/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Approximate the change in the vo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 33

Approximate the change in the volume of a right circular cylinder of fixed radius \(r=20 \mathrm{cm}\) when its height decreases from \(h=12 \mathrm{cm}\) to \(h=11.9 \mathrm{cm}\left(V(h)=\pi r^{2} h\right)\)

Short Answer

Expert verified
Answer: The approximate change in volume is a decrease of 96.02 cm鲁.

Step by step solution

01

List Given Information

We are given, - The radius of the cylinder (r) is fixed and is equal to 20 cm. - The initial height (h1) is 12 cm. - The final height (h2) is 11.9 cm. - The volume of the cylinder is given by \(V(h) = \pi r^2 h\).
02

Find the Initial Volume

First, we will find the initial volume of the cylinder using the volume formula and the initial height. \(V(h_1) = \pi r^2 h_1 = \pi(20)^2 (12)\) Calculate the value to get the initial volume, \(V(h_1) = 15067.964 \text{ cm}^3\) (approx.)
03

Find the Final Volume

Next, we will find the final volume of the cylinder using the volume formula and the final height. \(V(h_2) = \pi r^2 h_2 = \pi(20)^2 (11.9)\) Calculate the value to get the final volume, \(V(h_2) = 14971.944 \text{ cm}^3\) (approx.)
04

Compute the Change in Volume

Now, we will calculate the change in volume as the difference between the final volume and the initial volume. \(\Delta V = V(h_2) - V(h_1) = 14971.944 - 15067.964\) Computing the difference to find the change in volume, \(\Delta V = - 96.02 \text{ cm}^3\) (approx.)
05

Interpret the Result

The negative sign indicates a decrease in the volume, as expected. Therefore, the approximate change in the volume of the cylinder when its height decreases from 12cm to 11.9cm is a decrease of 96.02 cm鲁 (approximately).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the following indefinite integrals. Check your work by differentiation. $$\int\left(\csc ^{2} \theta+1\right) d \theta$$

Evaluate the following limits in terms of the parameters a and b, which are positive real numbers. In each case, graph the function for specific values of the parameters to check your results. $$\lim _{x \rightarrow 0} \frac{a^{x}-b^{x}}{x}$$

Suppose \(f(x)=\sqrt[3]{x}\) is to be approximated near \(x=8 .\) Find the linear approximation to \(f\) at 8 Then complete the following table, showing the errors in various approximations. Use a calculator to obtain the exact values. The percent error is \(100 \cdot |\) approximation \(-\) exact \(|/|\) exact \(| .\) Comment on the behavior of the errors as \(x\) approaches 8 .

Sketch the graph of a function that is continuous on \((-\infty, \infty)\) and satisfies the following sets of conditions. $$f(x)>f^{\prime}(x)>0 \text { for all } x ; f^{\prime \prime}(1)=0$$

Verify the following indefinite integrals by differentiation. $$\int \frac{x}{\left(x^{2}-1\right)^{2}} d x=-\frac{1}{2\left(x^{2}-1\right)}+C$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.