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CLEP Calculus : Functions

Study concepts, example questions & explanations for CLEP Calculus

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2 Practice Tests Question of the Day Flashcards

Example Questions

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:

820

1681

3362

164

4100

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:

$\frac{961}{4}$

1922

961

$\frac{961}{2}$

$\frac{961}{8}$

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:

4200

840

4418

4670

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 #600 : 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:

$\frac{9}{1682}$

$\frac{9}{841}$

$\frac{36}{841}$

$\frac{9}{3364}$

$\frac{18}{841}$

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:

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

$\phi =2r^2$

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

Report an Error

Example Question #601 : Rate Of Change

Possible Answers:

$\frac{7}{1369}$

$\frac{49}{1369}$

$\frac{7}{2738}$

$\frac{98}{1369}$

$\frac{49}{2738}$

Correct answer:

$\frac{49}{2738}$

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:

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

$\phi =2r^2$

$\phi =2(\frac{7}{74})^2=\frac{49}{2738}$

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Example Question #602 : Rate Of Change

Possible Answers:

$\frac{338}{81}$

$\frac{169}{81}$

$\frac{13}{9}$

$\frac{169}{648}$

$\frac{169}{324}$

Correct answer:

$\frac{169}{648}$

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:

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

$\phi =2r^2$

$\phi =2(\frac{13}{36})^2=\frac{169}{648}$

Report an Error

Example Question #603 : 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 160?

Possible Answers:

160

1280

2560

640

320

Correct answer:

640

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:

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

$\phi =4r$

$\phi =4(160)=640$

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Example Question #604 : 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 124?

Possible Answers:

961

248

496

3844

1922

Correct answer:

496

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:

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

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

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

$\phi =4r$

$\phi =4(124)=496$

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Example Question #691 : Rate

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 136?

Possible Answers:

272

289

1156

578

544

Correct answer:

544

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:

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

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

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

$\phi =4r$

$\phi =4(136)=544$

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Example Question #606 : 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 258?

Possible Answers:

1032

129

1548

516

774

Correct answer:

1032

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:

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

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

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

$\phi =4r$

$\phi =4(258)=1032$

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All CLEP Calculus Resources

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CLEP Calculus : Functions

Study concepts, example questions & explanations for CLEP Calculus

All CLEP Calculus Resources

Example Questions

Example Question #597 : Rate Of Change

Example Question #598 : Rate Of Change

Example Question #599 : Rate Of Change

Example Question #600 : Rate Of Change

Example Question #601 : Rate Of Change

Example Question #602 : Rate Of Change

Example Question #603 : Rate Of Change

Example Question #604 : Rate Of Change

Example Question #691 : Rate

Example Question #606 : Rate Of Change

All CLEP Calculus Resources

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