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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Briggs 3rd EditionCh. 3 - Derivatives문제 5b
3장, 문제 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.

검증된 단계별 안내
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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
08:29
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 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.
추천 영상:
가이드 코스
04:16
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 volume of water is increasing as the height of the water rises at a specified rate.
추천 영상:
가이드 코스
05:53
Finding Differentials
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