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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.

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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.

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  • $\begingroup$ "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

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