Question

Convert the parametric equation :
x=a (cos t)^3 , y = a (sin t)^3
to cartesian (rectangular form).


Please hellppp!

Answer 1

I had the following idea (not quite sure if it is exactly what you wanted):

\[x = a \cos(t)^3\]\[y = a \sin(t)^3\]

\[\sqrt[3]{x} = \sqrt[3]{a} \cos(t)\]\[\sqrt[3]{y} = \sqrt[3]{a} \sin(t)\]

\[(\sqrt[3]{x})^2 = (\sqrt[3]{a})^2 \cos(t)^2\]\[(\sqrt[3]{y})^2 = (\sqrt[3]{a})^2 \sin(t)^2\]

\[(\sqrt[3]{x})^2 + (\sqrt[3]{y})^2 = (\sqrt[3]{a})^2 * (\cos(t)^2 + \sin(t)^2) = (\sqrt[3]{a})^2 \]

Answer 2

We can see that\[\cos^3t=\frac{ x }{ a }\]So\[\cos t=\sqrt[3]{\frac{ x }{ a }}\]Also:\[\cos^2 t+\sin^2 t = 1 \Leftrightarrow \sin t = \pm \sqrt{1-\cos^2 t}\]Now combine the two results:\[\sin^3 t=\pm (\sqrt{1-\cos^2t})^3=\pm (1-\cos^2t)^{\frac{ 3 }{ 2 }}\]

Because y = a sin³t we now have:\[y=\pm a\left(1-\sqrt[3]{\left( \frac{ x }{ a } \right)^2}\right)^{\frac{ 3 }{ 2 } }\]

Answer 3

Thanks a ton!

Answer 4

yw!

Answer 5

you can see what this thing is looking at http://en.wikipedia.org/wiki/Astroid (it's for a = 1)