Applications of Derivatives: Related Rates
EXAMPLE 2: A balloon is launched straight up at a velocity of 8 ft/sec. An observer stands 150 ft away from the launch point on the ground. What is the rate of change of the distance from the balloon to the observer when the balloon reaches a height of 50 ft?
Solution: For this one, we need a sketch first. Draw and label the key distances.
NEED SKETCH
What are the known and unknown rates?
dh/dt = 8 ft/sec ds/dt = ? when h = 50 ft
Use the sketch to find an equation that relates the two variables, h and s.
1502 + h2 = s2
Now take the derivative with respect to time and solve.
0 + 2h dh/dt = 2s ds/dt
ds/dt = h/s dh/dt
We need to determine s when h = 50 ft.
1502 + (50)2 = s2 s = 158.1 ft
Then ds/dt = 50/158.1 (8 ft/sec) = 2.53 ft/sec
