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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 3 - DerivativesProblema 6a
Capitolo 3, Problema 6a

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.

Guida verificata passo dopo passo
1
Start by recalling the formula for the area of a rectangle, which is the product of its length and width: A = length × width.
According to the problem, the length of the rectangle is twice the width. Therefore, express the length in terms of the width: length = 2w.
Substitute the expression for the length into the area formula. This gives you A = 2w × w.
Simplify the equation to find the relationship between the area A and the width w: A = 2w^2.
This equation, A = 2w^2, now relates the area of the rectangle to its width, considering the given condition that the length is twice the width.

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Area of a Rectangle

The area A of a rectangle is calculated using the formula A = length × width. In this case, since the length is twice the width (l = 2w), the area can be expressed as A = 2w × w = 2w². Understanding this relationship is crucial for deriving the equation that relates area to width.
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Relationship Between Variables

In calculus, understanding how different variables relate to each other is essential. Here, the area A is a function of the width w, which means any change in w will affect A. This relationship is foundational for analyzing how the area changes as the width varies.
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Differentiation

Differentiation is a key concept in calculus that deals with the rate of change of a function. If the problem requires finding how the area changes with respect to time, we would differentiate the area function A(w) with respect to w and then apply the chain rule to relate it to time t, allowing us to analyze the dynamics of the rectangle's area.
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