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Chapter 2: Problem 30

Ball's changing volume What is the rate of change of the volume of a ball \(\left(V=(4 / 3) \pi r^{3}\right)\) with respect to the radius when the radius is \(r=2 ?\)

Short Answer

Expert verified
The rate of change of the volume at \( r = 2 \) is \( 16 \pi \).

Step by step solution

01

Identify the Given Variables and Formula

We are given the formula for the volume of a ball: \( V = \frac{4}{3} \pi r^3 \). We need to find the rate of change of volume with respect to the radius \( r \) at \( r = 2 \).
02

Derive the Volume Formula

To find the rate of change of volume with respect to the radius, take the derivative of \( V \) with respect to \( r \). This will give us \( \frac{dV}{dr} \).
03

Differentiate the Volume with Respect to Radius

Using the formula \( V = \frac{4}{3} \pi r^3 \), apply the power rule of differentiation. The derivative is \( \frac{dV}{dr} = 4 \pi r^2 \).
04

Substitute the Given Radius into the Derivative

Now that we have the derivative \( \frac{dV}{dr} = 4 \pi r^2 \), substitute \( r = 2 \) into this formula: \[ \frac{dV}{dr} \bigg|_{r=2} = 4 \pi (2)^2 = 4 \pi \times 4 = 16 \pi \].
05

Calculate the Rate of Change

Compute the value: \( 16 \pi \). This represents the rate of change of the volume of the ball when the radius is 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Rate of Change
The concept of 'rate of change' is crucial in understanding how one quantity varies with respect to another. In calculus, this is often represented by derivatives.
In the context of our exercise, the rate of change is referring to how the volume of a sphere changes as its radius changes. When we say "rate of change", we mean if we slightly increase the radius, how much does the volume increase by? This is represented as \( \frac{dV}{dr} \).
  • It helps quantify relationships between variables, such as volume and radius in this exercise.
  • Applications include physics, engineering, and economics to model dynamic systems.
In our case, finding the rate of change involves differentiating the volume formula, thereby giving us insight into how sensitive the volume is to changes in the radius.
Volume Differentiation Explained
Differentiation is a technique used in calculus to find the rate at which a quantity changes. Volume differentiation specifically involves finding how the volume changes concerning another variable, like the radius of a sphere in this exercise.
To perform volume differentiation, we use the derivative of the volume formula. Here, the volume \( V \) of a sphere is given by the formula \( V = \frac{4}{3} \pi r^3 \). Applying differentiation:
  • The power rule states that when differentiating \( r^n \), the derivative is \( nr^{n-1} \).
  • Applying this to \( \frac{4}{3} \pi r^3 \), the derivative becomes \( 4 \pi r^2 \).
By finding this derivative, we calculate how quickly the volume changes as the radius is slightly changed, essentially capturing the dynamic nature of the geometry involved.
Sphere Radius Derivative at a Specific Point
Now, let's delve into calculating the derivative of the sphere's volume at a specific radius, say \( r = 2 \). After finding that \( \frac{dV}{dr} = 4 \pi r^2 \), we need to determine what happens when the radius is exactly 2.
By substituting \( r = 2 \) into the derived formula:
  • We compute \( 4 \pi (2)^2 \).
  • This simplifies to \( 4 \pi \times 4 = 16 \pi \).
Thus, \( 16 \pi \) represents how quickly the volume of the sphere is expanding per unit change in radius when the radius is 2. It's a precise calculation showing the speed of change at that point in the sphere's size, which can be valuable in applications involving physical changes like inflating a balloon or measuring planetary growth.

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Most popular questions from this chapter

Graph the rational functions in Exercises \(27-38 .\) Include the graphs and equations of the asymptotes and dominant terms. $$ y=\frac{x^{2}+1}{x-1} $$

In Exercises \(11-18,\) find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$ h(t)=t^{3}, \quad(2,8) $$

Graphing Secant and Tangent Lines Use a CAS to perform the following steps for the functions in Exercises \(45-48 .\) a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\) b. Holding \(x_{0}\) fixed, the difference quotient $$ q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h} $$ at \(x_{0}\) becomes a function of the step size \(h .\) Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\) d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=3,2\) and \(1 .\) Graph them together with \(f\) and the tangent line over the interval in part (a). $$ f(x)=x^{3}+2 x, \quad x_{0}=0 $$

In Exercises \(39-42,\) sketch the graph of a function \(y=f(x)\) that satisfies the given conditions. No formulas are required- just label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer section.) $$ \begin{array}{l}{f(0)=0, \lim _{x \rightarrow \pm \infty} f(x)=0, \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow-1^{+}} f(x)=\infty} \\ {\lim _{x \rightarrow 1^{+}} f(x)=-\infty, \text { and } \lim _{x \rightarrow-1^{-}} f(x)=-\infty}\end{array} $$

In Exercises \(41-44,\) graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0 .\) If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$ f(x)=\frac{10^{|x|}-1}{x} $$
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