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Calculus 1 : How to find rate of change

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Example Questions

Example Question #631 : 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 $\frac{92}{95}$ $\displaystyle \frac{92}{95}$ ?

Possible Answers:

$\frac{46}{19}$ $\displaystyle \frac{46}{19}$

$\frac{46}{95}$ $\displaystyle \frac{46}{95}$

$\frac{23}{95}$ $\displaystyle \frac{23}{95}$

$\frac{23}{19}$ $\displaystyle \frac{23}{19}$

$\frac{92}{19}$ $\displaystyle \frac{92}{19}$

Correct answer:

$\frac{23}{95}$ $\displaystyle \frac{23}{95}$

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 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

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

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

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

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

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

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

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

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Example Question #3541 : 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 surface area when its sides have length $\frac{32}{41}$ $\displaystyle \frac{32}{41}$ ?

Possible Answers:

$\frac{8}{41}$ $\displaystyle \frac{8}{41}$

$\frac{64}{41}$ $\displaystyle \frac{64}{41}$

$\frac{16}{41}$ $\displaystyle \frac{16}{41}$

$\frac{32}{41}$ $\displaystyle \frac{32}{41}$

$\frac{4}{41}$ $\displaystyle \frac{4}{41}$

Correct answer:

$\frac{8}{41}$ $\displaystyle \frac{8}{41}$

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 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

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

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

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

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

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

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

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

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Example Question #3542 : 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 surface area when its sides have length $\frac{79}{80}$ $\displaystyle \frac{79}{80}$ ?

Possible Answers:

$\frac{79}{20}$ $\displaystyle \frac{79}{20}$

$\frac{79}{320}$ $\displaystyle \frac{79}{320}$

$\frac{79}{5}$ $\displaystyle \frac{79}{5}$

$\frac{79}{160}$ $\displaystyle \frac{79}{160}$

$\frac{316}{5}$ $\displaystyle \frac{316}{5}$

Correct answer:

$\frac{79}{320}$ $\displaystyle \frac{79}{320}$

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 $\displaystyle V=s^3$

A=6s^2 $\displaystyle A=6s^2$

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

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

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

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

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

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

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

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Example Question #3543 : 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 the area of a single face when its sides have length 93?

Possible Answers:

$\frac{279}{5}$ $\displaystyle \frac{279}{5}$

$\frac{31}{2}$ $\displaystyle \frac{31}{2}$

$\frac{31}{5}$ $\displaystyle \frac{31}{5}$

$\frac{93}{2}$ $\displaystyle \frac{93}{2}$

$\frac{279}{2}$ $\displaystyle \frac{279}{2}$

Correct answer:

$\frac{279}{2}$ $\displaystyle \frac{279}{2}$

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 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

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

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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 166?

Possible Answers:

$\displaystyle 83$

249 $\displaystyle 249$

41.5 $\displaystyle 41.5$

166 $\displaystyle 166$

332 $\displaystyle 332$

Correct answer:

249 $\displaystyle 249$

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 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

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

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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 404?

Possible Answers:

202 $\displaystyle 202$

404 $\displaystyle 404$

101 $\displaystyle 101$

606 $\displaystyle 606$

1212 $\displaystyle 1212$

Correct answer:

606 $\displaystyle 606$

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 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

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

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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:

$\displaystyle 72$

432 $\displaystyle 432$

216 $\displaystyle 216$

1296 $\displaystyle 1296$

5184 $\displaystyle 5184$

Correct answer:

1296 $\displaystyle 1296$

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 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

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

Report an Error

Example Question #721 : Rate

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:

642 $\displaystyle 642$

535 $\displaystyle 535$

214 $\displaystyle 214$

321 $\displaystyle 321$

107 $\displaystyle 107$

Correct answer:

642 $\displaystyle 642$

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 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

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

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Example Question #722 : Rate

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:

$\displaystyle 74$

222 $\displaystyle 222$

444 $\displaystyle 444$

148 $\displaystyle 148$

111 $\displaystyle 111$

Correct answer:

444 $\displaystyle 444$

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 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

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

Report an Error

Example Question #721 : 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 the area of a single face when its sides have length $\frac{13}{6}$ $\displaystyle \frac{13}{6}$ ?

Possible Answers:

$\frac{13}{2}$ $\displaystyle \frac{13}{2}$

$\frac{13}{24}$ $\displaystyle \frac{13}{24}$

$\frac{13}{6}$ $\displaystyle \frac{13}{6}$

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

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

Correct answer:

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

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 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

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

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Calculus 1 : How to find rate of change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #631 : Rate Of Change

Example Question #3541 : Calculus

Example Question #3542 : Calculus

Example Question #3543 : Calculus

Example Question #631 : Rate Of Change

Example Question #2519 : Functions

Example Question #631 : Rate Of Change

Example Question #721 : Rate

Example Question #722 : Rate

Example Question #721 : Rate

All Calculus 1 Resources

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