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>
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Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 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.
Verified step by step guidance1
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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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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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Related Practice
Textbook Question
Find the derivative the following ways:
Using the Product Rule or the Quotient Rule. Simplify your result.
f(x) = x(x-1)
Textbook Question
The volume V of a sphere of radius r changes over time t.
b. At what rate is the volume changing if the radius increases at 2 in/min when when the radius is 4 inches?
Textbook Question
Use differentiation to verify each equation.
d/dx(x / √1−x²) = 1 / (1−x²)^3/2.
Textbook Question
An equation of the line tangent to the graph of f at the point (2,7) is y = 4x−1. Find f(2) and f′(2).
Textbook Question
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.