Chapter 2.9, Problem 46E

To determine

The minimum distance between them and determine whether it occur before or after the first of them has reached the watering hole.

Answer to Problem 46E

The occurrence of minimum distance between them happens after cheetah has reacted the watering hole at t=3.33sec.

It happens before gazelle has reacted the watering hole at t=4sec.

Explanation of Solution

Given:

The speed of the Cheetah is 30m/s towards the popular watering hole and the misguided gazelle is running due east towards the same spot at rate of 20m/s.

Cheetah starts from 100m and gazelle starts from 80m.

The rate of change of the distance between them at t =0, t =2, t =3 and t =4.

Considering both cheetah and gazelle started at the same time and they travel for t seconds.

As, the cheetah is running in the south direction for the speed of 30m/s for t seconds , the distance travelled is 30t.

As, it begins from the distance of 100m, the position of the cheetah is given by the following equation:

  $y\left(t\right)=30t-100$

Also, the gazelle is running in the east direction at the speed of 20m/s for t seconds, the distance travelled is 20t.

As it begins from the distance of 80m, the position of gazelle:

  $x\left(t\right)=20t-80$

As, the both of them are moving in the direction which are perpendicular to each-other, the distance between them is calculated by Pythagoras theorem.

  $\begin{array}{c}r\left(t\right)=\sqrt{{\left(x\left(t\right)\right)}^2+{\left(y\left(t\right)\right)}^2}\\ =\sqrt{{\left(20t-80\right)}^2+{\left(30t-100\right)}^2}\mathrm{.............}\left(1\right)\end{array}$

Evaluating the rate of change of the distance between them by differentiating:

  $\begin{array}{c}\frac{d}{dt}r\left(t\right)=\frac{d}{dt}\sqrt{{\left(20t-80\right)}^2+{\left(30t-100\right)}^2}\\ =\frac{1}{2\sqrt{{\left(20t-80\right)}^2+{\left(30t-100\right)}^2}}\frac{d}{dt}\left[\left(20t-{80}^2\right)+\left(30t-100\right)\right]\\ =\frac{\left[2\left(20t-80\right)\frac{d}{dt}\left(20t-80\right)+2\left(30t-100\right)\frac{d}{dt}\left(30t-100\right)\right]}{2\sqrt{{\left(20t-80\right)}^2+{\left(30t-100\right)}^2}}\\ =\frac{20\left(20t-80\right)+30\left(30t-100\right)}{2\sqrt{{\left(20t-80\right)}^2+{\left(30t-100\right)}^2}}\end{array}$

Therefore, the rate of change of the distance between them at t=0 is calculated as follows:

  $\begin{array}{c}\frac{dr\left(0\right)}{dt}=\frac{20\left(0-80\right)+30\left(0-100\right)}{\sqrt{{\left(0-80\right)}^2+{\left(0-100\right)}^2}}\\ =\frac{20\left(-80\right)+30\left(-100\right)}{\sqrt{{\left(-80\right)}^2+{\left(-100\right)}^2}}\\ =-35.91\text{m/s}\end{array}$

The rate of change of the distance between them at t=2 is calculated as follows

  $\begin{array}{c}\frac{dr\left(2\right)}{dt}=\frac{20\left(40-80\right)+30\left(60-100\right)}{\sqrt{{\left(40-80\right)}^2+{\left(60-100\right)}^2}}\\ =\frac{20\left(40-80\right)+30\left(60-100\right)}{\sqrt{{\left(-40\right)}^2+{\left(-40\right)}^2}}\\ =-35.35\text{m/s}\end{array}$

The rate of change of the distance between them at t=3 is calculated as follows

  $\begin{array}{c}\frac{dr\left(3\right)}{dt}=\frac{20\left(60-80\right)+30\left(90-100\right)}{\sqrt{{\left(60-80\right)}^2+{\left(90-100\right)}^2}}\\ =\frac{20\left(-20\right)+30\left(-10\right)}{\sqrt{{\left(-20\right)}^2+{\left(-10\right)}^2}}\\ =-31.30\text{m/s}\end{array}$

The rate of change of the distance between them at t=4 is as follows:

  $\begin{array}{c}\frac{dr\left(4\right)}{dt}=\frac{20\left(80-80\right)+30\left(120-100\right)}{\sqrt{{\left(80-80\right)}^2+{\left(120-100\right)}^2}}\\ =\frac{20\left(0\right)+30\left(20\right)}{\sqrt{{\left(0\right)}^2+{\left(20\right)}^2}}\\ =30\text{m/s}\end{array}$

The minimum distance between the animals is calculated by equating the derivative to zero.

  $\begin{array}{l}\frac{20\left(20t-80\right)+30\left(30t-100\right)}{2\sqrt{{\left(20t-80\right)}^2+{\left(30t-100\right)}^2}}=0\\ 20\left(20t-80\right)+30\left(30t-100\right)=0\\ 200\left(2t-8\right)+300\left(3t-10\right)=0\end{array}$

It gives:

  $\begin{array}{l}\frac{\left(2t-8\right)}{\left(3t-10\right)}=-\frac{300}{200}\\ \frac{\left(2t-8\right)}{\left(3t-10\right)}=-\frac{3}{2}\\ 4t-16=-9t+30\\ t=\frac{46}{13}\end{array}$

The occurrence of minimum distance between them happens after cheetah has reacted the watering hole at t=3.33sec

But this happens before gazelle has reacted the watering hole at t=4sec.