Can someone tell me the steps to get the parametric form of a curve? For example:
$x^{2\over 3}$ +$y^{2\over 3}$ =1
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You could substitute $y = tx$ into the equation of the curve and solve it for $x$ in terms of $t$, then obtain an expression for $y$ from the relation $y=tx$. – ivanpenev May 30 '14 at 13:56
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i will try it,thanks – user128576 May 30 '14 at 14:25
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How would you parametrize $x^2 + y^2 = 1$? Recall the well-known identity $\cos^2 \theta + \sin^2 \theta = 1$, and the obvious parametrization is $x = \cos \theta$, $y = \sin \theta$. Can you think of something similar here?
The same parametrization doesn't work, because of the cube roots. Somehow you need to make the cube roots go away and get $\cos^2 \theta$ and $\sin^2 \theta$. Think and try to get it on your own, otherwise read further.
As you want $x^{2/3}$ to be $\cos^2 \theta$, obviously, put $x = \cos^3 \theta$. Similarly, $y = \sin^3 \theta$.
More generally, for the astroid $x^{2/3} + y^{2/3} = a^{2/3}$, the parametrization is $x = a\cos^3 \theta$, $y = a\sin^3 \theta$.
M. Vinay
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