# Hypocycloid

• Oct 20th 2008, 07:11 AM
magentarita
Hypocycloid
The hypocloid is a curve defined by the parametric equations x(t) = cos^3 (t), y(t) = sin^3 (t), where
0 < or = to t < or = to 2pi.

Given the information above, find rectangular equations of the hypocycloid.
• Oct 20th 2008, 12:43 PM
Soroban
Hello, magentarita!

Quote:

The hypocloid is a curve defined by the parametric equations: .**

. . $\begin{array}{ccc} x(t) &=& \cos^3\!t \\ y(t) &=& \sin^3\!t\end{array}\;\;\text{ where } 0 \leq t \leq 2\pi$

Given the information above, find rectangular equations of the hypocycloid.

**
.There are many many hypocycloids. .This is just one of them.
This requires a clever trick . . .

We have: . $\begin{array}{c}x\:=\:\cos^3\!t \\ y \:=\:\sin^3\!t \end{array}$

Raise both equations to the power $\tfrac{2}{3}$

. . $\begin{array}{ccccccc}x^{\frac{2}{3}} \;=\;\left(\cos^3\!t \right)^{\frac{2}{3}} & \Rightarrow & x^{\frac{2}{3}} \:=\:\cos^2\!t & {\color{blue}[1]}\\
y^{\frac{2}{3}} \;=\;\left(\sin^3\!t\right)^{\frac{2}{3}} & \Rightarrow & y^{\frac{2}{3}} \:=\:\sin^2\!t & {\color{blue}[2]} \end{array}$

$\text{Add {\color{blue}[1]} and {\color{blue}[2]}: }\;x^{\frac{2}{3}} + y^{\frac{2}{3}} \;=\;\underbrace{\cos^2\!t + \sin^2\!t}_{\text{This is 1}}$

Therefore: . $x^{\frac{2}{3}} + y^{\frac{2}{3}} \;=\;1$

• Oct 20th 2008, 09:56 PM
magentarita
Soroban...
Quote:

Originally Posted by Soroban
Hello, magentarita!

This requires a clever trick . . .

We have: . $\begin{array}{c}x\:=\:\cos^3\!t \\ y \:=\:\sin^3\!t \end{array}$

Raise both equations to the power $\tfrac{2}{3}$

. . $\begin{array}{ccccccc}x^{\frac{2}{3}} \;=\;\left(\cos^3\!t \right)^{\frac{2}{3}} & \Rightarrow & x^{\frac{2}{3}} \:=\:\cos^2\!t & {\color{blue}[1]}\\$ $
y^{\frac{2}{3}} \;=\;\left(\sin^3\!t\right)^{\frac{2}{3}} & \Rightarrow & y^{\frac{2}{3}} \:=\:\sin^2\!t & {\color{blue}[2]} \end{array}" alt="
y^{\frac{2}{3}} \;=\;\left(\sin^3\!t\right)^{\frac{2}{3}} & \Rightarrow & y^{\frac{2}{3}} \:=\:\sin^2\!t & {\color{blue}[2]} \end{array}" />

$\text{Add {\color{blue}[1]} and {\color{blue}[2]}: }\;x^{\frac{2}{3}} + y^{\frac{2}{3}} \;=\;\underbrace{\cos^2\!t + \sin^2\!t}_{\text{This is 1}}$

Therefore: . $x^{\frac{2}{3}} + y^{\frac{2}{3}} \;=\;1$