- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
4. Applications of Derivatives
Differentials
Problem 25
Textbook Question
As a result of a heavy rain, the volume of water in a reservoir increased by 1400 acre-ft in 24 hours. Show that at some instant during that period the reservoir’s volume was increasing at a rate in excess of 225,000 gal/min. (An acre-foot is 43,560 ft³, the volume that would cover 1 acre to the depth of 1 ft. A cubic foot holds 7.48 gal.)

1
First, convert the total increase in volume from acre-feet to cubic feet. Since 1 acre-foot is 43,560 cubic feet, multiply 1400 acre-feet by 43,560 to get the total volume increase in cubic feet.
Next, convert the total volume increase from cubic feet to gallons. Since 1 cubic foot holds 7.48 gallons, multiply the volume in cubic feet by 7.48 to get the total volume increase in gallons.
Now, calculate the average rate of increase in gallons per minute over the 24-hour period. There are 24 hours in a day and 60 minutes in an hour, so multiply 24 by 60 to find the total number of minutes in the period.
Divide the total volume increase in gallons by the total number of minutes to find the average rate of increase in gallons per minute.
Apply the Mean Value Theorem for integrals, which states that if a function is continuous over a closed interval, then there exists at least one point in the interval where the instantaneous rate of change (derivative) equals the average rate of change. Since the average rate is less than 225,000 gal/min, there must be an instant where the rate exceeds 225,000 gal/min.
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