Chapter 8.4, Problem 59E

(a)

To determine

To find: the parametric equations for the hypocycloid.

Answer to Problem 59E

  $\begin{array}{l}x=\left(a-b\right)\cos \theta +b\cos \left(\frac{a-b}{b}\theta \right)\\ y=\left(a-b\right)\sin \theta -b\sin \left(\frac{a-b}{b}\theta \right)\end{array}$

Explanation of Solution

Calculation:

Consider the following sketch of a hypocycloid.

  

The center of the smaller circle at point $Q$ has the following coordinates

  $\left(\left(a-b\right)\cos \theta ,\left(a-b\right)\sin \theta \right)$ .Next find the arc $\text{PS}$ on the circle $\text{C}$ which has the length of $\text{a}\theta$

Which is equal in length to the arc $\text{AS}$ . Therefore the angle $\angle \text{PQS=}\frac{a}{b}\theta$ and

  $\angle PQT=\frac{a}{b}\theta -\theta$

  $\begin{array}{l}x=\left(a-b\right)\cos \theta +b\cos \left(\angle PQT\right)\\ y=\left(a-b\right)\sin \theta -b\sin \left(\angle PQT\right)\end{array}$ Find the coordinates at point $Q$

  $\begin{array}{l}x=\left(a-b\right)\cos \theta +b\cos \left(\frac{a-b}{b}\theta \right)\\ y=\left(a-b\right)\sin \theta -b\sin \left(\frac{a-b}{b}\theta \right)\end{array}$ Plug in for the angle $\angle PQT$

Conclusion:

  $\begin{array}{l}x=\left(a-b\right)\cos \theta +b\cos \left(\frac{a-b}{b}\theta \right)\\ y=\left(a-b\right)\sin \theta -b\sin \left(\frac{a-b}{b}\theta \right)\end{array}$

(b)

To determine

To find: the parameter to obtain an equation for the asteroid in rectangular coordinates.

Answer to Problem 59E

  

Explanation of Solution

Calculation:

Plug in for $a=4b$

  $\begin{array}{l}x=\left(4b-b\right)\cos \theta +b\cos \left(\frac{4b-b}{b}\theta \right)\\ y=\left(4b-b\right)\sin \theta +b\sin \left(\frac{4b-b}{b}\theta \right)\end{array}$ plug in for $a=4b$

  $\begin{array}{l}x=3b\cos \theta +b\cos \left(3\theta \right)\\ y=3b\sin \theta -b\sin \left(3\theta \right)\end{array}$ Simplify

  $\begin{array}{l}x=\frac{3a}{4}\cos \theta +\frac{a}{4}\cos \left(3\theta \right)\\ y=\frac{3a}{4}\sin \theta +\frac{a}{4}\sin \left(3\theta \right)\end{array}$ Plug in for $\text{b=}\frac{a}{4}$

  $\begin{array}{l}x=\frac{3a}{4}\cos \theta +\frac{a}{4}\left(4{\cos }^3\theta -3\cos \theta \right)\\ y=\frac{3a}{4}\sin \theta +\frac{a}{4}\left(3\sin \theta -4{\sin }^3\theta \right)\end{array}$ Recall $\begin{array}{l}\cos 3\theta =4{\cos }^3\theta -3\cos \theta \\ \sin 3\theta =3\sin \theta -4{\sin }^3\theta \end{array}$

  $\begin{array}{l}x=\frac{3a}{4}\cos \theta +\frac{a}{4}{\cos }^3\theta -\frac{3a}{4}\cos \theta \\ y=\frac{3a}{4}\sin \theta -\frac{3a}{4}\sin \theta +a{\sin }^3\theta \end{array}$ Distribute

  $\begin{array}{l}x=a{\cos }^3\theta \\ y=a{\sin }^3\theta \end{array}$ Simplify

Now eliminate the parameter to find the equation of the function

  $\begin{array}{l}\frac{x}{a}={\cos }^3\theta \\ \frac{y}{a}={\sin }^3\theta \end{array}$

  $\begin{array}{l}{\left(\frac{x}{a}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}={\cos }^2\theta \\ {\left(\frac{y}{a}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}={\sin }^2\theta \end{array}$ Take the each side to the $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ power

  ${\left(\frac{x}{a}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+{\left(\frac{y}{a}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=1$ Use the property ${\sin }^2\theta +{\cos }^2\theta =1$

Now sketch the graph by plotting points to obtain

  

Conclusion: