Parametric Equations and Curves — Question 2

Question 2 – Solution

Solution

1. Begin with $$x=\cos^3 t, \qquad y=\sin^3 t.$$ Taking each coordinate to the power $2/3$ gives $$x^{2/3}=\cos^2 t, \qquad y^{2/3}=\sin^2 t.$$ Here $x^{2/3}$ means $(\sqrt[3]{x})^2$, so the calculation is valid even when a coordinate is negative.

2. Add the two equations and use $\cos^2t+\sin^2t=1$: $$\boxed{x^{2/3}+y^{2/3}=1}.$$ This curve is an astroid.

3. The curve is least circle-like where it has cusps. These occur when one trigonometric coordinate is $0$ and the other is $\pm1$.

4. Evaluating the parametrization at the corresponding parameter values gives $$\begin{array}{c|c} t & (x,y)\\ \hline 0 & (1,0)\\ \pi/2 & (0,1)\\ \pi & (-1,0)\\ 3\pi/2 & (0,-1) \end{array}$$

5. Therefore, the four sharp points are $$\boxed{(1,0),\ (0,1),\ (-1,0),\ (0,-1)}.$$