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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 #527 : How To Find Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 316?

Possible Answers:

158

316

39.5

632

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{316}{4}=79$

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Example Question #528 : How To Find Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 432?

Possible Answers:

360

216

108

180

Correct answer:

108

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{432}{4}=108$

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Example Question #522 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 520?

Possible Answers:

325

130

650

260

Correct answer:

130

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{520}{4}=130$

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Example Question #521 : How To Find Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 924?

Possible Answers:

204

308

616

231

Correct answer:

231

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{924}{4}=231$

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Example Question #531 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its surface area when its sides have length $\frac{1}{7}$ ?

Possible Answers:

$\frac{1}{14}$

$\frac{1}{7}$

$\frac{4}{7}$

$\frac{1}{56}$

$\frac{1}{28}$

Correct answer:

$\frac{1}{28}$

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{\frac{1}{7}}{4}=\frac{1}{28}$

Report an Error

Example Question #3442 : Calculus

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its surface area when its sides have length $\frac{1}{20}$ ?

Possible Answers:

$\frac{1}{5}$

$\frac{1}{10}$

$\frac{1}{40}$

$\frac{1}{20}$

$\frac{1}{80}$

Correct answer:

$\frac{1}{80}$

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{\frac{1}{20}}{4}=\frac{1}{80}$

Report an Error

Example Question #3443 : Calculus

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its surface area when its sides have length $\frac{1}{64}$ ?

Possible Answers:

$\frac{1}{8}$

$\frac{1}{64}$

$\frac{1}{16}$

$\frac{1}{128}$

$\frac{1}{256}$

Correct answer:

$\frac{1}{256}$

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{\frac{1}{64}}{4}=\frac{1}{256}$

Report an Error

Example Question #3441 : Calculus

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its surface area when its sides have length $\frac{1}{128}$ ?

Possible Answers:

$\frac{1}{4096}$

$\frac{1}{8192}$

$\frac{1}{512}$

$\frac{1}{1024}$

$\frac{1}{2048}$

Correct answer:

$\frac{1}{512}$

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{\frac{1}{128}}{4}=\frac{1}{512}$

Report an Error

Example Question #3445 : Calculus

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its surface area when its sides have length $\frac{4}{5}$ ?

Possible Answers:

$\frac{1}{20}$

$\frac{2}{5}$

$\frac{16}{5}$

$\frac{1}{5}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{5}$

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{\frac{4}{5}}{4}=\frac{1}{5}$

Report an Error

Example Question #531 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its surface area when its sides have length $\frac{28}{27}$ ?

Possible Answers:

$\frac{7}{27}$

$\frac{14}{9}$

$\frac{14}{27}$

$\frac{112}{27}$

$\frac{7}{9}$

Correct answer:

$\frac{7}{27}$

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

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

$\phi=\frac{s}{4}$

$\phi =\frac{\frac{28}{27}}{4}=\frac{7}{27}$

Report an Error

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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 #527 : How To Find Rate Of Change

Example Question #528 : How To Find Rate Of Change

Example Question #522 : Rate Of Change

Example Question #521 : How To Find Rate Of Change

Example Question #531 : Rate Of Change

Example Question #3442 : Calculus

Example Question #3443 : Calculus

Example Question #3441 : Calculus

Example Question #3445 : Calculus

Example Question #531 : Rate Of Change

All CLEP Calculus Resources

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