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Calculus 1 : Rate of Change

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Example Questions

Example Question #431 : Rate Of Change

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the radius when the radius is $\frac{1}{8}$ ?

Possible Answers:

$\frac{\pi}{16}$

$\frac{\pi}{64}$

$\frac{\pi}{4}$

$\frac{\pi}{8}$

$\frac{\pi}{128}$

Correct answer:

$\frac{\pi}{16}$

Explanation:

Let's begin by writing the equation for the volume of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

The rate of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and radius, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)\frac{dr}{dt}$

$\phi =4\pi r^2$

$\phi =4\pi(\frac{1}{8})^2=\frac{\pi}{16}$

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Example Question #432 : Rate Of Change

Possible Answers:

$\frac{9\pi}{4}$

$\frac{36}{\pi^2}$

$\frac{9}{4\pi^2}$

$\frac{36}{\pi}$

$\frac{9}{4\pi}$

Correct answer:

$\frac{36}{\pi}$

Explanation:

Let's begin by writing the equation for the volume of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

The rate of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and radius, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)\frac{dr}{dt}$

$\phi =4\pi r^2$

$\phi =4\pi(\frac{3}{\pi})^2=\frac{36}{\pi}$

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Example Question #433 : Rate Of Change

Possible Answers:

$\pi$

$\frac{1}{\pi^2}$

$\pi^2$

$2\pi^2$

$\pi^3$

Correct answer:

$\pi^3$

Explanation:

Let's begin by writing the equation for the volume of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

The rate of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and radius, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)\frac{dr}{dt}$

$\phi =4\pi r^2$

$\phi =4\pi(\frac{\pi}{2})^2=\pi^3$

Report an Error

Example Question #434 : Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the circumference when the radius is $\frac{1}{4}$ ?

Possible Answers:

$\frac{1}{16}$

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{8}$

Explanation:

Let's begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$C=2\pi r^$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and circumference, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =2r^2$

$\phi =2(\frac{1}{4})^2=\frac{1}{8}$

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Example Question #435 : Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the circumference when the radius is 20?

Possible Answers:

800

400

160

Correct answer:

800

Explanation:

Let's begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$C=2\pi r^$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and circumference, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =2r^2$

$\phi =2(20)^2=800$

Report an Error

Example Question #436 : Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the circumference when the radius is $\frac{1}{2\pi}$ ?

Possible Answers:

$\frac{1}{4\pi^2}$

$\frac{1}{2\pi^2}$

$4\pi^2$

$\pi^2$

$2\pi$

Correct answer:

$\frac{1}{2\pi^2}$

Explanation:

Let's begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$C=2\pi r^$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and circumference, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =2r^2$

$\phi =2(\frac{1}{2\pi})^2=\frac{1}{2\pi^2}$

Report an Error

Example Question #437 : Rate Of Change

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the circumference when the radius is $\frac{1}{16}$ ?

Possible Answers:

$\frac{1}{64}$

$\frac{1}{128}$

$\frac{1}{32}$

$\frac{1}{256}$

$\frac{1}{512}$

Correct answer:

$\frac{1}{128}$

Explanation:

Let's begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$C=2\pi r^$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and circumference, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =2r^2$

$\phi =2(\frac{1}{16})^2=\frac{1}{128}$

Report an Error

Example Question #438 : Rate Of Change

Possible Answers:

$2\sqrt{2}$

Correct answer:

Explanation:

Let's begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$C=2\pi r^$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and circumference, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =2r^2$

$\phi =2(cos(\frac{\pi}{4})+sin(\frac{\pi}{4}))^2=4$

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Example Question #439 : Rate Of Change

A spherical balloon is deflating, although it retains a spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the circumference when the radius is 107?

Possible Answers:

428

107

22898

11449

214

Correct answer:

22898

Explanation:

Let's begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$C=2\pi r^$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and circumference, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =2r^2$

$\phi =2(107)^2=22898$

Report an Error

Example Question #440 : Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the surface area of the sphere to the rate of growth of the circumference when the radius is $\frac{\pi}{2}$ ?

Possible Answers:

$4\pi$

$2\pi$

$\frac{\pi}{2}$

$\pi$

$\frac{\pi}{8}$

Correct answer:

$2\pi$

Explanation:

Let's begin by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

$A=4\pi r^2$

$C=2\pi r$

The rates of change can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the surface area and circumference, divide:

$8\pi r \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =4r$

$\phi =4(\frac{\pi}{2})=2\pi$

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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #431 : Rate Of Change

Example Question #432 : Rate Of Change

Example Question #433 : Rate Of Change

Example Question #434 : Rate Of Change

Example Question #435 : Rate Of Change

Example Question #436 : Rate Of Change

Example Question #437 : Rate Of Change

Example Question #438 : Rate Of Change

Example Question #439 : Rate Of Change

Example Question #440 : Rate Of Change

All Calculus 1 Resources

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