Question

A balloon is rising vertically above a​ level, straight road at a constant rate of 4 ft/sec. Just when the balloon is 80 ft above the​ ground, a bicycle moving at a constant rate of 13 ft/sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 6 seconds​ later?

Answer 1

Answer:

The distance s(t) is increasing at 11 ft/sec

Step-by-step explanation:

Since the balloon is moving vertically and the bicycle is moving horizontally, the distance between them is the hypotenuse of the triangle that is shown in the diagram attached at the bottom of this answer. By Pythagoras:

s(t)²=a(t)²+b(t)²

being a(t) the speed of the balloon at any given time t and b(t) the speed of the bicycle at any given time t.

$s(t)=\sqrt{(80+4t)^2+(13t)^2}=\sqrt{185 t^2 + 640 t + 6400}$

To know how fast this distance is changing we need to find the derivative s'(t):

$s'(t)=\frac{37t+64}{\sqrt{\frac{37t^2}{5}+\frac{128t}{5}+256}}$

Evaluating it at t = 6 sec we get:

$s'(6)=\frac{37(6)+64}{\sqrt{\frac{37(6)^2}{5}+\frac{128(6)}{5}+256}}=11ft/sec$