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Calculus 1 : Rate of Change

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Example Questions

Example Question #651 : 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 $2\sqrt{19}$ ?

Possible Answers:

$38\sqrt{6}$

$38\sqrt{3}$

$19\sqrt{6}$

$19\sqrt{3}$

$76\sqrt{3}$

Correct answer:

$76\sqrt{3}$

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

$d=s\sqrt{3}$

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:

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

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

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

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Example Question #652 : 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 $\frac{1}{5}$ ?

Possible Answers:

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

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

$\frac{3}{5}$

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

$\frac{3}{25}$

Correct answer:

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

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

$d=s\sqrt{3}$

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:

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

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

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

Report an Error

Example Question #651 : 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 $\frac{3}{4}$ ?

Possible Answers:

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

$\frac{9}{4}$

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

$\frac{3}{2}$

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

Correct answer:

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

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

$d=s\sqrt{3}$

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:

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

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

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

Report an Error

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 $\frac{5}{16}$ ?

Possible Answers:

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

$\frac{75}{256}$

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

$\frac{75}{128}$

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

Correct answer:

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

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

$d=s\sqrt{3}$

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:

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

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

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

Report an Error

Example Question #655 : 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 $\frac{7}{11}$ ?

Possible Answers:

$\frac{21}{121}$

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

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

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

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

Correct answer:

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

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

$d=s\sqrt{3}$

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:

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

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

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

Report an Error

Example Question #656 : 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 $\frac{2}{13}$ ?

Possible Answers:

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

$\frac{12}{169}$

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

$\frac{12}{13}$

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

Correct answer:

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

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

$d=s\sqrt{3}$

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:

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

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

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

Report an Error

Example Question #2541 : 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 $\frac{1}{14}$ ?

Possible Answers:

$\frac{3}{98}$

$\frac{3}{196}$

$\frac{\sqrt{3}}{196}$

$\frac{3}{14}$

$\frac{\sqrt{3}}{14}$

Correct answer:

$\frac{\sqrt{3}}{196}$

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:

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

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

$\phi =(\frac{1}{14})^2\sqrt{3}=\frac{\sqrt{3}}{196}$

Report an Error

Example Question #3571 : 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 sides when its sides have length 110?

Possible Answers:

1210

108900

36300

12100

990

Correct answer:

36300

Explanation:

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

V=s^3

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

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

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

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

$\phi=3s^2$

$\phi =3(110)^2=36300$

Report an Error

Example Question #652 : 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 sides when its sides have length 130?

Possible Answers:

5070

1170

50700

16900

1690

Correct answer:

50700

Explanation:

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

V=s^3

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

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

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

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

$\phi=3s^2$

$\phi =3(130)^2=50700$

Report an Error

Example Question #3573 : 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 sides when its sides have length 47?

Possible Answers:

19881

141

6627

423

2209

Correct answer:

6627

Explanation:

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

V=s^3

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

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

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

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

$\phi=3s^2$

$\phi =3(47)^2=6627$

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 #651 : Rate Of Change

Example Question #652 : Rate Of Change

Example Question #651 : Rate Of Change

Example Question #651 : How To Find Rate Of Change

Example Question #655 : Rate Of Change

Example Question #656 : Rate Of Change

Example Question #2541 : Functions

Example Question #3571 : Calculus

Example Question #652 : How To Find Rate Of Change

Example Question #3573 : Calculus

All Calculus 1 Resources

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