Chapter 4.4, Problem 25E

To determine

To Find:The coordinates of the leftmost and the rightmost point of the loop.

Answer to Problem 25E

Rightmost point is $\left(\frac{2\sqrt{3c}}{9},\frac{c}{3}\right)$

Leftmost point is $\left(-\frac{2\sqrt{3c}}{9},\frac{c}{3}\right)$

Explanation of Solution

Given:Investigate the family of curves given by the parametric equations $x=t^3-ct,y=t^2$ . In a particular determine the value of $c$ for which there is a loop and find the point where the curve intersect itself. What happens to the loop as $c$ increases? Find the coordinates of the leftmost and the rightmost point of the loop.

Given: $x=t^3-ct,y=t^2$

The graph for the families of parametric equation for all the values of c

  

Here

  $\begin{array}{l}c>0\\ c\in (0,\infty )\end{array}$

From the graph it is seen that when the $c$ is positive the graph intersect itself at a point on $y$ axis having the coordinates $(0,c)$

The leftmost and the rightmost both have vertical tangentshaving slope $k=\infty$

Then

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\infty \\ \frac{dx}{dt}=0\\ \frac{d(t^3-ct)}{dt}=3t^2-c=0\\ t=\pm \sqrt{\frac{c}{3}}\end{array}$

Now for $t=\sqrt{\frac{c}{3}}$

  $\begin{array}{l}x=2{\left(\sqrt{\frac{c}{3}}\right)}^3-c\left(\sqrt{\frac{c}{3}}\right)=\frac{2\sqrt{3c}}{9}\\ y=\frac{c}{3}\\ (x,y)=\left(\frac{2\sqrt{3c}}{9},\frac{c}{3}\right)\end{array}$

Thus the rightmost point is $\left(\frac{2\sqrt{3c}}{9},\frac{c}{3}\right)$

Similarly

Now for $t=-\sqrt{\frac{c}{3}}$

  $\begin{array}{l}x=2{\left(-\sqrt{\frac{c}{3}}\right)}^3-c\left(-\sqrt{\frac{c}{3}}\right)=-\frac{2\sqrt{3c}}{9}\\ y=\frac{c}{3}\\ (x,y)=\left(-\frac{2\sqrt{3c}}{9},\frac{c}{3}\right)\end{array}$

Thus the leftmost point is $\left(-\frac{2\sqrt{3c}}{9},\frac{c}{3}\right)$