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Calculus 1 : Rate of Change

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Example Questions

Example Question #591 : 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 17?

Possible Answers:

$1156\pi$ $\displaystyle 1156\pi$

$578\pi$ $\displaystyle 578\pi$

1156 $\displaystyle 1156$

289 $\displaystyle 289$

$289\pi$ $\displaystyle 289\pi$

Correct answer:

$1156\pi$ $\displaystyle 1156\pi$

Explanation:

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

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

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

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

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:

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

$\phi =4\pi r^2$ $\displaystyle \phi =4\pi r^2$

$\phi =4\pi(17)^2=1156\pi$ $\displaystyle \phi =4\pi(17)^2=1156\pi$

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Example Question #681 : Rate

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 34?

Possible Answers:

$534\pi$ $\displaystyle 534\pi$

$4624\pi$ $\displaystyle 4624\pi$

$1068\pi$ $\displaystyle 1068\pi$

$136\pi$ $\displaystyle 136\pi$

$1156\pi$ $\displaystyle 1156\pi$

Correct answer:

$4624\pi$ $\displaystyle 4624\pi$

Explanation:

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

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

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

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

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:

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

$\phi =4\pi r^2$ $\displaystyle \phi =4\pi r^2$

$\phi =4\pi(34)^2=4624\pi$ $\displaystyle \phi =4\pi(34)^2=4624\pi$

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Example Question #682 : Rate

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 28?

Possible Answers:

$3136\pi$ $\displaystyle 3136\pi$

$112\pi$ $\displaystyle 112\pi$

$784\pi$ $\displaystyle 784\pi$

$6272\pi$ $\displaystyle 6272\pi$

$448\pi$ $\displaystyle 448\pi$

Correct answer:

$3136\pi$ $\displaystyle 3136\pi$

Explanation:

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

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

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

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

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:

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

$\phi =4\pi r^2$ $\displaystyle \phi =4\pi r^2$

$\phi =4\pi(28)^2=3136\pi$ $\displaystyle \phi =4\pi(28)^2=3136\pi$

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Example Question #3501 : Calculus

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 radius when the radius is $\frac{1}{34}$ $\displaystyle \frac{1}{34}$?

Possible Answers:

$\frac{16\pi}{289}$ $\displaystyle \frac{16\pi}{289}$

$\frac{2\pi}{289}$ $\displaystyle \frac{2\pi}{289}$

$\frac{4\pi}{289}$ $\displaystyle \frac{4\pi}{289}$

$\frac{8\pi}{289}$ $\displaystyle \frac{8\pi}{289}$

$\frac{\pi}{289}$ $\displaystyle \frac{\pi}{289}$

Correct answer:

$\frac{\pi}{289}$ $\displaystyle \frac{\pi}{289}$

Explanation:

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

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

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

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

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:

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

$\phi =4\pi r^2$ $\displaystyle \phi =4\pi r^2$

$\phi =4\pi(\frac{1}{34})^2=\frac{\pi}{289}$ $\displaystyle \phi =4\pi(\frac{1}{34})^2=\frac{\pi}{289}$

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Example Question #3502 : Calculus

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 radius when the radius is $\frac{1}{60}$ $\displaystyle \frac{1}{60}$?

Possible Answers:

$\frac{4\pi}{225}$ $\displaystyle \frac{4\pi}{225}$

$\frac{\pi}{450}$ $\displaystyle \frac{\pi}{450}$

$\frac{\pi}{900}$ $\displaystyle \frac{\pi}{900}$

$\frac{\pi}{225}$ $\displaystyle \frac{\pi}{225}$

$\frac{2\pi}{225}$ $\displaystyle \frac{2\pi}{225}$

Correct answer:

$\frac{\pi}{900}$ $\displaystyle \frac{\pi}{900}$

Explanation:

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

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

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

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

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:

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

$\phi =4\pi r^2$ $\displaystyle \phi =4\pi r^2$

$\phi =4\pi(\frac{1}{60})^2=\frac{\pi}{900}$ $\displaystyle \phi =4\pi(\frac{1}{60})^2=\frac{\pi}{900}$

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Example Question #3503 : Calculus

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 radius when the radius is $\frac{1}{71}$ $\displaystyle \frac{1}{71}$?

Possible Answers:

$\frac{8\pi}{5041}$ $\displaystyle \frac{8\pi}{5041}$

$\frac{2\pi}{5041}$ $\displaystyle \frac{2\pi}{5041}$

$\frac{\pi}{5041}$ $\displaystyle \frac{\pi}{5041}$

$\frac{4\pi}{5041}$ $\displaystyle \frac{4\pi}{5041}$

$\frac{\pi}{10082}$ $\displaystyle \frac{\pi}{10082}$

Correct answer:

$\frac{4\pi}{5041}$ $\displaystyle \frac{4\pi}{5041}$

Explanation:

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

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

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

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

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:

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

$\phi =4\pi r^2$ $\displaystyle \phi =4\pi r^2$

$\phi =4\pi(\frac{1}{71})^2=\frac{4\pi}{5041}$ $\displaystyle \phi =4\pi(\frac{1}{71})^2=\frac{4\pi}{5041}$

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Example Question #681 : Rate

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 41?

Possible Answers:

820 $\displaystyle 820$

1681 $\displaystyle 1681$

4100 $\displaystyle 4100$

164 $\displaystyle 164$

3362 $\displaystyle 3362$

Correct answer:

3362 $\displaystyle 3362$

Explanation:

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

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

$C=2\pi r^$ $\displaystyle C=2\pi r^$

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

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

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

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:

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

$\phi =2r^2$ $\displaystyle \phi =2r^2$

$\phi =2(41)^2=3362$ $\displaystyle \phi =2(41)^2=3362$

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Example Question #2481 : Functions

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 $\frac{31}{2}$ $\displaystyle \frac{31}{2}$?

Possible Answers:

1922 $\displaystyle 1922$

961 $\displaystyle 961$

$\frac{961}{2}$ $\displaystyle \frac{961}{2}$

$\frac{961}{4}$ $\displaystyle \frac{961}{4}$

$\frac{961}{8}$ $\displaystyle \frac{961}{8}$

Correct answer:

$\frac{961}{2}$ $\displaystyle \frac{961}{2}$

Explanation:

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

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

$C=2\pi r^$ $\displaystyle C=2\pi r^$

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

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

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

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:

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

$\phi =2r^2$ $\displaystyle \phi =2r^2$

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

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Example Question #591 : How To Find 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 47?

Possible Answers:

840 $\displaystyle 840$

4670 $\displaystyle 4670$

4418 $\displaystyle 4418$

2209 $\displaystyle 2209$

4200 $\displaystyle 4200$

Correct answer:

4418 $\displaystyle 4418$

Explanation:

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

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

$C=2\pi r^$ $\displaystyle C=2\pi r^$

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

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

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

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:

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

$\phi =2r^2$ $\displaystyle \phi =2r^2$

$\phi =2(47)^2=4418$ $\displaystyle \phi =2(47)^2=4418$

Report an Error

Example Question #2482 : Functions

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 $\frac{3}{29}$ $\displaystyle \frac{3}{29}$?

Possible Answers:

$\frac{18}{841}$ $\displaystyle \frac{18}{841}$

$\frac{9}{841}$ $\displaystyle \frac{9}{841}$

$\frac{9}{3364}$ $\displaystyle \frac{9}{3364}$

$\frac{9}{1682}$ $\displaystyle \frac{9}{1682}$

$\frac{36}{841}$ $\displaystyle \frac{36}{841}$

Correct answer:

$\frac{18}{841}$ $\displaystyle \frac{18}{841}$

Explanation:

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

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

$C=2\pi r^$ $\displaystyle C=2\pi r^$

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

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

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

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:

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

$\phi =2r^2$ $\displaystyle \phi =2r^2$

$\phi =2(\frac{3}{29})^2=\frac{18}{841}$ $\displaystyle \phi =2(\frac{3}{29})^2=\frac{18}{841}$

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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 #591 : Rate Of Change

Example Question #681 : Rate

Example Question #682 : Rate

Example Question #3501 : Calculus

Example Question #3502 : Calculus

Example Question #3503 : Calculus

Example Question #681 : Rate

Example Question #2481 : Functions

Example Question #591 : How To Find Rate Of Change

Example Question #2482 : Functions

All Calculus 1 Resources

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