows (and should be a known result): from $$\mathrm{d}(r\vec e_r)=(\mathrm{d}r)\vec e_r+r\mathrm{d}\vec e_r=r'(\theta)\vec e_r\mathrm{d}\theta+r(\theta)\vec e_\theta\mathrm{d}\theta$$ we obtain $$\mathrm{d}\ell=\bigl\lvert\mathrm{d}(r\vec e_r)\bigr\rvert=\sqrt{r(\theta)^2+r'(\theta)^2}\mathrm{d}\theta$$ since $(\vec e_r,\vec e_\theta)$ is an orthonormal frame.

Now the length of $(\Omega)$ is: $$\mathscr{L}=\int_{(\Omega)}\mathrm{d}\ell=\int_0^{2\pi}\sqrt{r(\theta)^2+r'(\theta)^2}\,\mathrm{d}\theta=\int_0^{2\pi}\sqrt{a^2\sin^2(2\theta)+4a^2\cos^2(2\theta)}\,\mathrm{d}\theta$$.

Now a straightforward substitution ($t=2\theta$) yields: $$\mathscr{L}=\int_0^{4\pi}\sqrt{a^2\sin^2(t)+4a^2\cos^2(t)}\,\frac{\mathrm{d}t}2$$

Because we're integrating a function that is periodic of period $2\pi$ we also have:

$$\mathscr{L}=2\int_0^{2\pi}\sqrt{a^2\sin^2(t)+4a^2\cos^2(t)}\,\frac{\mathrm{d}t}2.$$ Simplify the $2$'s and you obtain the length of the curve $(\Gamma)$.

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Great, this method goes around plotting AND integrating. I already know the length element for a polar curve but where $r(t)$ and $\theta (t)$ instead of $r(\theta)$. Making the right substitutions yields the same. Thanks for the insight. – GeorgSaliba Apr 13 '14 at 15:32

$S$ is term inside square root.

$S_1=1+3 \sin^2 x$

$S_2=\sin^4 x+\cos^4 x-\sin^2 x \cos^2 x$

$=1-3\sin^2 x \cos^2 x$

Bring that 2 inside

$=4-3\sin^2 2x$

Put $2x=t \implies 2dx=dt$

Limit will go to $4\pi$ but $dt/2$ shall counter it due to periodicity.

$=4-3\sin^2 t$


Plotting both functions in a rough graph will lead to believe that they have the same area.

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This is a pretty compelling answer! But what I don't understand is the last part, how come they represent the length of those two curves? – GeorgSaliba Apr 13 '14 at 9:59
length of $f(x)=\int \sqrt{1+f'^2(x)}dx$ within proper limits – evil999man Apr 13 '14 at 10:00
Yes exactly! How could've I forgotten? We also use this in Mechanics. Thanks a lot! – GeorgSaliba Apr 13 '14 at 10:03
Just one last verification if you may : shouldn't they represent the lengths of $-\sqrt{3} \cos x and \sqrt{3} \sin x $ respectively since $y'^2 = 3(sinx)^2$ for the first and $y'^2= 3(cosx)^2$ for the second ?? – GeorgSaliba Apr 13 '14 at 10:22
@GeorgSaliba Right. My bad. Will edit. – evil999man Apr 13 '14 at 10:23

I do not know how much this could help $$4 \sin ^2(t)+\cos ^2(t)-4 \left(\sin ^6(t)+\cos ^6(t)\right)=-\frac{3}{2} (\cos (2 t)+\cos (4 t))$$

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Please read again. There's a square root, so I don't think that'll work. – GeorgSaliba Apr 13 '14 at 9:45
Yes, I know. I was just wondering is this simplification could bring something. – Claude Leibovici Apr 13 '14 at 9:59

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