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Please look at this webpage for reference: http://mathworld.wolfram.com/Hypocycloid.html

Go to line 7 to 15 on that webpage. Line 7 and 8 show 2 parametric equations. These have been rewritten as in line 12 and 13. I understand how this has been done. But I want to know how these parametric equations (line 12 and 13) have been used to get the arc length and the area equations (line 14 and 15 respectively).

I have tried using the arc length formula:

I rearranged to get the radius in terms of S and theta. Then I substitute that into the area equation:

But that gives me the area in terms of theta. However, on the webpage, there is no angle or theta.

How should I do this? If there is something I am mission or these is another, better way to do this, please tell me.

Sorry for my incompetence and thanks in advance.

EDIT

By "LINE" I mean the numbers that are on the right side of the page in brackets

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Lakshya Goyal
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1 Answers1

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Note that hypocycloid is simply connected when $\displaystyle \frac{a}{b}=n=3,4,5, \ldots$

Now \begin{align*} x &= \frac{a}{n}[(n-1)\cos t+\cos (n-1)t] \\ y &= \frac{a}{n}[(n-1)\sin t-\sin (n-1)t] \\ x' &= \frac{a(n-1)}{n} [-\sin t-\sin (n-1)t] \\ y' &= \frac{a(n-1)}{n} [\cos t-\cos (n-1)t] \\ x'^2+y'^2 &= \frac{a^2(n-1)^2}{n^2} (2-2\cos nt)] \\ &= \frac{4a^2(n-1)^2}{n^2} \sin^2 \frac{nt}{2} \\ ds &= \frac{2a(n-1)}{n} \left| \sin \frac{nt}{2} \right| dt \\ P &= n\int_{0}^{2\pi/n} \frac{2a(n-1)}{n} \sin \frac{nt}{2} \, dt \\ &= 2a(n-1) \left[ -\frac{2}{n} \cos \frac{nt}{2} \right]_{0}^{2\pi/n} \\ &= \frac{8a(n-1)}{n} \\ A &= \oint_{C} x\, dy \\ &= \frac{a^2(n-1)}{n^2} \times \pi [(n-1)-1] \quad \quad \text{(see the note below)} \\ &= \frac{\pi a^2(n-1)(n-2)}{n^2} \end{align*}

Note that $$x\, dy = \frac{a^2(n-1)}{n^2} [(n-1)\cos^2 t+(2-n)\cos t \cos (n-1)t-\cos^2 (n-1)t] \, dt$$

By $$\int_{0}^{2\pi} \cos nt \, \cos mt \, dt = \pi \delta_{n,m}$$ only the square terms in $x\, dy$ survive after integration.

For the area of hypocycloid with other values of $n$, see another post here.

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Ng Chung Tak
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    Interesting answer. Besides, I had a look at your answer dated July 15th. May I ask you how you do you realize your animations ? – Jean Marie Aug 01 '16 at 15:46
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    @JeanMarie, The animation was made by Geogebra which is a freeware and free to download. Thanks to your visit anyways. – Ng Chung Tak Aug 01 '16 at 15:53
  • Thank you very much. I didn't know that Geogebra permit animations... – Jean Marie Aug 01 '16 at 15:57
  • @NgChungTak It's a very good answer but can you explain what is "ds" and "P". Also, that integral sign with the circle and the letter C (for the area), what is that? Sorry, I don't know what that symbol means or what its called. – Lakshya Goyal Aug 02 '16 at 02:24
  • @NgChungTak Also, why did you square and then add the x and y parametric equations? What is the reason for doing this? I looked online and some of them said do this: (Go to think link)

    http://latex.codecogs.com/gif.latex?A=\int_{0}^{2\pi}Y(\theta)(\frac{d}{d\theta}X(\theta))d\theta

    Does this work?

     – Lakshya Goyal Aug 02 '16 at 02:44
  • $ds$ is differential of arclength, $$ds=\sqrt{(dx)^2+(dy)^2}=\sqrt{x'^2+y'^2} dt$$ $P$ is perimeter, $$P=\int ds=\int \sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt$$ $C$ is a close contour/curve, usinig Green's theorem, $$A=\oint_{C} x dy = -\oint_{C} y dx = \frac{1}{2} \oint_{C} x dy-y dx$$ – Ng Chung Tak Aug 02 '16 at 07:43
  • @NgChungTak Why did you square root to get ds. Can't you just get the derivative of http://latex.codecogs.com/gif.latex?{x}'^{2}+{y}'^{2}. Also, did you need to find the perimeter to get the Area? – Lakshya Goyal Aug 02 '16 at 12:50
  • @NgChungTak Also, why do you use the close curve integral instead of normal integral? – Lakshya Goyal Aug 02 '16 at 13:18
  • Either $x(y)$ or $y(x)$ is multi-valued. A useful link – Ng Chung Tak Aug 02 '16 at 13:32
  • There're no correlation among the perimeter of arbitrary close curve $C$ and its area. However ,the area is maximized when $C$ is a circle, please see this. Detail for arc length, please read this. By the way, do borrow and read more reference books or texts on calculus in your school library. – Ng Chung Tak Aug 02 '16 at 13:49
  • @NgChungTak Thanks for this so far but is there a way to get to the Area equation without using Green's Theorem? I am not at that high level of mathematics to be able to understand that. Online I found way which differentiated the y equation and multiplied it by the x equation. I would then integrate that from 0 to 2pi. Here is that website:http://tutorial.math.lamar.edu/Classes/CalcII/ParaArea.aspx – Lakshya Goyal Aug 02 '16 at 14:19
  • @NgChungTak OR even better would be: if I could use the original parametric equations (Line 7 and 8 on the wolfram website) and then use the method shown here (differentiat the y equation and multiply it by the x equation): http://tutorial.math.lamar.edu/Classes/CalcII/ParaArea.aspx Doing that, is there some way to get to the Area equation? – Lakshya Goyal Aug 02 '16 at 14:24