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Calculus 1 : Rate of Change

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Example Questions

Example Question #471 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's surface area to the rate of growth of its sides when its sides have length 12?

Possible Answers:

144

132

Correct answer:

144

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:

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

$\phi=12s$

$\phi =12(12)=144$

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Example Question #472 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's surface area to the rate of growth of its sides when its sides have length 6?

Possible Answers:

108

288

144

Correct answer:

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:

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

$\phi=12s$

$\phi =12(6)=72$

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Example Question #473 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's surface area to the rate of growth of its sides when its sides have length 9?

Possible Answers:

108

216

162

Correct answer:

108

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:

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

$\phi=12s$

$\phi =12(9)=108$

Report an Error

Example Question #474 : Rate Of Change

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 24?

Possible Answers:

432

144

216

288

Correct answer:

288

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:

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

$\phi=12s$

$\phi =12(24)=288$

Report an Error

Example Question #475 : Rate Of Change

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 30?

Possible Answers:

120

180

240

360

Correct answer:

360

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:

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

$\phi=12s$

$\phi =12(30)=360$

Report an Error

Example Question #476 : Rate Of Change

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 22?

Possible Answers:

132

264

484

2904

Correct answer:

264

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:

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

$\phi=12s$

$\phi =12(22)=264$

Report an Error

Example Question #471 : How To Find Rate Of Change

A cube is growing in size. What is the length of the diagonal of the cube at the time that the rate of growth of the cube's volume is equal to $\sqrt{3}$ times the rate of growth of its surface area?

Possible Answers:

$4\sqrt{6}$

$6\sqrt{3}$

$4\sqrt{3}$

Correct answer:

Explanation:

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

V=s^3

A=6s^2

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to $\sqrt{3}$ times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(\sqrt{3})12s\frac{ds}{dt}$

$s=(\sqrt{3})4=4\sqrt{3}$

The diagonal of a cube is given by

$d=s\sqrt{3}$

d=12

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Example Question #472 : How To Find Rate Of Change

A cube is growing in size. What is the length of the sides of the cube at the time that the rate of growth of the cube's volume is equal to 30 times the rate of growth of its surface area?

Possible Answers:

360

720

540

120

Correct answer:

120

Explanation:

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

V=s^3

A=6s^2

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to 30 times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(30)12s\frac{ds}{dt}$

s=(30)4=120

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Example Question #473 : How To Find Rate Of Change

A cube is growing in size. What is the volume of the cube at the time that the rate of growth of the cube's volume is equal to $\frac{1}{8}$ times the rate of growth of its surface area?

Possible Answers:

$\frac{1}{32}$

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{16}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{8}$

Explanation:

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

V=s^3

A=6s^2

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to $\frac{1}{8}$ times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(\frac{1}{8})12s\frac{ds}{dt}$

$s=(\frac{1}{8})4=\frac{1}{2}$

Now to find the volume

V=s^3

$V=(\frac{1}{2})^3=\frac{1}{8}$

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Example Question #474 : How To Find Rate Of Change

A cube is growing in size. What is the surface area of the cube at the time that the rate of growth of the cube's volume is equal to $\frac{2}{3}$ times the rate of growth of its surface area?

Possible Answers:

$\frac{128}{3}$

$\frac{8}{3}$

$\frac{64}{3}$

$\frac{32}{3}$

$\frac{16}{3}$

Correct answer:

$\frac{128}{3}$

Explanation:

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

V=s^3

A=6s^2

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to $\frac{2}{3}$ times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(\frac{2}{3})12s\frac{ds}{dt}$

$s=(\frac{2}{3})4=\frac{8}{3}$

Then to find the surface area:

A=6s^2

$A=6(\frac{8}{3})^2=\frac{128}{3}$

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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 #471 : Rate Of Change

Example Question #472 : Rate Of Change

Example Question #473 : Rate Of Change

Example Question #474 : Rate Of Change

Example Question #475 : Rate Of Change

Example Question #476 : Rate Of Change

Example Question #471 : How To Find Rate Of Change

Example Question #472 : How To Find Rate Of Change

Example Question #473 : How To Find Rate Of Change

Example Question #474 : How To Find Rate Of Change

All Calculus 1 Resources

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