A balloon is at a height of 50 m and is rising at a constant rate of 5 m/s. A cyclist passes beneath it, traveling in a straight line at a constant speed of 10 m/s. How fast is the distance between the cyclist and the balloon increasing 2 seconds later?
Let height of balloon at any instant of time $= h$ meter
Rate of change of height of balloon $= \dfrac{dh}{dt} = 5 \; m/s$
Initial height of balloon $= 50$ meter
$\therefore$ Height of balloon after 2 sec $= h = 50 \; m + 5 \; m/s \times 2 \; s = 60 \; m$
Let distance covered by the cyclist at any instant of time $= x$ meter
Speed of cyclist $= \dfrac{dx}{dt} = 10 \; m/s$
$\therefore$ Distance covered by the cyclist after 2 seconds $= x = 10 \; m/s \times 2 \; s = 20 \; m$
Let the distance between the cyclist and the balloon $= L$ meter
Then, $L = \sqrt{h^2 + x^2}$ $\;\; \cdots$ (1)
Differentiating equation (1) with respect to time t gives
$\dfrac{dL}{dt} = \dfrac{1}{2 \sqrt{h^2 + x^2}} \times \left(2h \dfrac{dh}{dt}+ 2x \dfrac{dx}{dt}\right)$
i.e. $\dfrac{dL}{dt} = \dfrac{1}{\sqrt{h^2 + x^2}} \times \left(h \dfrac{dh}{dt} + x \dfrac{dx}{dt}\right)$ $\;\; \cdots$ (2)
After 2 seconds, $x = 20 \; m$, $h = 60 \; m$, $\dfrac{dx}{dt} = 10 \; m/s$ and $\dfrac{dh}{dt} = 5 \; m/s$
$\therefore$ $\dfrac{dL}{dt} = \dfrac{1}{\sqrt{60^2 + 20^2}} \times \left(60 \times 5 + 20 \times 10\right) = \dfrac{500}{\sqrt{4000}} = \dfrac{500}{20\sqrt{10}} = \dfrac{5\sqrt{10}}{2}$
$\therefore$ After 2 seconds, the distance between the cyclist and the balloon is increasing at the rate of $\dfrac{5 \sqrt{10}}{2} \; m/s$.