Question

A balloon is 200 ft. off the ground and rising vertically at the constant rate of 15 ft/sec. An automobile passes beneath it traveling along a straight road at the constant rate of 45 mph or 66 ft/sec. How fast is the distance between them changing 1 sec. later?

Answer 1

After 1 second has passed, the balloon will be 215 ft above the ground, and the automobile will be 66 ft away from its initial position directly below the balloon, forming a right triangle whose hypotenuse has length $\sqrt{215^2+66^2}=\sqrt{50,581}$ ft.

The distance between the balloon and automobile $d$ is related to the height of the balloon $b$ and the distance $a$ the automobile has moved away from its starting position by the Pythagorean formula,

$d^2=a^2+b^2$

Differentiate both sides with respect to time $t$:

$2d\dfrac{\mathrm dd}{\mathrm dt}=2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}$

Plug in everything you know and solve for the rate you want:

$2(\sqrt{50,581}\,\mathrm{ft})\dfrac{\mathrm dd}{\mathrm dt}=2(66\,\mathrm{ft})\left(66\dfrac{\rm ft}{\rm s}\right)+2(215\,\mathrm{ft})\left(15\dfrac{\rm ft}{\rm s}\right)$

$\implies\dfrac{\mathrm dd}{\mathrm dt}\approx33.71\dfrac{\rm ft}{\rm s}$