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Consider the parametric curve: $$x=6\cos^3(t), y=6\sin^3(t)$$ Write it in Cartesian form.

I am really struggling with the solution for this. I've been trying to find $t$ from $x$, and then plugging that into $y$, but it ends up being too complicated.

It is asked to be put in the form $F(x,y)=c$, for some function $F,$ and some constant $c$.

Any help at all would be appreciated, thank you.

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    $\begingroup$ Could you check what you wrote in the title? You have defined $x$ twice. $\endgroup$
    – Empy2
    May 18, 2015 at 11:06
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    $\begingroup$ yes my bad, one of the x's is supposed to be y=6sin^3(t) $\endgroup$
    – AidanLeith
    May 18, 2015 at 11:10

3 Answers 3

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solving for $t,$ you get $$\cos t = \left(\frac x6\right)^{1/3} , \, \sin t = \left(\frac y6\right)^{1/3} $$ now use the fact $$\sin^2 t + \cos ^2 t = 1 \to \left(\frac x6\right)^{2/3} + \left(\frac y6\right)^{2/3} = 1$$

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    $\begingroup$ Greatly appreciated, thankyou. $\endgroup$
    – AidanLeith
    May 18, 2015 at 11:25
  • $\begingroup$ @AidanLeith, you are welcome. $\endgroup$
    – abel
    May 18, 2015 at 11:29
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From $$x=6\cos^3(t), y=6\sin^3(t)$$ we get $$\cos^2t=\left(\frac{x}{6}\right)^{2/3}$$ and $$\sin^2t=\left(\frac{y}{6}\right)^{2/3}$$ From the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$ then we get $$\left(\frac{x}{6}\right)^{2/3}+\left(\frac{y}{6}\right)^{2/3}=1$$ or $$x^{2/3}+y^{2/3}=6^{2/3}$$

This is a very useful technique for finding the Cartesian form of a parametric curve given in terms of sines and cosines (etc), namely use some sort of trigonometric identity to relate the sines and cosines and then substitute your $x$s and $y$s.

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  • $\begingroup$ Greatly appreciated thank you. $\endgroup$
    – AidanLeith
    May 18, 2015 at 11:25
  • $\begingroup$ no worries! you should choose a particular answer you want to accept, not necessarily mine but whichever you found answered your question the best :) $\endgroup$
    – danimal
    May 18, 2015 at 11:46
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Perhaps the problem was $x=6\cos^3 t,y=6\sin^3t$.
Write $\cos^2t$ as a function of $x$, and $\sin^2t$ as a function of $y$.

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  • $\begingroup$ Yes it was supposed to be that, thank you :) $\endgroup$
    – AidanLeith
    May 18, 2015 at 11:11
  • $\begingroup$ I get cos^2(t)=x/6cos(t), and sin^2(t)=y/6sin(t), but that still doesn't eliminate to parameter, do I still do x/6cos(t) + y/6sin(t) = 1, or am I missing something? $\endgroup$
    – AidanLeith
    May 18, 2015 at 11:17

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