Parametrization of a hypocycloid

Question

How do I prove that a hypocycloid, which has equation $$x^{2/3} + y^{2/3} = a^{2/3}$$ can be parameterized by $$x = a\cos^3(\theta),\qquad y = a\sin^3(\theta)$$?

The problem assumes that it is true, but I'm not quite sure how to go about proving it. How do I proceed?

Answer 1

Consider the more general equation $x^p+y^p=r^p$ where $p>0$. Compare this to the identity $\sin^2(\theta)+\cos^2(\theta)=1$. Rewriting the first equation we have.

$$\frac{x^p}{r^p}+\frac{y^p}{r^p}=1$$

$$\left(\frac{x}{r}\right)^p+\left(\frac{y}{r}\right)^p=1$$

$$\left[\left(\frac{x}{r}\right)^{p/2}\right]^2+\left[\left(\frac{y}{r}\right)^{p/2}\right]^2=1$$

Using the trig identity

$$\left(\frac x r\right)^{p/2}=\cos(\theta)\quad\text{and}\quad\left(\frac y r\right)^{p/2}=\sin(\theta)$$

Solving for $x$ and $y$ we get

$$x=r\cos^{2/p}(\theta)\quad\text{and}\quad y=r\sin^{2/p}(\theta)$$ The hypocycloid is for the case $p=2/3$.

Answer 2

Upon substituting it in, you can quickly see that it is true.$$x^{\frac23}+y^{\frac23}=a^{\frac23}$$$$[a\cos^3(\theta)]^{\frac23}+[a\sin^3(\theta)]^{\frac23}=a^{\frac23}$$Distribute the exponents.$$a^{\frac23}\cos^2(\theta)+a^{\frac23}\sin^2(\theta)=a^{\frac23}$$Divide by $a^{\frac23}$.$$\cos^2(\theta)+\sin^2(\theta)=1$$Which is a true statement.

Answer 3

You only need to find two functions, $x=f(t)$ and $y=g(t)$, such that $$x^{\frac{2}{3}}+y^{\frac{2}{3}}=f(t)^{\frac{2}{3}}+g(t)^{\frac{2}{3}}=a^{\frac{2}{3}} \dots(1)$$ Then $f(\theta)=a \cos^3(\theta)$ and $g(\theta)=a \sin^3(\theta)$ give us $$(a\cos^3(\theta))^{\frac{2}{3}}+(a \sin^3(\theta))^{\frac{2}{3}}=a^{\frac{2}{3}}(\cos^2(\theta)+\sin^2(\theta))=a^{\frac{2}{3}}$$ So for that $f(\theta)$ and $g(\theta)$ it's a parametrization of the hypocycloide. Note that $f$ and $g$ can be any functions that satisfy the condition (1)