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Calculus 1 : Rate

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Example Questions

Example Question #771 : Rate Of Change

A cube is growing in size. What is the rate of growth of one of the cube's faces if its sides have a length of 22 and a rate of growth of 2?

Possible Answers:

968 $\displaystyle 968$

$\displaystyle 88$

$\displaystyle 44$

528 $\displaystyle 528$

1936 $\displaystyle 1936$

Correct answer:

$\displaystyle 88$

Explanation:

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

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

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

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

Once we have the rate equation for the area of the face, we can use what we know about the cube, specifically that its sides have a length of 22 and a rate of growth of 2

$\frac{da}{dt}=2(22)(2)=88$ $\displaystyle \frac{da}{dt}=2(22)(2)=88$

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Example Question #3682 : Calculus

A cube is growing in size. What is the rate of growth of one of the cube's faces if its sides have a length of 2 and a rate of growth of 23?

Possible Answers:

529 $\displaystyle 529$

1104 $\displaystyle 1104$

2116 $\displaystyle 2116$

$\displaystyle 92$

552 $\displaystyle 552$

Correct answer:

$\displaystyle 92$

Explanation:

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

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

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

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

Once we have the rate equation for the area of the face, we can use what we know about the cube, specifically that its sides have a length of 2 and a rate of growth of 23:

$\frac{da}{dt}=2(2)(23)=92$ $\displaystyle \frac{da}{dt}=2(2)(23)=92$

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Example Question #3683 : Calculus

A cube is growing in size. What is the rate of growth of one of the cube's faces if its sides have a length of 3 and a rate of growth of 25?

Possible Answers:

3750 $\displaystyle 3750$

1875 $\displaystyle 1875$

150 $\displaystyle 150$

900 $\displaystyle 900$

450 $\displaystyle 450$

Correct answer:

150 $\displaystyle 150$

Explanation:

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

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

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

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

Once we have the rate equation for the area of the face, we can use what we know about the cube, specifically that its sides have a length of 3 and a rate of growth of 25:

$\frac{da}{dt}=2(3)(25)=150$ $\displaystyle \frac{da}{dt}=2(3)(25)=150$

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Example Question #3684 : Calculus

A cube is growing in size. What is the rate of growth of the cube's volume if its sides have a length of 1 and a rate of growth of 40?

Possible Answers:

2400 $\displaystyle 2400$

4800 $\displaystyle 4800$

1600 $\displaystyle 1600$

$\displaystyle 40$

120 $\displaystyle 120$

Correct answer:

120 $\displaystyle 120$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

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

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

Now with this known, we can solve for the rate of change of the volume of the cube knowing the condition of cube, in particular that its sides have a length of 1 and a rate of growth of 40:

$\frac{dV}{dt}=3(1)^2(40)=120$ $\displaystyle \frac{dV}{dt}=3(1)^2(40)=120$

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Example Question #3685 : Calculus

A cube is growing in size. What is the rate of growth of the cube's volume if its sides have a length of 2 and a rate of growth of 39?

Possible Answers:

234 $\displaystyle 234$

468 $\displaystyle 468$

$\displaystyle 78$

$\displaystyle 39$

156 $\displaystyle 156$

Correct answer:

468 $\displaystyle 468$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

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

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

Now with this known, we can solve for the rate of change of the volume of the cube knowing the condition of cube, in particular that its sides have a length of 2 and a rate of growth of 39:

$\frac{dV}{dt}=3(2)^2(39)=468$ $\displaystyle \frac{dV}{dt}=3(2)^2(39)=468$

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Example Question #861 : Rate

A cube is growing in size. What is the rate of growth of the cube's volume if its sides have a length of 3 and a rate of growth of 38?

Possible Answers:

1026 $\displaystyle 1026$

171 $\displaystyle 171$

$\displaystyle 57$

114 $\displaystyle 114$

342 $\displaystyle 342$

Correct answer:

1026 $\displaystyle 1026$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

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

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

Now with this known, we can solve for the rate of change of the volume of the cube knowing the condition of cube, in particular that its sides have a length of 3 and a rate of growth of 38:

$\frac{dV}{dt}=3(3)^2(38)=1026$ $\displaystyle \frac{dV}{dt}=3(3)^2(38)=1026$

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Example Question #771 : How To Find Rate Of Change

A cube is growing in size. What is the rate of growth of the cube's volume if its sides have a length of 4 and a rate of growth of 37?

Possible Answers:

1776 $\displaystyle 1776$

296 $\displaystyle 296$

2368 $\displaystyle 2368$

4736 $\displaystyle 4736$

592 $\displaystyle 592$

Correct answer:

1776 $\displaystyle 1776$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

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

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

Now with this known, we can solve for the rate of change of the volume of the cube knowing the condition of cube, in particular that its sides have a length of 4 and a rate of growth of 37:

$\frac{dV}{dt}=3(4)^2(37)=1776$ $\displaystyle \frac{dV}{dt}=3(4)^2(37)=1776$

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Example Question #772 : How To Find Rate Of Change

A cube is growing in size. What is the rate of growth of the cube's volume if its sides have a length of 5 and a rate of growth of 36?

Possible Answers:

540 $\displaystyle 540$

180 $\displaystyle 180$

900 $\displaystyle 900$

2700 $\displaystyle 2700$

8100 $\displaystyle 8100$

Correct answer:

2700 $\displaystyle 2700$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

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

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

Now with this known, we can solve for the rate of change of the volume of the cube knowing the condition of cube, in particular that its sides have a length of 5 and a rate of growth of 36:

$\frac{dV}{dt}=3(5)^2(36)=2700$ $\displaystyle \frac{dV}{dt}=3(5)^2(36)=2700$

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Example Question #773 : How To Find Rate Of Change

A cube is growing in size. What is the rate of growth of the cube's volume if its sides have a length of 6 and a rate of growth of 35?

Possible Answers:

3780 $\displaystyle 3780$

7560 $\displaystyle 7560$

1260 $\displaystyle 1260$

630 $\displaystyle 630$

210 $\displaystyle 210$

Correct answer:

3780 $\displaystyle 3780$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

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

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

Now with this known, we can solve for the rate of change of the volume of the cube knowing the condition of cube, in particular that its sides have a length of 6 and a rate of growth of 35:

$\frac{dV}{dt}=3(6)^2(35)=3780$ $\displaystyle \frac{dV}{dt}=3(6)^2(35)=3780$

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Example Question #781 : How To Find Rate Of Change

A cube is growing in size. What is the rate of growth of the cube's volume if its sides have a length of 7 and a rate of growth of 34?

Possible Answers:

1666 $\displaystyle 1666$

238 $\displaystyle 238$

714 $\displaystyle 714$

4998 $\displaystyle 4998$

169932 $\displaystyle 169932$

Correct answer:

4998 $\displaystyle 4998$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

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

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

Now with this known, we can solve for the rate of change of the volume of the cube knowing the condition of cube, in particular that its sides have a length of 7 and a rate of growth of 34

$\frac{dV}{dt}=3(7)^2(34)=4998$ $\displaystyle \frac{dV}{dt}=3(7)^2(34)=4998$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #771 : Rate Of Change

Example Question #3682 : Calculus

Example Question #3683 : Calculus

Example Question #3684 : Calculus

Example Question #3685 : Calculus

Example Question #861 : Rate

Example Question #771 : How To Find Rate Of Change

Example Question #772 : How To Find Rate Of Change

Example Question #773 : How To Find Rate Of Change

Example Question #781 : How To Find Rate Of Change

All Calculus 1 Resources

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