Chapter 3.8 Exercise 33

Back to MAT 194F

Chapter 3.8 Exercise 33

Problem

A balloon is released 500 feet away from an observer. If the balloon rises vertically at the rate of 100 feet per minute and at the same time the wind is carrying it horizontally away from the observer at the rate of 75 feet per minute, at what rate is the angle of inclination of the observer's line of sight changing 6 minutes after the balloon has been released?

Solution

First name our variables:

theta := angle in radians of inclination of observer's line of sight
h := height in feet of balloon
x := horizontal distance in feet of balloon from observer
t := elapsed time in minutes, starting when balloon is released

Then express the mathematical relations among them:

tan theta = h / x
x = 500 + 75 * t
h = 100 * t

Finally calculate the desired value:

(sec**2 theta) d theta / dt
= (x dh/dt - h dx/dt) / x**2
= (100 x - 75 h) / x**2.

If t = 6, then x = 950, h = 600, tan theta = 600/950 and sec**2 theta = 1 + tan**2 theta = 505/361.

The angle in question is therefore increasing at a rate of (100*950 - 75*600)/(950**2 * 505/361) = 4/101 radians per second.