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Calculus 1 : Functions

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Example Questions

Example Question #763 : How To Find 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 3 and a rate of growth of 15?

Possible Answers:

$\displaystyle 90$

2025 $\displaystyle 2025$

270 $\displaystyle 270$

$\displaystyle 45$

4050 $\displaystyle 4050$

Correct answer:

$\displaystyle 90$

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 15:

$\frac{da}{dt}=2(3)(15)=90$ $\displaystyle \frac{da}{dt}=2(3)(15)=90$

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Example Question #2651 : Functions

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 9?

Possible Answers:

648 $\displaystyle 648$

$\displaystyle 18$

324 $\displaystyle 324$

$\displaystyle 72$

$\displaystyle 36$

Correct answer:

$\displaystyle 36$

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 9:

$\frac{da}{dt}=2(2)(9)=36$ $\displaystyle \frac{da}{dt}=2(2)(9)=36$

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Example Question #769 : How To Find 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 1 and a rate of growth of 31?

Possible Answers:

124 $\displaystyle 124$

248 $\displaystyle 248$

$\displaystyle 62$

1922 $\displaystyle 1922$

961 $\displaystyle 961$

Correct answer:

$\displaystyle 62$

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 1 and a rate of growth of 31:

$\frac{da}{dt}=2(1)(31)=62$ $\displaystyle \frac{da}{dt}=2(1)(31)=62$

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Example Question #761 : 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 2 and a rate of growth of 14?

Possible Answers:

$\displaystyle 56$

$\displaystyle 28$

392 $\displaystyle 392$

112 $\displaystyle 112$

784 $\displaystyle 784$

Correct answer:

$\displaystyle 56$

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 14:

$\frac{da}{dt}=2(2)(14)=56$ $\displaystyle \frac{da}{dt}=2(2)(14)=56$

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Example Question #3681 : 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 4 and a rate of growth of 10?

Possible Answers:

320 $\displaystyle 320$

$\displaystyle 80$

3200 $\displaystyle 3200$

$\displaystyle 20$

800 $\displaystyle 800$

Correct answer:

$\displaystyle 80$

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 4 and a rate of growth of 10:

$\frac{da}{dt}=2(4)(10)=80$ $\displaystyle \frac{da}{dt}=2(4)(10)=80$

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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:

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

Report an Error

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:

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

Report an Error

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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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #763 : How To Find Rate Of Change

Example Question #2651 : Functions

Example Question #769 : How To Find Rate Of Change

Example Question #761 : Rate Of Change

Example Question #3681 : Calculus

Example Question #771 : Rate Of Change

Example Question #3682 : Calculus

Example Question #3683 : Calculus

Example Question #3684 : Calculus

Example Question #3685 : Calculus

All Calculus 1 Resources

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