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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #881 : Rate

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

Possible Answers:

56644

679728

2856

97104

1666

Correct answer:

2856

Explanation:

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

A=6s^2

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

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

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 7 and a rate of growth of 34:

$\frac{dA}{dt}=12(7)(34)=2856$

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Example Question #882 : Rate

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

Possible Answers:

22050

7350

44100

2520

1260

Correct answer:

2520

Explanation:

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

A=6s^2

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

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

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 6 and a rate of growth of 35:

$\frac{dA}{dt}=12(6)(35)=2520$

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Example Question #883 : Rate

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

Possible Answers:

10800

180

2160

5400

900

Correct answer:

2160

Explanation:

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

A=6s^2

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

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

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 5 and a rate of growth of 36:

$\frac{dA}{dt}=12(5)(36)=2160$

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Example Question #884 : Rate

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

Possible Answers:

1184

592

296

14208

1776

Correct answer:

1776

Explanation:

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

A=6s^2

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

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

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 4 and a rate of growth of 37:

$\frac{dA}{dt}=12(4)(37)=1776$

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Example Question #885 : Rate

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

Possible Answers:

1368

342

4104

2052

684

Correct answer:

1368

Explanation:

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

A=6s^2

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

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

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 3 and a rate of growth of 38:

$\frac{dA}{dt}=12(3)(38)=1368$

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Example Question #886 : Rate

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

Possible Answers:

936

312

Correct answer:

936

Explanation:

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

A=6s^2

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

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

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 2 and a rate of growth of 39:

$\frac{dA}{dt}=12(2)(39)=936$

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Example Question #887 : Rate

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

Possible Answers:

240

480

960

Correct answer:

480

Explanation:

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

A=6s^2

The rate of change of the surface area can be found by taking the derivative of each side of the equation with respect to time:

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

Now that we have a relationship between the surface area and the side parameters, we can use what we were told about the cube, in particular that its sides have a length of 1 and a rate of growth of 40:

$\frac{dA}{dt}=12(1)(40)=480$

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Example Question #888 : Rate

Find the rate of change of y=sec(2x) from $x:(0, \pi)$

Possible Answers:

$\frac{-2}{\pi}$

$\frac{1}{\pi}$

$\frac{2}{\pi}$

Correct answer:

Explanation:

To do rate of change , remember it is equivalent to finding slope.

$m=\frac{f(\pi)-f(0)}{\pi-0}=\frac{sec(2\pi)-sec(0)}{\pi-0} =\frac{1-1}{\pi-0} = 0$

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Example Question #889 : Rate

The position of a car is defined by the equation $r(t)=5t^{2}-7t+6$ . What is the average velocity of the car between t=2 and t=4 ?

Possible Answers:

$v_{average}=20$

$v_{average}=0$

$v_{average}=29$

$v_{average}=58$

Correct answer:

$v_{average}=29$

Explanation:

The average velocity of an object between t=a and t=b is given by the equation

$v_{average}=\frac{\Delta y}{\Delta x}=\frac{r(b)-r(a)}{b-a}$

In this problem,

$r(4)=5*(4)^{2}-7(4)+6=80-28+6=58$

$r(2)=5*(2)^{2}-7(2)+6=20-14+6=0$

$v_{average}=\frac{r(4)-r(2)}{4-2}=\frac{58-0}{4-2}=29$

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Example Question #801 : Rate Of Change

The position of a car is defined by the equation $r(t)=5t^{2}-7t+6$ . What is the instantaneous velocity of the car at t=4 ?

Possible Answers:

$v_{ins}=0$

$v_{ins}=30$

$v_{ins}=-4$

$v_{ins}=33$

Correct answer:

$v_{ins}=33$

Explanation:

The instantaneous velocity of the car is the first derivative of the position at a given point.

In this problem,

$r(t)=5t^{2}-7t+6$

$v_{ins}=r'(t)=10t-7$

$v_{ins}=r'(4)=10*4-7=40-7=33$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #881 : Rate

Example Question #882 : Rate

Example Question #883 : Rate

Example Question #884 : Rate

Example Question #885 : Rate

Example Question #886 : Rate

Example Question #887 : Rate

Example Question #888 : Rate

Example Question #889 : Rate

Example Question #801 : Rate Of Change

All Calculus 1 Resources

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