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

Question

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.

Answer 1

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 $y$.

Answer 2

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 $$\sin^2 t + \cos ^2 t = 1 \to \left(\frac x6\right)^{2/3} + \left(\frac y6\right)^{2/3} = 1$$

Answer 3

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.