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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #3601 : Calculus

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its sides when its sides have length $\frac{31}{21}$ ?

Possible Answers:

$\frac{124}{21}$

$\frac{124}{7}$

$\frac{248}{21}$

$\frac{31}{7}$

$\frac{93}{7}$

Correct answer:

$\frac{124}{7}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

A=6s^2

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

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the surface area and diagonal:

$12s\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=12s$

$\phi =12(\frac{31}{21})=\frac{124}{7}$

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Example Question #3602 : Calculus

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its sides when its sides have length $\frac{5}{39}$ ?

Possible Answers:

$\frac{10}{39}$

$\frac{10}{13}$

$\frac{20}{13}$

$\frac{20}{39}$

$\frac{40}{39}$

Correct answer:

$\frac{20}{13}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

A=6s^2

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

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the surface area and diagonal:

$12s\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=12s$

$\phi =12(\frac{5}{39})=\frac{20}{13}$

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Example Question #3603 : Calculus

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its sides when its sides have length $\frac{2}{87}$ ?

Possible Answers:

$\frac{2}{29}$

$\frac{16}{29}$

$\frac{8}{29}$

$\frac{1}{29}$

$\frac{4}{29}$

Correct answer:

$\frac{8}{29}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its surface area in terms of the length of its sides:

A=6s^2

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

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the surface area and diagonal:

$12s\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=12s$

$\phi =12(\frac{2}{87})=\frac{8}{29}$

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Example Question #771 : 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 surface area when the radius is 206?

Possible Answers:

103

824

1648

3296

412

Correct answer:

103

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{206}{2}=103$

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Example Question #3605 : 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 surface area when the radius is 276?

Possible Answers:

138

184

Correct answer:

138

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{276}{2}=138$

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Example Question #3606 : 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 surface area when the radius is 2112?

Possible Answers:

176

704

528

1056

264

Correct answer:

1056

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{2112}{2}=1056$

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Example Question #3607 : 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 surface area when the radius is 2376?

Possible Answers:

108

1188

594

792

396

Correct answer:

1188

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{2376}{2}=1188$

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Example Question #3608 : 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 surface area when the radius is 3112?

Possible Answers:

6224

12448

1556

778

4149

Correct answer:

1556

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{3112}{2}=1556$

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Example Question #3609 : 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 surface area when the radius is 1480?

Possible Answers:

740

2220

370

2960

1110

Correct answer:

740

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{1480}{2}=740$

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Example Question #3610 : Calculus

A spherical balloon is deflating, although it maintains its spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the surface area when the radius is $\frac{1}{74}$ ?

Possible Answers:

$\frac{1}{296}$

$\frac{3}{296}$

$\frac{1}{148}$

$\frac{1}{37}$

$\frac{2}{37}$

Correct answer:

$\frac{1}{148}$

Explanation:

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

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

$A=4\pi r^2$

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

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

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

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 surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{\frac{1}{74}}{2}=\frac{1}{148}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3601 : Calculus

Example Question #3602 : Calculus

Example Question #3603 : Calculus

Example Question #771 : Rate

Example Question #3605 : Calculus

Example Question #3606 : Calculus

Example Question #3607 : Calculus

Example Question #3608 : Calculus

Example Question #3609 : Calculus

Example Question #3610 : Calculus

All Calculus 1 Resources

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