Textbook Question
Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.
ƒ(x) = x²/₃(4 -x²) on -2,2
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Ch. 4 - Applications of the Derivative
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 4, Problem 4.6.57
Approximating changes
Approximate the change in the volume of a right circular cylinder of fixed radius r = 20 cm when its height decreases from h = 12 to h = 11.9 cm (V(h) = πr²h).
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Identify the formula for the volume of a right circular cylinder: \( V(h) = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height.
Since the radius \( r \) is fixed at 20 cm, substitute \( r = 20 \) into the volume formula to get \( V(h) = \pi (20)^2 h = 400\pi h \).
To approximate the change in volume as the height changes from 12 cm to 11.9 cm, use the concept of differentials. The differential \( dV \) is given by \( dV = \frac{dV}{dh} \cdot dh \).
Calculate the derivative \( \frac{dV}{dh} \) of the volume function with respect to \( h \). Since \( V(h) = 400\pi h \), the derivative is \( \frac{dV}{dh} = 400\pi \).
Substitute \( dh = 11.9 - 12 = -0.1 \) cm into the differential formula \( dV = 400\pi \cdot (-0.1) \) to find the approximate change in volume.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Differential Calculus
Differential calculus focuses on the concept of the derivative, which measures how a function changes as its input changes. In this context, it helps us understand how small changes in the height of the cylinder affect its volume. The derivative of the volume function with respect to height will provide the rate of change of volume as height varies.
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Volume of a Cylinder
The volume of a right circular cylinder is given by the formula V(h) = πr²h, where r is the radius and h is the height. This formula is essential for calculating the volume based on the height and radius. In this problem, we are specifically interested in how the volume changes when the height decreases slightly.
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Linear Approximation
Linear approximation is a method used to estimate the value of a function near a given point using the tangent line at that point. By applying this concept, we can approximate the change in volume when the height of the cylinder changes from 12 cm to 11.9 cm. This involves using the derivative to find the slope of the tangent line at the height of 12 cm.
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