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Calculus 1 : Calculus

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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 $4\sqrt{2}$ ?

Possible Answers:

$16\sqrt{6}$

$8\sqrt{3}$

$32\sqrt{6}$

$32\sqrt{3}$

$16\sqrt{3}$

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

Possible Answers:

$25\sqrt{21}$

$175\sqrt{21}$

$25\sqrt{3}$

$35\sqrt{3}$

$175\sqrt{3}$

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

Possible Answers:

$16\sqrt{33}$

$176\sqrt{3}$

$22\sqrt{3}$

$176\sqrt{33}$

$44\sqrt{6}$

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:

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

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

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

Report an Error

Example Question #3562 : 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:

$19\sqrt{3}$

$38\sqrt{6}$

$76\sqrt{3}$

$38\sqrt{3}$

$19\sqrt{6}$

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:

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

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

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

Report an Error

Example Question #3563 : 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 diagonal when its sides have length $\frac{1}{5}$ ?

Possible Answers:

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

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

$\frac{3}{25}$

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

$\frac{3}{5}$

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 #741 : Rate

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:

$\frac{3}{2}$

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

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

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

$\frac{9}{4}$

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 #3564 : 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 diagonal when its sides have length $\frac{5}{16}$ ?

Possible Answers:

$\frac{75}{128}$

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

$\frac{75}{256}$

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

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

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 #3565 : 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 diagonal when its sides have length $\frac{7}{11}$ ?

Possible Answers:

$\frac{21}{121}$

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

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

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

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

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 #3566 : 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 diagonal when its sides have length $\frac{2}{13}$ ?

Possible Answers:

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

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

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

$\frac{12}{169}$

$\frac{12}{13}$

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:

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

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

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

Report an Error

Example Question #742 : Rate

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 $\frac{1}{14}$ ?

Possible Answers:

$\frac{3}{14}$

$\frac{3}{196}$

$\frac{3}{98}$

$\frac{\sqrt{3}}{14}$

$\frac{\sqrt{3}}{196}$

Correct answer:

$\frac{\sqrt{3}}{196}$

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:

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

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

$\phi =(\frac{1}{14})^2\sqrt{3}=\frac{\sqrt{3}}{196}$

Report an Error

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3561 : Calculus

Example Question #641 : How To Find Rate Of Change

Example Question #643 : Rate Of Change

Example Question #3562 : Calculus

Example Question #3563 : Calculus

Example Question #741 : Rate

Example Question #3564 : Calculus

Example Question #3565 : Calculus

Example Question #3566 : Calculus

Example Question #742 : Rate

All Calculus 1 Resources

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