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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 12a

Shrinking square The sides of a square decrease in length at a rate of 1 m/s.
a. At what rate is the area of the square changing when the sides are 5 m long?

Verified step by step guidance
1
Identify the relationship between the side length of the square and its area. The area \( A \) of a square with side length \( s \) is given by \( A = s^2 \).
Recognize that the problem involves rates of change, which suggests using derivatives. We need to find \( \frac{dA}{dt} \), the rate of change of the area with respect to time.
Apply the chain rule to differentiate the area with respect to time: \( \frac{dA}{dt} = \frac{d}{dt}(s^2) = 2s \frac{ds}{dt} \).
Substitute the given values into the differentiated equation. We know \( \frac{ds}{dt} = -1 \) m/s (since the side is decreasing) and \( s = 5 \) m.
Calculate \( \frac{dA}{dt} \) using the substituted values: \( \frac{dA}{dt} = 2 \times 5 \times (-1) \). This will give the rate at which the area is changing when the sides are 5 m long.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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 area of the square changes as the length of its sides decreases. This requires applying the chain rule from calculus to relate the rates of change of the side length and the area.
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Area of a Square

The area of a square is calculated using the formula A = s², where s is the length of a side. As the side length changes, the area will also change. Understanding this relationship is crucial for determining how the area is affected by the rate at which the side length is decreasing.
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Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. In this context, it allows us to express the rate of change of the area with respect to time by relating it to the rate of change of the side length. This is essential for solving the problem and finding the rate at which the area is changing.
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