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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 #736 : Rate

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 diagonal when its sides have length 17?

Possible Answers:

289

$289\sqrt{3}$

867

$17\sqrt{3}$

Correct answer:

$289\sqrt{3}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(17)^2\sqrt{3}=289\sqrt{3}$

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Example Question #737 : Rate

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 diagonal when its sides have length $4\sqrt{2}$ ?

Possible Answers:

$16\sqrt{6}$

$32\sqrt{3}$

$16\sqrt{3}$

$8\sqrt{3}$

$32\sqrt{6}$

Correct answer:

$32\sqrt{3}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(4\sqrt{2})^2\sqrt{3}=32\sqrt{3}$

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Example Question #738 : Rate

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 diagonal when its sides have length $5\sqrt{7}$ ?

Possible Answers:

$25\sqrt{21}$

$175\sqrt{21}$

$35\sqrt{3}$

$175\sqrt{3}$

$25\sqrt{3}$

Correct answer:

$175\sqrt{3}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(5\sqrt{7})^2\sqrt{3}=175\sqrt{3}$

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Example Question #739 : Rate

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 diagonal when its sides have length $4\sqrt{11}$ ?

Possible Answers:

$22\sqrt{3}$

$16\sqrt{33}$

$176\sqrt{33}$

$176\sqrt{3}$

$44\sqrt{6}$

Correct answer:

$176\sqrt{3}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(4\sqrt{11})^2\sqrt{3}=176\sqrt{3}$

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Example Question #3561 : Calculus

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 diagonal when its sides have length $2\sqrt{19}$ ?

Possible Answers:

$19\sqrt{6}$

$76\sqrt{3}$

$38\sqrt{6}$

$19\sqrt{3}$

$38\sqrt{3}$

Correct answer:

$76\sqrt{3}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(2\sqrt{19})^2\sqrt{3}=76\sqrt{3}$

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Example Question #652 : 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 diagonal when its sides have length $\frac{1}{5}$ ?

Possible Answers:

$\frac{\sqrt{3}}{5}$

$\frac{\sqrt{3}}{75}$

$\frac{3}{25}$

$\frac{3}{5}$

$\frac{\sqrt{3}}{25}$

Correct answer:

$\frac{\sqrt{3}}{25}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(\frac{1}{5})^2\sqrt{3}=\frac{\sqrt{3}}{25}$

Report an Error

Example Question #653 : 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 diagonal when its sides have length $\frac{3}{4}$ ?

Possible Answers:

$\frac{9\sqrt{3}}{16}$

$\frac{9}{4}$

$\frac{9\sqrt{3}}{4}$

$\frac{3}{2}$

$\frac{3\sqrt{3}}{4}$

Correct answer:

$\frac{9\sqrt{3}}{16}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(\frac{3}{4})^2\sqrt{3}=\frac{9\sqrt{3}}{16}$

Report an Error

Example Question #654 : 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 diagonal when its sides have length $\frac{5}{16}$ ?

Possible Answers:

$\frac{75}{128}$

$\frac{25\sqrt{3}}{128}$

$\frac{5\sqrt{3}}{16}$

$\frac{25\sqrt{3}}{256}$

$\frac{75}{256}$

Correct answer:

$\frac{25\sqrt{3}}{256}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(\frac{5}{16})^2\sqrt{3}=\frac{25\sqrt{3}}{256}$

Report an Error

Example Question #655 : 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 diagonal when its sides have length $\frac{7}{11}$ ?

Possible Answers:

$\frac{49\sqrt{3}}{121}$

$\frac{49\sqrt{3}}{11}$

$\frac{21}{121}$

$\frac{7\sqrt{3}}{11}$

$\frac{7\sqrt{3}}{121}$

Correct answer:

$\frac{49\sqrt{3}}{121}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(\frac{7}{11})^2\sqrt{3}=\frac{49\sqrt{3}}{121}$

Report an Error

Example Question #656 : 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 diagonal when its sides have length $\frac{2}{13}$ ?

Possible Answers:

$\frac{12}{13}$

$\frac{4\sqrt{3}}{13}$

$\frac{2\sqrt{3}}{169}$

$\frac{4\sqrt{3}}{169}$

$\frac{12}{169}$

Correct answer:

$\frac{4\sqrt{3}}{169}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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

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

$\phi=s^2\sqrt{3}$

$\phi =(\frac{2}{13})^2\sqrt{3}=\frac{4\sqrt{3}}{169}$

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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 #736 : Rate

Example Question #737 : Rate

Example Question #738 : Rate

Example Question #739 : Rate

Example Question #3561 : Calculus

Example Question #652 : Rate Of Change

Example Question #653 : Rate Of Change

Example Question #654 : Rate Of Change

Example Question #655 : Rate Of Change

Example Question #656 : Rate Of Change

All CLEP Calculus Resources

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