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Q15E

Expert-verifiedFound in: Page 133

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.

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.

**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)..**

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.

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