Chapter 10, Problem 4PS

To determine

To show: The parametric equations $x=r\left(\cos \theta +\theta \sin \theta \right)$ and $y=r\left(\sin \theta -\theta \sin \theta \right)$ represent the involutes of a circle.

Answer to Problem 4PS

Yes, the given parametric equations $x=r\left(\cos \theta +\theta \sin \theta \right)$ and $y=r\left(\sin \theta -\theta \sin \theta \right)$ represent the involutes of a circle.

Explanation of Solution

Given information:

  

The parametric equations $x=r\left(\cos \theta +\theta \sin \theta \right)$ and $y=r\left(\sin \theta -\theta \sin \theta \right)$ .

Calculation:

From the figure, we can understand that $QP$ is a tangent.

So let’s draw a line $QR$ and name the angle $\angle PQR$ as $t$ . Now, we have a right angled triangle $PQR$ .

Then, draw the line $QO\text{and}QT$ and make another right angled triangle.

  

Relationship between the angle $t$ and angle $\theta$ is given by the equation,

  $\begin{array}{c}t+{90}^{\circ }=\theta \\ t=\theta -{90}^{\circ }\end{array}$

  $QP$ is equal to the arc length corresponding to the central angle $\theta$ with the length of,

  $QP=r\theta$

Now, let’s solve for the parametric equation

  $\begin{array}{c}x=QR-QT\\ \cos t=\frac{QR}{QP}\to \text{from the larger triangle}\\ \cos t=\left(\theta -{90}^{\circ }\right)=\frac{QR}{r\theta }\\ QR=r\theta \cos \left(\theta -{90}^{\circ }\right)\\ QR=r\theta \sin \theta \\ \sin t=\frac{QT}{r}\to \text{from the smaller triangle}\\ \sin t\left(\theta -{90}^{\circ }\right)=\frac{QT}{r}\\ QT=r\sin \left(\theta -{90}^{\circ }\right)\\ QT=-r\cos \theta \end{array}$

So,

  $\begin{array}{l}x=r\theta \sin \theta -\left(-r\cos \theta \right)\\ x=r\theta \sin \theta +\cos \theta \\ x=r\left(\cos \theta +\theta \sin \theta \right)\end{array}$

Similarly for the parametric equation of y,

  $\begin{array}{c}y=PR-TO\\ \sin t=\frac{PR}{QP}\to \text{from the larger triangle}\\ \sin t=\left(\theta -{90}^{\circ }\right)=\frac{PR}{r\theta }\\ QR=r\theta \sin \left(\theta -{90}^{\circ }\right)\\ QR=-r\theta \cos \theta \\ \cos t=\frac{TO}{r}\to \text{from the smaller triangle}\\ \sin t\left(\theta -{90}^{\circ }\right)=\frac{TO}{r}\\ TO=r\cos \left(\theta -{90}^{\circ }\right)\\ QT=r\sin \theta \end{array}$

Therefore,

  $\begin{array}{l}y=-\left(-r\cos \theta \right)+r\theta \sin \theta \\ y=-r\theta \cos \theta +\sin \theta \\ y=r\left(\sin \theta -\theta \cos \theta \right)\end{array}$

Thus we can show that the given parametric equations represent the involutes of a circle.