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Chapter 6: Problem 5

Air is being pumped into a spherical balloon at the rate of \(1000 \mathrm{~cm}^{3} / \mathrm{min}\). At what rate is the radius of the balloon increasing when the volume is \(3000 \mathrm{~cm}^{3} ?\) Note: \(V(t)=\frac{4}{3} \pi r^{3}(t)\).

Short Answer

Expert verified
The radius increases at a rate of approximately 0.092 cm/min when the volume is 3000 cm鲁.

Step by step solution

01

Identify Known Quantities and Formula

We are given the rate at which the volume is increasing, \( \frac{dV}{dt} = 1000 \, \mathrm{cm}^3/\mathrm{min} \), and need to find \( \frac{dr}{dt} \), the rate of change of the radius. We use the volume formula for a sphere: \( V = \frac{4}{3} \pi r^3 \).
02

Differentiate the Volume Formula

Differentiate both sides of the volume formula with respect to time \( t \): \( \frac{d}{dt} (V) = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) \). This gives \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \) using the chain rule for differentiation.
03

Substitute Known Values

We know \( \frac{dV}{dt} = 1000 \mathrm{~cm}^3/\mathrm{min} \) and want \( \frac{dr}{dt} \) when \( V = 3000 \mathrm{~cm}^3 \). First, calculate the radius \( r \) when \( V = 3000 \mathrm{~cm}^3 \) using the volume formula: \( 3000 = \frac{4}{3} \pi r^3 \). Solve this equation for \( r \).
04

Solve for the Radius

In order to find \( r \), rearrange the volume equation: \( r^3 = \frac{3 \cdot 3000}{4\pi} \). Calculate \( r = \left( \frac{9000}{4\pi} \right)^{1/3} \).
05

Calculate \( \frac{dr}{dt} \)

Substitute \( \frac{dV}{dt} = 1000 \mathrm{~cm}^3/\mathrm{min} \) and the found \( r \) into \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \) to solve for \( \frac{dr}{dt} \). Rearrange to find \( \frac{dr}{dt} = \frac{1000}{4 \pi r^2} \) and compute the value using the radius from Step 4.

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Key Concepts

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

Volume of a Sphere
Understanding the volume of a sphere is crucial when dealing with related rates problems involving spherical objects, like balloons. The mathematical formula used for calculating the volume of a sphere is given by:\[ V = \frac{4}{3} \pi r^3 \]where:
  • \(V\) is the volume of the sphere.
  • \(r\) is the radius of the sphere.
  • \(\pi\) is the constant Pi, approximately equal to 3.14159.
The volume directly depends on the cube of the radius, hence small changes in the radius will have a significant impact on the volume. This is why, when air is pumped into a balloon, even a slight increase in its radius results in a large increase in its volume. Recognizing this relationship is important when solving questions about how fast the radius is changing as air is added.
Chain Rule for Differentiation
The chain rule is a fundamental principle in calculus used to find the derivative of composite functions. It is especially useful in related rates problems where different rates of change need to be connected.In our context, when differentiating the volume of a sphere with respect to time to find how its radius changes over time, we use the chain rule:\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \frac{dr}{dt} \]Here's how it works:
  • \(r^3\) is a function of time \(r(t)\), so its derivative requires the chain rule.
  • First, differentiate \(r^3\) with respect to \(r\) (giving us \(3r^2\)), then multiply by the derivative of \(r\) with respect to \(t\) (\(\frac{dr}{dt}\)).
  • This gives the term \(3r^2 \cdot \frac{dr}{dt}\), which is then multiplied by \(\frac{4}{3} \pi\) to obtain the final formula \( 4 \pi r^2 \frac{dr}{dt} \).\
The chain rule links the rate of change of volume \(\frac{dV}{dt}\) to the rate of change of the radius \(\frac{dr}{dt}\), making it possible to solve for unknown rates if certain values are provided.
Rate of Change
In calculus, the rate of change is a key concept that describes how one quantity changes in relation to another. Understanding this helps in solving many applied problems, including related rates. Related rates problems often involve determining the speed at which one quantity is changing by using the rate at which another quantity is changing. For our problem:
  • The rate of volume change \(\frac{dV}{dt}\) is given as \(1000 \textrm{ cm}^3/\textrm{min}\).
  • We aim to find the rate of change of the radius \(\frac{dr}{dt}\) when the volume \(V\) is \(3000 \textrm{ cm}^3\).
Through the steps, once the radius is known at a particular volume, substituting it back into the differentiated formula connects these rates:\[\frac{dr}{dt} = \frac{1000}{4 \pi r^2}\]This formula shows how the rate of volume growth influences the rate at which the radius expands. This understanding is crucial in practical applications like inflating balloons, where managing size and pressure is necessary.

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Most popular questions from this chapter

A bird searches bushes in a field for insects. The total weight of insects found after \(t\) minutes of searching a single bush is given by \(w(t)=\frac{2 t}{t+4}\) grams. Draw a graph of \(w .\) From your graph, does it appear that a bird should search a single bush for more than 10 minutes? It takes the bird one minute to move from one bush to another. How long should the bird search each bush in order to harvest the most insects in an hour of feeding?

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An examination of 1000 people showed that 41 were carriers (heterozygotic) of the gene for cystic fibrosis. Let \(p\) be the proportion of all people who are carriers of cystic fibrosis. We can not say with certainty that \(p=41 / 1000\). For any number \(p\) in [0,1] , let \(L(p)\) be the likelihood of the event that 41 of 1000 people are carriers of cystic fibrosis given that the probability of being a carrier is \(p\). Then $$ L(p)=\left(\begin{array}{c} 1000 \\ 41 \end{array}\right) p^{41} \times(1-p)^{959} $$ where \(\left(\begin{array}{c}1000 \\ 41\end{array}\right)\) is a constant \(^{1}\) approximately equal to \(1.3 \times 10^{73}\). a. Compute \(L^{\prime}(p)\). b. Find the value \(\hat{p}\) of \(p\) for which \(L^{\prime}(p)=0\) and compute \(L(\hat{p})\). The value \(L(\hat{p})\) is the maximum value of \(L(p)\) and \(\hat{p}\) is called the maximum likelihood estimator of \(p\).

Differentiate a. \(\quad P(t)=t^{4}+e^{2 t}\) b. \(\quad P(t)=t^{4} e^{2 t}\) c. \(\quad P(t)=\frac{t^{4}}{e^{2 t}}\) d. \(P(t)=\left(e^{2 t}\right)^{4}\) e. \(P(t)=e^{2 t^{4}}\) f. \(P(t)=\left(t^{2}+1\right)^{4}(5 t+1)^{7}\) g. \(\quad P(t)=(\ln t)^{3}\) h. \(\quad P(t)=\left(e^{3 t} \ln 2 t\right)^{4}\) i. \(\quad P(t)=\frac{\ln t}{t}\) j. \(\quad P(t)=t \ln t-t\) k. \(P(t)=\frac{1}{\ln t}\) l. \(P(t)=e^{(\ln t)}\) m. \(P(t)=\frac{5 t^{2}-2 t-7}{t^{2}+1}\) n. \(P(t)=\frac{(t+2)^{2}}{t^{2}+2}\)

Exercise 6.2 .4 The Doppler effect. You are standing 100 meters south of a straight train track on which a train is traveling from west to east at the speed 30 meters per second. See Figure 6.2 .4 . Let \(y(t)\) be the distance from the train to you and \(|x(t)|\) be the distance from the train to the point on the track nearest you; \(x(t)\) is negative when the train is west of the point on the track nearest you. a. Write \(y(t)\) in terms of \(x(t)\). b. Find \(y^{\prime}(t)\) for a time \(t\) at which \(x(t)=-200\) c. Time is measured so that \(x(0)=0 .\) Write an equation for \(x(t)\). d. Write an equation for \(y^{\prime}(t)\) in terms of \(t\). e. The whistle from the train projects sound waves at frequency \(f\) cycles per second. The frequency, \(f_{L},\) of the sound reaching your ear is $$ f_{L}(t)=\frac{331.4}{331.4+y^{\prime}(t)} f \frac{\text { cycles }}{\text { second }} $$ \(331.4 \mathrm{~m} / \mathrm{s}\) is the speed of sound in air. Draw a graph of \(f_{L}(t)\) assuming \(f=500\). Derivation of Equation 6.18 for the Doppler effect. A sound of frequency \(f\) traveling in still air has wave length \((331.4 \mathrm{~m} / \mathrm{s}) /(f\) cycles \(/ \mathrm{s})=(331.4 / \mathrm{f}) \mathrm{m} /\) cycle. If the source of the sound is moving at a velocity \(v\) with respect to a listener, the wave length of the sound reaching the listener is \(((331.4+v) / f) \mathrm{m} /\) cycle. These waves travel at \(331.4 \mathrm{~m} / \mathrm{sec},\) and the frequency \(f_{L}\) of these waves reaching the listener is $$ f_{L}=\frac{331.4 \mathrm{~m} / \mathrm{s}}{((331.4+v) / f) \mathrm{m} / \text { cycle }}=\frac{331.4}{331.4+v} f \frac{\text { cycles }}{\text { second }} $$ High frequency sound waves may be used to measure the rate of blood flow in an artery. A high frequency sound is introduced on the skin surface above the artery, and the frequency of the waves reflected from the arterial flow is measured. The difference in frequencies emitted and received is used to measure blood velocity. A train and track with listener location. As drawn, \(x(t)\) is negative
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