Page 83 - Calculus Demystified

CHAPTER 2
Foundations of Calculus
70

x

Fig. 2.9

SOLUTION
2
Let ψ(t) = 75 − 10t + t. Of course the rate of loss of air is given by ψ (1).

We therefore calculate
ψ(1 + h) − ψ(1)

ψ (1) = lim
h→0 h
2 2
[75 − 10(1 + h) + (1 + h)]−[75 − 10 · 1 + 1]
= lim
h→0 h
2
[75 − (10 + 20h + 10h ) + (1 + h)]−[66]
= lim
h→0 h
−19h − 10h 2
= lim
h→0 h
= lim −19 − 10h
h→0
=−19.

In conclusion, the rate of air loss in the balloon at time t = 1is ψ (1) =
3
−19 ft /sec. Observe that the negative sign in this answer indicates that the
change is negative, i.e., that the quantity is decreasing.
2
You Try It: The amount of water in a leaky tank is given by W(t) = 50 − 5t + t
gallons. What is the rate of leakage of the water at time t = 2?

Math Note: We have noted that the derivative may be used to describe a rate of
change and also to denote the slope of the tangent line to a graph. These are really
two different manifestations of the same thing, for a slope is the rate of change of
rise with respect to run (see Section 1.4 on the slope of a line).