0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ May refer to my answer here $\endgroup$ – Ng Chung Tak Aug 1 '16 at 13:06
  • $\begingroup$ @NgChungTak I don't think that answers my question. I already have the parametric equations of the hypocycloid. They are on line 7 and 8. I want to know how the equation in line 14 and 15 came to be. $\endgroup$ – Lakshya Goyal Aug 1 '16 at 13:18
  • $\begingroup$ Line $7$ reads $$\theta=\frac{a-b}{b} \phi$$ which is equation $(2).$ Please specify which equations you don't understand instead of which lines, THAT'S HELP! $\endgroup$ – Ng Chung Tak Aug 1 '16 at 13:52
  • $\begingroup$ @NgChungTak Look at my edit $\endgroup$ – Lakshya Goyal Aug 1 '16 at 14:02
  • $\begingroup$ @NgChungTak Sorry for not making it clear before. $\endgroup$ – Lakshya Goyal Aug 2 '16 at 2:25
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Interesting answer. Besides, I had a look at your answer dated July 15th. May I ask you how you do you realize your animations ? $\endgroup$ – Jean Marie Aug 1 '16 at 15:46
  • 1
    $\begingroup$ @JeanMarie, The animation was made by Geogebra which is a freeware and free to download. Thanks to your visit anyways. $\endgroup$ – Ng Chung Tak Aug 1 '16 at 15:53
  • $\begingroup$ Thank you very much. I didn't know that Geogebra permit animations... $\endgroup$ – Jean Marie Aug 1 '16 at 15:57
  • $\begingroup$ @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. $\endgroup$ – Lakshya Goyal Aug 2 '16 at 2:24
  • $\begingroup$ @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? $\endgroup$ – Lakshya Goyal Aug 2 '16 at 2:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.