Chapter 4.1, Problem 15E

To determine

To find: The rate at which the diffrence between two cars increasing after 2 hours.

Answer to Problem 15E

The distance betwqeen the car is increasing with the rate of $65$ miles/hr.

Explanation of Solution

Given information:

Given that the two cars moving from same point from which one is going at $60$ mi/h to south and other at $25$ mi/h to west direction.

Calculation:

  

Figure $\left(1\right)$

Let both the cars start moving from the point P

After $t$ hours the first car is at point A and second car is at B

Speed of first car is $60$ mi/hr, then $\frac{dx}{dt}=60$ mi/h

Speed of first car is $25$ mi/hr, then $\frac{dy}{dt}=25$ mi/h

After 2 hours the distannce covered by the first car is $x=120$ mi

And the distance covered by the second car is $y=50$ mi

Let the distance between the cars be $z$ at the time $t$

Then, by pythagoreas theorem

  $z^2=x^2+y^2$

When $x=120$ miles and $y=50$ miles

Then $\begin{array}{l}{z}^2={\left(120\right)}^2+{\left(50\right)}^2\\ z^2=16900\\ z=130\end{array}$

Differentiating $\left(1\right)$ with respect to $t$ implicity

  $\begin{array}{c}2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\\ \frac{dz}{dt}=\frac{1}{z}\left[x\frac{dx}{dt}+y\frac{dy}{dt}\right]\end{array}$

Putting the values if $x,y,z,\frac{dx}{dt}+\frac{dy}{dt}$ , the rate of change vof the distance between car is

  $\begin{array}{c}\frac{dz}{dt}=\frac{1}{130}\left[120\left(60\right)+50\left(25\right)\right]\\ =\frac{1}{130}\left[120\left(60\right)+50\left(25\right)\right]\\ \frac{dz}{dt}=65\end{array}$

Therefore the distance betwqeen the car is increasing with the rate of $65$ miles/hr.