Basically I have proved that the parametric for epicycloid is $$x=(a+b)\cos t-b\cos (\frac{a+b}{b}t)$$ and $$y=(a+b)\sin t-b\sin (\frac{a+b}{b}t)$$ So, if $b=a$ this leads to $$x=2a\cos t-a\cos 2t$$ and $$y=2a\sin t-a\sin 2t$$ where a is the radius of bigger circle and b is the radius of smaller circle.

I need to prove that that the new $x$ and $y$ gives cardioid $r=2a(1-\cos \theta)$. The hint in the book says that I need to eliminate the parameter, which I think in this case is $t$. After eliminating the paratemer, I then need to prove that $x=r\cos \theta + a$ and $y=r\sin \theta$. Then show $r=2a(1-\cos \theta)$. So, I guess I need to convert to polar coordinate.

I am not sure how to eliminate the parameter here. Basically what I have done so far is using the identity $\sin 2t = 2\sin t \cos t$. So I get, \begin{align}y&=2a\sin t-a\sin 2t\\ &=2a\sin t-a(2\sin t \cos t)\\ &=2a\sin t-2a\sin t \cos t\\ y&=2a\sin t(1-\cos t). \end{align} I don't know how to isolate the $t$. In fact, I don't really know if this is even a correct approach. So, any help appreciated.


Note that the polar formula differs from the Cartesian one by a translation of $a$ units in the $-x$ direction. With that in mind, going the other way from polar to Cartesian proves the formulas: $$x=r\cos\theta+a=2a(1-\cos\theta)\cos\theta+a=2a\cos\theta-2a\cos^2\theta+a=2a\cos\theta-a(\cos2\theta+1)+a=2a\cos\theta-a\cos2\theta$$ $$y=r\sin\theta=2a(1-\cos\theta)\sin\theta=2a\sin\theta-2a\sin\theta\cos\theta=2a\sin\theta-a\sin2\theta$$ So $t$ in the epicycloid functions exactly corresponds to $\theta$ in the cardioid function.

  • $\begingroup$ "Note that the polar formula differs from the Cartesian one by a translation of $a$ units in the $-x$ direction.' Could you elaborate more on this please?. If we replace $a$ by $a-a\cos \theta$, then it is true. But I am not sure why that statement is true. Thanks for the quick reply though. $\endgroup$ – aaaaaa Dec 3 '17 at 4:45
  • $\begingroup$ @ardhemist I used Desmos to verify graphically that $x=r\cos\theta+a$. The translation comes from there. $\endgroup$ – Parcly Taxel Dec 3 '17 at 4:47
  • $\begingroup$ ah, I graph it and it is true indeed. Do you have an idea on how to isolate eliminate the parameter $t$ since my teacher is pretty strict? Thanks for the solution. $\endgroup$ – aaaaaa Dec 3 '17 at 5:29
  • $\begingroup$ @ardhemist $t=\theta$. That's it. $\endgroup$ – Parcly Taxel Dec 3 '17 at 5:30

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.