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Calculus 1 : Rate of Change

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Example Questions

Example Question #531 : 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 surface area when its sides have length $\frac{1}{7}$ ?

Possible Answers:

$\frac{1}{28}$

$\frac{4}{7}$

$\frac{1}{14}$

$\frac{1}{7}$

$\frac{1}{56}$

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:

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

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

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

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Example Question #532 : 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 surface area when its sides have length $\frac{1}{20}$ ?

Possible Answers:

$\frac{1}{80}$

$\frac{1}{20}$

$\frac{1}{10}$

$\frac{1}{5}$

$\frac{1}{40}$

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 #533 : 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 surface area when its sides have length $\frac{1}{64}$ ?

Possible Answers:

$\frac{1}{8}$

$\frac{1}{64}$

$\frac{1}{128}$

$\frac{1}{256}$

$\frac{1}{16}$

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 #534 : 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 surface area when its sides have length $\frac{1}{128}$ ?

Possible Answers:

$\frac{1}{2048}$

$\frac{1}{512}$

$\frac{1}{8192}$

$\frac{1}{4096}$

$\frac{1}{1024}$

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 #535 : 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 surface area when its sides have length $\frac{4}{5}$ ?

Possible Answers:

$\frac{16}{5}$

$\frac{1}{10}$

$\frac{1}{5}$

$\frac{2}{5}$

$\frac{1}{20}$

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 #536 : 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 surface area when its sides have length $\frac{28}{27}$ ?

Possible Answers:

$\frac{14}{27}$

$\frac{7}{9}$

$\frac{7}{27}$

$\frac{14}{9}$

$\frac{112}{27}$

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

Example Question #537 : Rate Of Change

A cube is growing in size. What is the length of the diagonal of the cube at the time that the rate of growth of the cube's surface area is equal to 32 times the rate of growth of its diagonal?

Possible Answers:

$\frac{8\sqrt{3}}{3}$

$\frac{8}{3}$

$8\sqrt{3}$

Correct answer:

Explanation:

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

A=6s^2

$d=s\sqrt{3}$

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

$\frac{dA}{dt}=12s\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the time that the rate of growth of the cube's surface area is equal to 32 times the rate of growth of its diagonal:

$12s\frac{ds}{dt}=(32)\sqrt{3}\frac{ds}{dt}$

$s=(32)\frac{\sqrt{3}}{12}=\frac{8\sqrt{3}}{3}$

The diagonal is then:

$d=(\frac{8\sqrt{3}}{3})\sqrt{3}=8$

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Example Question #538 : Rate Of Change

A cube is growing in size. What is the length of the sides of the cube at the time that the rate of growth of the cube's surface area is equal to 15 times the rate of growth of its diagonal?

Possible Answers:

$15\sqrt{3}$

$\frac{4\sqrt{3}}{5}$

$\frac{15\sqrt{3}}{4}$

$\frac{5\sqrt{3}}{2}$

$\frac{5\sqrt{3}}{4}$

Correct answer:

$\frac{5\sqrt{3}}{4}$

Explanation:

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

A=6s^2

$d=s\sqrt{3}$

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

$\frac{dA}{dt}=12s\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's surface area is equal to 15 times the rate of growth of its diagonal:

$12s\frac{ds}{dt}=(15)\sqrt{3}\frac{ds}{dt}$

$s=(15)\frac{\sqrt{3}}{12}=\frac{5\sqrt{3}}{4}$

Report an Error

Example Question #539 : Rate Of Change

A cube is growing in size. What is the surface area of the cube at the time that the rate of growth of the cube's surface area is equal to 4 times the rate of growth of its diagonal?

Possible Answers:

$\sqrt{3}$

Correct answer:

Explanation:

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

A=6s^2

$d=s\sqrt{3}$

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

$\frac{dA}{dt}=12s\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's surface area is equal to 4 times the rate of growth of its diagonal

$12s\frac{ds}{dt}=(4)\sqrt{3}\frac{ds}{dt}$

$s=(4)\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{3}$

The surface area is then:

$A=6(\frac{\sqrt{3}}{3})^2=2$

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Example Question #540 : Rate Of Change

A cube is growing in size. What is the volume of the cube at the time that the rate of growth of the cube's surface area is equal to 24 times the rate of growth of its diagonal?

Possible Answers:

$24\sqrt{3}$

$12\sqrt{3}$

$8\sqrt{3}$

Correct answer:

$24\sqrt{3}$

Explanation:

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

A=6s^2

$d=s\sqrt{3}$

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

$\frac{dA}{dt}=12s\frac{ds}{dt}$

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's surface area is equal to 24 times the rate of growth of its diagonal

$12s\frac{ds}{dt}=(24)\sqrt{3}\frac{ds}{dt}$

$s=(24)\frac{\sqrt{3}}{12}=2\sqrt{3}$

Then to find the volume:

V=s^3

$V=(2\sqrt{3})^3=24\sqrt{3}$

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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 #531 : How To Find Rate Of Change

Example Question #532 : How To Find Rate Of Change

Example Question #533 : How To Find Rate Of Change

Example Question #534 : How To Find Rate Of Change

Example Question #535 : How To Find Rate Of Change

Example Question #536 : How To Find Rate Of Change

Example Question #537 : Rate Of Change

Example Question #538 : Rate Of Change

Example Question #539 : Rate Of Change

Example Question #540 : Rate Of Change

All Calculus 1 Resources

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