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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 5b
Chapter 3, Problem 5b

A rectangular swimming pool 10 ft wide by 20 ft long and of uniform depth is being filled with water.
b. At what rate is the volume of the water increasing if the water level is rising at 1/4ft/min.

Verified step by step guidance
1
First, identify the dimensions of the swimming pool: it is 10 ft wide and 20 ft long. The depth of the water is changing as the pool is being filled.
The volume of the water in the pool can be expressed as a function of the water depth, h, using the formula for the volume of a rectangular prism: V = length × width × height. In this case, V = 20 × 10 × h.
Since the water level is rising at a rate of 1/4 ft/min, we need to find the rate at which the volume is increasing. This involves differentiating the volume with respect to time, t, using the chain rule.
Differentiate the volume function V = 200h with respect to time t. This gives us dV/dt = 200(dh/dt), where dh/dt is the rate at which the water level is rising.
Substitute the given rate of change of the water level, dh/dt = 1/4 ft/min, into the differentiated equation to find dV/dt, the rate at which the volume is increasing.

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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 dimensions of the pool are given, and the height corresponds to the water level. Understanding this formula is essential for determining how the volume changes as the water level rises.
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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 the volume of water in the pool changes as the height of the water increases. This requires applying the concept of derivatives to relate the rates of change of volume and height.
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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 volume of water is increasing as the height of the water rises at a specified rate.
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Related Practice
Textbook Question

At all times, the length of a rectangle is twice the width w of the rectangleas the area of the rectangle changes with respect to time t.

a. Find an equation relating A to w.

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Textbook Question

The table gives the position s(t)of an object moving along a line at time t, over a two-second interval. Find the average velocity of the object over the following intervals. <IMAGE>


c. [0,1][0, 1]

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Textbook Question

Derivatives using tables Let h(x)=f(g(x))h(x)=f(g(x)) and p(x)=g(f(x))p(x)=g(f(x)). Use the table to compute the following derivatives.

<IMAGE>

e. h′(5)h^{\(\prime\)}\(\left\)(5\(\right\))

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Textbook Question

The table gives the position s(t)of an object moving along a line at time t, over a two-second interval. Find the average velocity of the object over the following intervals. <IMAGE>


a. [0,2][0, 2]

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Textbook Question

Use differentiation to verify each equation.

d/dx(x / √1−x²) = 1 / (1−x²)^3/2.

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Textbook Question

Let f(x) = sin x. What is the value of f′(π)?

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