Question

A balloon is rising vertically above a​ level, straight road at a constant rate of 4 ft divided by sec4 ft/sec. Just when the balloon is 7272 ft above the​ ground, a bicycle moving at a constant rate of 1212 ft divided by secft/sec passes under it. How fast is the distance s (t )s(t) between the bicycle and balloon increasing 66 seconds​ later?

Answer 1

12.27 ft/s

Step-by-step explanation:

At 72 ft above the ground, the balloons height increases at a rate of 4ft/s. For 66s, vertical distance moved, y = 4ft/s × 66 s = 264 ft. When the balloon is at 72 ft above the ground, just below it, the bicycle which moves at a rate of 12 ft/s in 66 s, covers a horizontal distance, x = 12ft/s 66 = 792 ft.

The distance between the bicycle and the balloon 66 s later is given by

s = √(x² + (y + 72)²) = √(792² + (264 + 72)²) = √(792² + 336²) = √740160 ft = 860.33 ft

From calculus

The rate of change of the distance between the balloon and bicycle s is obtained by differentiating s with respect to t. So,

ds/dt = (1/s)(xdx/dt + ydy/dt)

dx/dt = 12 ft/s, x = 792 ft, dy/dt = 4 ft/s, y = 264 ft, s = 860.33. These are the values of the variables at t = 66 s.

So, substituting these values into ds/dt, we have

ds/dt = (1/860.33)(792 ft × 12 ft/s + 264 ft × 4ft/s) = (1/860.33)(9504 + 1056) = 10560/860.33 = 12.27 ft/s