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Calculus 1 : Rate of Change

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Example Questions

Example Question #2515 : 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 surface area when its sides have length $\frac{92}{95}$ ?

Possible Answers:

$\frac{46}{95}$

$\frac{23}{95}$

$\frac{46}{19}$

$\frac{23}{19}$

$\frac{92}{19}$

Correct answer:

$\frac{23}{95}$

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:

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

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

$\phi =\frac{(\frac{92}{95})}{4}=\frac{23}{95}$

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Example Question #2516 : 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 surface area when its sides have length $\frac{32}{41}$ ?

Possible Answers:

$\frac{64}{41}$

$\frac{8}{41}$

$\frac{32}{41}$

$\frac{16}{41}$

$\frac{4}{41}$

Correct answer:

$\frac{8}{41}$

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:

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

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

$\phi =\frac{(\frac{32}{41})}{4}=\frac{8}{41}$

Report an Error

Example Question #2517 : 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 surface area when its sides have length $\frac{79}{80}$ ?

Possible Answers:

$\frac{79}{5}$

$\frac{79}{20}$

$\frac{79}{320}$

$\frac{79}{160}$

$\frac{316}{5}$

Correct answer:

$\frac{79}{320}$

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:

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

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

$\phi =\frac{(\frac{79}{80})}{4}=\frac{79}{320}$

Report an Error

Example Question #2518 : 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 the area of a single face when its sides have length 93?

Possible Answers:

$\frac{31}{2}$

$\frac{93}{2}$

$\frac{279}{2}$

$\frac{279}{5}$

$\frac{31}{5}$

Correct answer:

$\frac{279}{2}$

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:

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

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

$\phi =\frac{3(93)}{2}=\frac{279}{2}$

Report an Error

Example Question #2519 : 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 the area of a single face when its sides have length 166?

Possible Answers:

41.5

166

332

249

Correct answer:

249

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:

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

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

$\phi =\frac{3(166)}{2}=249$

Report an Error

Example Question #2520 : 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 the area of a single face when its sides have length 404?

Possible Answers:

101

404

1212

202

606

Correct answer:

606

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:

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

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

$\phi =\frac{3(404)}{2}=606$

Report an Error

Example Question #631 : 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 the area of a single face when its sides have length 864?

Possible Answers:

432

216

1296

5184

Correct answer:

1296

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:

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

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

$\phi =\frac{3(864)}{2}=1296$

Report an Error

Example Question #631 : 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 the area of a single face when its sides have length 428?

Possible Answers:

535

321

107

642

214

Correct answer:

642

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:

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

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

$\phi =\frac{3(428)}{2}=642$

Report an Error

Example Question #634 : 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 the area of a single face when its sides have length 296?

Possible Answers:

444

222

148

111

Correct answer:

444

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:

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

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

$\phi =\frac{3(296)}{2}=444$

Report an Error

Example Question #3551 : 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 the area of a single face when its sides have length $\frac{13}{6}$ ?

Possible Answers:

$\frac{13}{4}$

$\frac{13}{2}$

$\frac{13}{6}$

$\frac{13}{12}$

$\frac{13}{24}$

Correct answer:

$\frac{13}{4}$

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:

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

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

$\phi =\frac{3(\frac{13}{6})}{2}=\frac{13}{4}$

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 #2515 : Functions

Example Question #2516 : Functions

Example Question #2517 : Functions

Example Question #2518 : Functions

Example Question #2519 : Functions

Example Question #2520 : Functions

Example Question #631 : Rate Of Change

Example Question #631 : How To Find Rate Of Change

Example Question #634 : Rate Of Change

Example Question #3551 : Calculus

All Calculus 1 Resources

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