# 10 - A boy on a bike

**Problem 10**

A boy on a bike rides north 5 mi, then turns east (Fig. 47). If he rides 10 mi/hr, at what rate does his distance to the starting point S changing 2 hour after he left that point?

**Solution 10**

For 5 miles:

$5 = 10t$

$5 = 10t$

$t = 0.5 ~ \text{hr}$

$d^2 = 5^2 + 10^2(t - 0.5)^2$

$d = \sqrt{25 + 100(t - 0.5)^2}$

$\dfrac{dd}{dt} = \dfrac{200(t - 0.5)}{2\sqrt{25 + 100(t - 0.5)^2}}$

$\dfrac{dd}{dt} = \dfrac{100(t - 0.5)}{\sqrt{25 + 100(t - 0.5)^2}}$

when t = 2 hrs

$\dfrac{dd}{dt} = \dfrac{100(2 - 0.5)}{\sqrt{25 + 100(2 - 0.5)^2}}$

$\dfrac{dd}{dt} = \dfrac{150}{\sqrt{250}} = \dfrac{150}{5\sqrt{10}} \times \dfrac{\sqrt{10}}{\sqrt{10}}$

$\dfrac{dd}{dt} = 3\sqrt{10} \, \text{ mi/hr}$ *answer*

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