Table of contents
- 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 Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
4. Applications of Derivatives
Related Rates
Problem 3.11.5c
Textbook Question
A rectangular swimming pool 10 ft wide by 20 ft long and of uniform depth is being filled with water.
c. At what rate is the water level rising if the pool is filled at a rate of 10ft³/min?

1
First, identify the variables involved in the problem. Let h be the height of the water level in the pool, which is changing over time.
The volume V of the water in the pool can be expressed as V = length × width × height, or V = 20 × 10 × h. This simplifies to V = 200h.
Since the pool is being filled at a rate of 10 ft³/min, this is the rate of change of the volume with respect to time, denoted as dV/dt = 10 ft³/min.
To find the rate at which the water level is rising, we need to find dh/dt. Use the relationship between the rates: dV/dt = 200 × dh/dt.
Solve for dh/dt by dividing both sides of the equation by 200: dh/dt = (dV/dt) / 200. Substitute dV/dt = 10 ft³/min into the equation to find dh/dt.

This video solution was recommended by our tutors as helpful for the problem above
Video duration:
3mPlay a video:
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Volume of a Rectangular Prism
The volume of a rectangular prism, such as a swimming pool, is calculated using the formula V = length × width × height. In this case, the pool's dimensions are given, and understanding this formula is essential to relate the volume of water being added to the change in water level.
Recommended video:
Example 5: Packaging Design
Related Rates
Related rates involve finding the rate at which one quantity changes in relation to another. In this problem, we need to determine how fast the water level (height) is rising as the volume of water (10 ft³/min) is added, which requires applying the concept of derivatives to relate the rates of change.
Recommended video:
Intro To Related Rates
Differentiation
Differentiation is a fundamental concept in calculus that deals with finding the rate of change of a function. In this context, we will differentiate the volume formula with respect to time to find the rate at which the height of the water is increasing as the pool is filled.
Recommended video:
Finding Differentials
Related Videos
Related Practice