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Calculus 1 : Rate of Change

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Example Questions

Example Question #711 : 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 $\frac{4}{31}$ ?

Possible Answers:

$\frac{64}{961}$

$\frac{32}{961}$

$\frac{16}{961}$

$\frac{8}{31}$

$\frac{8}{961}$

Correct answer:

$\frac{32}{961}$

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:

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

$\phi =2r^2$

$\phi =2(\frac{4}{31})^2=\frac{32}{961}$

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Example Question #711 : Rate Of Change

Possible Answers:

$\frac{16}{1089}$

$\frac{32}{1089}$

$\frac{64}{1089}$

$\frac{128}{1089}$

$\frac{8}{1089}$

Correct answer:

$\frac{128}{1089}$

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:

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

$\phi =2r^2$

$\phi =2(\frac{8}{33})^2=\frac{128}{1089}$

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Example Question #711 : Rate Of Change

Possible Answers:

$\frac{75}{64}$

$\frac{75}{128}$

$\frac{225}{256}$

$\frac{225}{64}$

$\frac{225}{128}$

Correct answer:

$\frac{225}{128}$

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:

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

$\phi =2r^2$

$\phi =2(\frac{15}{16})^2=\frac{225}{128}$

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Example Question #713 : Rate Of Change

Possible Answers:

$\frac{169}{961}$

$\frac{338}{961}$

$\frac{676}{961}$

$\frac{169}{31}$

$\frac{13}{961}$

Correct answer:

$\frac{338}{961}$

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:

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

$\phi =2r^2$

$\phi =2(\frac{13}{31})^2=\frac{338}{961}$

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Example Question #714 : Rate Of Change

A spherical balloon is being filled with air. What is the rate of growth of the sphere's volume when the radius is 4 and the rate of change of the radius is 5?

Possible Answers:

$80\pi$

$320\pi$

$400\pi$

$1600\pi$

$160\pi$

Correct answer:

$320\pi$

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:

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

Now given our problem conditions, the radius is 4 and the rate of change of the radius is 5, we can solve for the rate of change of our volume:

$\frac{dV}{dt}=4\pi (4)^2 (5)=320\pi$

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Example Question #711 : Rate Of Change

A spherical balloon is being filled with air. What is the rate of growth of the sphere's volume when the radius is 3 and the rate of change of the radius is 1?

Possible Answers:

$3\pi$

$12\pi$

$108\pi$

$9\pi$

$36\pi$

Correct answer:

$36\pi$

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:

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

Now given our problem conditions, the radius is 3 and the rate of change of the radius is 1, we can solve for the rate of change of our volume:

$\frac{dV}{dt}=4\pi (3)^2 (1)=36\pi$

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Example Question #712 : Rate Of Change

A spherical balloon is being filled with air. What is the rate of growth of the sphere's volume when the radius is $\frac{1}{2}$ and the rate of change of the radius is 4?

Possible Answers:

$\frac{2\pi}{3}$

$\pi$

$\frac{\pi}{6}$

$\frac{4\pi}{3}$

$4\pi$

Correct answer:

$4\pi$

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:

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

Now given our problem conditions, the radius is $\frac{1}{2}$ and the rate of change of the radius is 4, we can solve for the rate of change of our volume:

$\frac{dV}{dt}=4\pi (\frac{1}{2})^2 (4)=4\pi$

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Example Question #711 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is the rate of growth of the sphere's volume when the radius is 1 and the rate of change of the radius is 5?

Possible Answers:

$25\pi$

$80\pi$

$20\pi$

$5\pi$

$16\pi$

Correct answer:

$20\pi$

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:

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

Now given our problem conditions, the radius is 1 and the rate of change of the radius is 5, we can solve for the rate of change of our volume:

$\frac{dV}{dt}=4\pi (1)^2 (5)=20\pi$

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Example Question #717 : Rate Of Change

A spherical balloon is being filled with air. What is the rate of growth of the sphere's volume when the radius is 2 and the rate of change of the radius is 2?

Possible Answers:

$\frac{16\pi}{3}$

$16\pi$

$\frac{8\pi}{3}$

$\frac{32\pi}{3}$

$32\pi$

Correct answer:

$32\pi$

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:

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

Now given our problem conditions, the radius is 2 and the rate of change of the radius is 2, we can solve for the rate of change of our volume:

$\frac{dV}{dt}=4\pi (2)^2 (2)=32\pi$

Report an Error

Example Question #713 : Rate Of Change

A spherical balloon is being filled with air. What is the rate of growth of the sphere's volume when the radius is $\frac{1}{3}$ and the rate of change of the radius is 3?

Possible Answers:

$\frac{2}{3}\pi$

$\frac{8}{9}\pi$

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

$\frac{4}{3}\pi$

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

Correct answer:

$\frac{4}{3}\pi$

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:

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

Now given our problem conditions, the radius is $\frac{1}{3}$ and the rate of change of the radius is 3, we can solve for the rate of change of our volume:

$\frac{dV}{dt}=4\pi (\frac{1}{3})^2 (3)=\frac{4}{3}\pi$

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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 #711 : Rate Of Change

Example Question #711 : Rate Of Change

Example Question #711 : Rate Of Change

Example Question #713 : Rate Of Change

Example Question #714 : Rate Of Change

Example Question #711 : Rate Of Change

Example Question #712 : Rate Of Change

Example Question #711 : How To Find Rate Of Change

Example Question #717 : Rate Of Change

Example Question #713 : Rate Of Change

All Calculus 1 Resources

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