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Calculus 1 : Calculus

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Example Questions

Example Question #682 : 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 $\frac{31}{2}$ ?

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 #591 : 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

4200

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

Possible Answers:

$\frac{18}{841}$

$\frac{9}{841}$

$\frac{9}{3364}$

$\frac{9}{1682}$

$\frac{36}{841}$

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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Example Question #685 : Rate

Possible Answers:

$\frac{98}{1369}$

$\frac{7}{2738}$

$\frac{49}{1369}$

$\frac{7}{1369}$

$\frac{49}{2738}$

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 #2483 : Functions

Possible Answers:

$\frac{13}{9}$

$\frac{169}{648}$

$\frac{169}{81}$

$\frac{338}{81}$

$\frac{169}{324}$

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 #692 : 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 160?

Possible Answers:

1280

640

320

160

2560

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 #693 : 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 124?

Possible Answers:

248

3844

1922

496

961

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 #694 : 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:

289

1156

544

272

578

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:

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

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

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

$\phi =4r$

$\phi =4(136)=544$

Report an Error

Example Question #695 : 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 258?

Possible Answers:

516

129

1548

1032

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:

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

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

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

$\phi =4r$

$\phi =4(258)=1032$

Report an Error

Example Question #601 : 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 372?

Possible Answers:

496

1984

992

744

1488

Correct answer:

1488

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:

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

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

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

$\phi =4r$

$\phi =4(372)=1488$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #682 : Rate

Example Question #591 : Rate Of Change

Example Question #2482 : Functions

Example Question #685 : Rate

Example Question #2483 : Functions

Example Question #692 : Rate

Example Question #693 : Rate

Example Question #694 : Rate

Example Question #695 : Rate

Example Question #601 : Rate Of Change

All Calculus 1 Resources

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