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Calculus 1 : Functions

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Example Questions

Example Question #3551 : 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 17?

Possible Answers:

$\displaystyle 51$

867 $\displaystyle 867$

289 $\displaystyle 289$

$17\sqrt{3}$ $\displaystyle 17\sqrt{3}$

$289\sqrt{3}$ $\displaystyle 289\sqrt{3}$

Correct answer:

$289\sqrt{3}$ $\displaystyle 289\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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Example Question #641 : 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 $4\sqrt{2}$ $\displaystyle 4\sqrt{2}$?

Possible Answers:

$16\sqrt{6}$ $\displaystyle 16\sqrt{6}$

$32\sqrt{6}$ $\displaystyle 32\sqrt{6}$

$8\sqrt{3}$ $\displaystyle 8\sqrt{3}$

$16\sqrt{3}$ $\displaystyle 16\sqrt{3}$

$32\sqrt{3}$ $\displaystyle 32\sqrt{3}$

Correct answer:

$32\sqrt{3}$ $\displaystyle 32\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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Example Question #2531 : Functions

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 $5\sqrt{7}$ $\displaystyle 5\sqrt{7}$?

Possible Answers:

$175\sqrt{21}$ $\displaystyle 175\sqrt{21}$

$35\sqrt{3}$ $\displaystyle 35\sqrt{3}$

$175\sqrt{3}$ $\displaystyle 175\sqrt{3}$

$25\sqrt{21}$ $\displaystyle 25\sqrt{21}$

$25\sqrt{3}$ $\displaystyle 25\sqrt{3}$

Correct answer:

$175\sqrt{3}$ $\displaystyle 175\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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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 $4\sqrt{11}$ $\displaystyle 4\sqrt{11}$?

Possible Answers:

$16\sqrt{33}$ $\displaystyle 16\sqrt{33}$

$176\sqrt{3}$ $\displaystyle 176\sqrt{3}$

$22\sqrt{3}$ $\displaystyle 22\sqrt{3}$

$176\sqrt{33}$ $\displaystyle 176\sqrt{33}$

$44\sqrt{6}$ $\displaystyle 44\sqrt{6}$

Correct answer:

$176\sqrt{3}$ $\displaystyle 176\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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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 $2\sqrt{19}$ $\displaystyle 2\sqrt{19}$?

Possible Answers:

$19\sqrt{3}$ $\displaystyle 19\sqrt{3}$

$38\sqrt{6}$ $\displaystyle 38\sqrt{6}$

$76\sqrt{3}$ $\displaystyle 76\sqrt{3}$

$38\sqrt{3}$ $\displaystyle 38\sqrt{3}$

$19\sqrt{6}$ $\displaystyle 19\sqrt{6}$

Correct answer:

$76\sqrt{3}$ $\displaystyle 76\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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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 $\frac{1}{5}$ $\displaystyle \frac{1}{5}$?

Possible Answers:

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

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

$\frac{3}{25}$ $\displaystyle \frac{3}{25}$

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

$\frac{3}{5}$ $\displaystyle \frac{3}{5}$

Correct answer:

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

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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Example Question #2531 : Functions

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 $\frac{3}{4}$ $\displaystyle \frac{3}{4}$?

Possible Answers:

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

$\frac{9}{4}$ $\displaystyle \frac{9}{4}$

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

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

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

Correct answer:

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

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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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 $\frac{5}{16}$ $\displaystyle \frac{5}{16}$?

Possible Answers:

$\frac{75}{128}$ $\displaystyle \frac{75}{128}$

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

$\frac{75}{256}$ $\displaystyle \frac{75}{256}$

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

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

Correct answer:

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

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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Example Question #651 : How To Find 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 $\frac{7}{11}$ $\displaystyle \frac{7}{11}$?

Possible Answers:

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

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

$\frac{21}{121}$ $\displaystyle \frac{21}{121}$

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

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

Correct answer:

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

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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Example Question #651 : How To Find 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 $\frac{2}{13}$ $\displaystyle \frac{2}{13}$?

Possible Answers:

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

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

$\frac{12}{13}$ $\displaystyle \frac{12}{13}$

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

$\frac{12}{169}$ $\displaystyle \frac{12}{169}$

Correct answer:

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

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3551 : Calculus

Example Question #641 : Rate Of Change

Example Question #2531 : Functions

Example Question #643 : Rate Of Change

Example Question #3562 : Calculus

Example Question #3563 : Calculus

Example Question #2531 : Functions

Example Question #3564 : Calculus

Example Question #651 : How To Find Rate Of Change

Example Question #651 : How To Find Rate Of Change

All Calculus 1 Resources

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