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Calculus 1 : Rate of Change

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Example Questions

Example Question #3501 : Calculus

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:

289

1156

$1156\pi$

$578\pi$

$289\pi$

Correct answer:

$1156\pi$

Explanation:

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

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

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:

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

$\phi =4\pi r^2$

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

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Example Question #592 : How To Find 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 34?

Possible Answers:

$1068\pi$

$4624\pi$

$1156\pi$

$136\pi$

$534\pi$

Correct answer:

$4624\pi$

Explanation:

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

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

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:

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

$\phi =4\pi r^2$

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

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Example Question #3503 : Calculus

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:

$3136\pi$

$784\pi$

$448\pi$

$6272\pi$

$112\pi$

Correct answer:

$3136\pi$

Explanation:

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

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

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:

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

$\phi =4\pi r^2$

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

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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 $\frac{1}{34}$ ?

Possible Answers:

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

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

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

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

$\frac{\pi}{289}$

Correct answer:

$\frac{\pi}{289}$

Explanation:

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

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

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:

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

$\phi =4\pi r^2$

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

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Example Question #595 : 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 radius when the radius is $\frac{1}{60}$ ?

Possible Answers:

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

$\frac{\pi}{225}$

$\frac{\pi}{900}$

$\frac{\pi}{450}$

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

Correct answer:

$\frac{\pi}{900}$

Explanation:

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

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

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:

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

$\phi =4\pi r^2$

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

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Example Question #592 : 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 radius when the radius is $\frac{1}{71}$ ?

Possible Answers:

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

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

$\frac{\pi}{5041}$

$\frac{\pi}{10082}$

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

Correct answer:

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

Explanation:

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

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

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:

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

$\phi =4\pi r^2$

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

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Example Question #597 : 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 41?

Possible Answers:

3362

1681

820

4100

164

Correct answer:

3362

Explanation:

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

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

$C=2\pi r^$

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:

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

$\phi =2r^2$

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

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Example Question #598 : 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 $\frac{31}{2}$ ?

Possible Answers:

$\frac{961}{8}$

1922

961

$\frac{961}{4}$

$\frac{961}{2}$

Correct answer:

$\frac{961}{2}$

Explanation:

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

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

$C=2\pi r^$

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:

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

$\phi =2r^2$

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

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Example Question #599 : 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:

4670

4418

4200

840

2209

Correct answer:

4418

Explanation:

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

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

$C=2\pi r^$

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:

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

$\phi =2r^2$

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

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Example Question #591 : How To Find 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 $\frac{3}{29}$ ?

Possible Answers:

$\frac{9}{1682}$

$\frac{9}{3364}$

$\frac{9}{841}$

$\frac{36}{841}$

$\frac{18}{841}$

Correct answer:

$\frac{18}{841}$

Explanation:

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

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

$C=2\pi r^$

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:

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

$\phi =2r^2$

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

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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 #3501 : Calculus

Example Question #592 : How To Find Rate Of Change

Example Question #3503 : Calculus

Example Question #3502 : Calculus

Example Question #595 : Rate Of Change

Example Question #592 : Rate Of Change

Example Question #597 : Rate Of Change

Example Question #598 : Rate Of Change

Example Question #599 : Rate Of Change

Example Question #591 : How To Find Rate Of Change

All Calculus 1 Resources

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