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Expert-verified Found in: Page 133 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # Two cars start moving from the same point. One travels south at $$60$$ mi/h and the other travels west at $$25$$mi/h. At what rate is the distance between the cars increasing two hours later?

Distance between the two cars after two hours is changing at a rate of $$65$$miles/hr.

See the step by step solution

## Step 1: Given Information.

The given speed of one car is $$60$$ mi/h towards south and the given speed of another car is $$25$$ mi/h towards west. Rate of increasing of distance two hours later between the cars is to be found out.

## Step 2: Definition of Derivative

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value)..

## Step 3: Find rate of increasing distance

Let $$x =$$distance travelled by the car towards west after time $$t$$

$$y =$$distance travelled by the car towards south after time $$t$$

$$s =$$distance travelled by the car towards south after time $$t$$ From the figure, use Pythagoras theorem

$${x^2} + {y^2} = {s^2}$$

Differentiate with respect to $$t$$

$$2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}} = 2s\frac{{ds}}{{dt}} \cdots \left( 1 \right)$$

Now, it is given in the question

$$\frac{{dx}}{{dt}} = 25$$

$$\frac{{dy}}{{dt}} = 60$$

Since the speed of two cars is constant, so distance travelled by two cars in two hours

$$\begin{array}{c}x = 25\left( 2 \right)\\ = 50\end{array}$$

$$\begin{array}{c}y = 60\left( 2 \right)\\ = 120\end{array}$$

Put the values in $$\left( 1 \right)$$

$$\begin{array}{l}2\left( {50} \right)\left( {25} \right) + 2\left( {120} \right)\left( {60} \right) = 2\left( {130} \right)\frac{{ds}}{{dt}}\\\left( {260} \right)\frac{{ds}}{{dt}} = 16900\\\frac{{ds}}{{dt}} = 65\end{array}$$

Therefore, distance between the two cars after two hours is changing at a rate of $$65$$miles/hr. ### Want to see more solutions like these? 