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Calculus 1 : Rate of Change

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Example Questions

Example Question #641 : 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 the area of a single face when its sides have length $\frac{14}{9}$ ?

Possible Answers:

$\frac{7}{9}$

$\frac{7}{3}$

$\frac{14}{3}$

$\frac{7}{6}$

Correct answer:

$\frac{7}{3}$

Explanation:

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

V=s^3

a=s^2

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

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

$\frac{dA}{dt}=2s\frac{ds}{dt}$

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

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

$\phi=\frac{3s}{2}$

$\phi =\frac{3(\frac{14}{9})}{2}=\frac{7}{3}$

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Example Question #642 : 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 the area of a single face when its sides have length $\frac{16}{3}$ ?

Possible Answers:

$\frac{8}{3}$

$\frac{2}{9}$

$\frac{4}{3}$

$\frac{16}{9}$

Correct answer:

Explanation:

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

V=s^3

a=s^2

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

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

$\frac{dA}{dt}=2s\frac{ds}{dt}$

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

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

$\phi=\frac{3s}{2}$

$\phi =\frac{3(\frac{16}{3})}{2}=8$

Report an Error

Example Question #643 : 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 the area of a single face when its sides have length $\frac{31}{6}$ ?

Possible Answers:

$\frac{31}{4}$

$\frac{93}{4}$

$\frac{93}{8}$

$\frac{31}{2}$

$\frac{31}{24}$

Correct answer:

$\frac{31}{4}$

Explanation:

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

V=s^3

a=s^2

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

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

$\frac{dA}{dt}=2s\frac{ds}{dt}$

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

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

$\phi=\frac{3s}{2}$

$\phi =\frac{3(\frac{31}{6})}{2}=\frac{31}{4}$

Report an Error

Example Question #644 : 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 the area of a single face when its sides have length $\frac{64}{27}$ ?

Possible Answers:

$\frac{16}{9}$

$\frac{16}{3}$

$\frac{32}{9}$

$\frac{4}{3}$

$\frac{8}{9}$

Correct answer:

$\frac{32}{9}$

Explanation:

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

V=s^3

a=s^2

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

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

$\frac{dA}{dt}=2s\frac{ds}{dt}$

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

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

$\phi=\frac{3s}{2}$

$\phi =\frac{3(\frac{64}{27})}{2}=\frac{32}{9}$

Report an Error

Example Question #2521 : 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 the area of a single face when its sides have length $\frac{22}{45}$ ?

Possible Answers:

$\frac{13}{90}$

$\frac{11}{90}$

$\frac{11}{15}$

$\frac{3}{5}$

$\frac{13}{15}$

Correct answer:

$\frac{11}{15}$

Explanation:

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

V=s^3

a=s^2

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

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

$\frac{dA}{dt}=2s\frac{ds}{dt}$

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

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

$\phi=\frac{3s}{2}$

$\phi =\frac{3(\frac{22}{45})}{2}=\frac{11}{15}$

Report an Error

Example Question #2522 : 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 $3\sqrt{5}$ ?

Possible Answers:

$15\sqrt{3}$

$9\sqrt{15}$

$15\sqrt{15}$

$45\sqrt{3}$

$3\sqrt{15}$

Correct answer:

$45\sqrt{3}$

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:

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

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

$\phi =(3\sqrt{5})^2\sqrt{3}=45\sqrt{3}$

Report an Error

Example Question #646 : 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 17?

Possible Answers:

289

$289\sqrt{3}$

$17\sqrt{3}$

867

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:

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

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

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

Report an Error

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 $4\sqrt{2}$ ?

Possible Answers:

$32\sqrt{6}$

$16\sqrt{6}$

$8\sqrt{3}$

$32\sqrt{3}$

$16\sqrt{3}$

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:

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

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

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

Report an Error

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:

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

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

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

Report an Error

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

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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 #641 : Rate Of Change

Example Question #642 : Rate Of Change

Example Question #643 : Rate Of Change

Example Question #644 : Rate Of Change

Example Question #2521 : Functions

Example Question #2522 : Functions

Example Question #646 : Rate Of Change

Example Question #641 : Rate Of Change

Example Question #641 : How To Find Rate Of Change

Example Question #643 : Rate Of Change

All Calculus 1 Resources

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