Chapter 8.4, Problem 65E

To determine

To find: the rectangular equation for cycloid by eliminating the parameter from the given equations.

Answer to Problem 65E

The given curve has the rectangular coordinate equation,

  $y=a-a\cos \left(\frac{x+\sqrt{2ya-y^2}}{a}\right)$ .

Explanation of Solution

Calculation:

Find the rectangular equation for cycloid $x=a\left(\theta -\sin \theta \right)$ and $y=a\left(1-\cos \theta \right)$ by eliminating the parameter $\theta$ .

From the given equations,

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

And,

  $\begin{array}{l}y=a\left(1-\cos \theta \right)\\ \frac{y}{a}=1-\cos \theta \\ \cos \theta =1-\frac{y}{a}\end{array}$

Substituting the value of $\sin \theta$ in ${\sin }^2\theta +{\cos }^2\theta =1$ ,

  $\begin{array}{l}{\sin }^2\theta +{\cos }^2\theta =1\\ {\left(\theta -\frac{x}{a}\right)}^2+{\cos }^2\theta =1\\ {\cos }^2\theta =1-{\left({\theta }^2-\frac{x^2}{a^2}-2\theta \frac{x}{a}\right)}^2\end{array}$

Substituting the value of $\cos \theta$ in ${\sin }^2\theta +{\cos }^2\theta =1$ ,

  $\begin{array}{l}{\sin }^2\theta +{\left(1-\frac{y}{a}\right)}^2=1\\ {\sin }^2\theta =1-{\left(1-\frac{y}{a}\right)}^2\\ =1-\left(1+\frac{y^2}{a^2}-2\frac{y}{a}\right)\\ =1-1-\frac{y^2}{a^2}+2\frac{y}{a}\\ \sin \theta =\sqrt{\frac{2ya-y^2}{a^2}}\\ =\frac{\sqrt{2ya-y^2}}{a}\end{array}$

Substituting the value of $\sin \theta$ , in $x=a\left(\theta -\sin \theta \right)$ ,

  $\begin{array}{l}x=a\left(\theta -\sin \theta \right)\\ =a\left(\theta -\frac{\sqrt{2ya-y^2}}{a}\right)\\ =a\theta -\sqrt{2ya-y^2}\\ x+\sqrt{2ya-y^2}=a\theta \end{array}$

Solving the values of $\theta$ , in the above equation,

  $\theta =\frac{x+\sqrt{2ya-y^2}}{a}$

Substituting the value of $\theta$ , in $y=a\left(1-\cos \theta \right)$ ,

  $\begin{array}{l}y=a\left(1-\cos \theta \right)\\ =a-a\cos \left(\frac{x+\sqrt{2ya-y^2}}{a}\right)\end{array}$

Therefore, the given curve has the rectangular coordinate equation,

  $y=a-a\cos \left(\frac{x+\sqrt{2ya-y^2}}{a}\right)$ .

Conclusion:

Therefore, the given curve has the rectangular coordinate equation,

  $y=a-a\cos \left(\frac{x+\sqrt{2ya-y^2}}{a}\right)$ .