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Calculus 1 : Calculus

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Example Questions

Example Question #318 : Rate Of Change

Differentiate the following function:

$f(x)=e^{x^2+cosx}$

Possible Answers:

$(2x+sinx)e^{x}$

$(2x+cosx)e^{x^{2}-sinx}$

$e^{2x-sinx}$

$(2x-sinx)e^{x^{2}+cosx}$

$(2x+cosx)e^{x}$

Correct answer:

$(2x-sinx)e^{x^{2}+cosx}$

Explanation:

To take the derivative of a composite function of the form f(x)=f(g(x)) one must use the chain rule which states: f'(x)=g'(x)f'(g(x))

The derivative of the 'inner function' for this question would be 2x-Sinx while the outer function stays the same since the derivative of $e^{x}$ is itself.

From this we find that the derivative is

$(2x-Sinx)e^{x^2+Cosx}$

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Example Question #319 : Rate Of Change

A cube is growing in size. What is the length of the sides of the cube at the time that the rate of growth of the cube's volume is equal to 0.56 times the rate of growth of its surface area?

Possible Answers:

5.60

4.48

3.36

2.24

1.12

Correct answer:

2.24

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

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:

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

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to 0.56 times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(0.56)12s\frac{ds}{dt}$

s=(0.56)4=2.24

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Example Question #320 : Rate Of Change

A cube is diminishing in size. What is the length of the diagonal of the cube at the time that the rate of shrinkage of the cube's volume is equal to 0.34 times the rate of shrinkage of its surface area?

Possible Answers:

1.36

2.80

4.12

1.97

2.36

Correct answer:

2.36

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

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:

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

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's volume is equal to 0.34 times the rate of shrinkage of its surface area:

$3s^2\frac{ds}{dt}=(0.34)12s\frac{ds}{dt}$

s=(0.34)4=1.36

The diagonal is then found from this:

$d=s\sqrt{3}$

$d=1.36\sqrt{3}=2.36$

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Example Question #321 : Rate Of Change

A cube is diminishing in size. What is the volume of the cube at the time that the rate of shrinkage of the cube's volume is equal to 9.3 times the rate of shrinkage of its diagonal?

Possible Answers:

9.72

3.10

12.44

6.18

11.38

Correct answer:

12.44

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, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's volume is equal to 9.3 times the rate of shrinkage of its diagonal:

$3s^2\frac{ds}{dt}=(9.3)\sqrt{3}\frac{ds}{dt}$

$s^2=(9.3)\frac{\sqrt{3}}{3}$

$s=\sqrt{3.1\sqrt{3}}}=2.317$

Then to find the volume:

V=s^3

V=2.317^3=12.44

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Example Question #322 : Rate Of Change

A cube is growing in size. What is the surface area of the cube at the time that the rate of growth of the cube's volume is equal to 1.91 times the rate of growth of its diagonal?

Possible Answers:

6.24

4.33

6.62

5.88

6.91

Correct answer:

6.62

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, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to 1.91 times the rate of growth of its diagonal:

$3s^2\frac{ds}{dt}=(1.91)\sqrt{3}\frac{ds}{dt}$

$s^2=(1.91)\frac{\sqrt{3}}{3}$

$s=\sqrt{\frac{1.91\sqrt{3}}{3}}=1.050$

The surface area is then:

A=6s^2

A=6(1.050)^2=6.62

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Example Question #323 : Rate Of Change

A cube is growing in size. What is the length of the sides of the cube at the time that the rate of growth of the cube's surface area is equal to 4.3 times the rate of growth of its diagonal?

Possible Answers:

5.19

4.41

3.12

1.08

0.62

Correct answer:

0.62

Explanation:

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

A=6s^2

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's surface area is equal to 4.3 times the rate of growth of its diagonal:

$12s\frac{ds}{dt}=(4.3)\sqrt{3}\frac{ds}{dt}$

$s=(4.3)\frac{\sqrt{3}}{12}=0.62$

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Example Question #324 : Rate Of Change

A cube is diminishing in size. What is the area of a face of the cube at the time that the rate of shrinkage of the cube's surface area is equal to 36 times the rate of shrinkage of its diagonal?

Possible Answers:

162

Correct answer:

Explanation:

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

A=6s^2

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's surface area is equal to 36 times the rate of shrinkage of its diagonal:

$12s\frac{ds}{dt}=(36)\sqrt{3}\frac{ds}{dt}$

$s=(36)\frac{\sqrt{3}}{12}=3\sqrt{3}$

Finding the area of a face:

a=s^2

a=27

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Example Question #325 : 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 surface area when its sides have length 0.72?

Possible Answers:

0.18

0.54

0.36

0.27

0.09

Correct answer:

0.18

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

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:

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

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

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

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

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

$\phi=\frac{0.72}{4}$

$\phi =0.18$

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Example Question #326 : 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 14.4?

Possible Answers:

9.6

3.6

4.8

7.2

14.4

Correct answer:

3.6

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

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:

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

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

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

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

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

$\phi=\frac{14.4}{4}$

$\phi =3.6$

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Example Question #327 : 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 $\sqrt[4]{27}$ ?

Possible Answers:

$3\sqrt{3}$

$\sqrt{3}$

$9\sqrt{3}$

Correct answer:

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:

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

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

$\phi =(\sqrt[4]{27})^2\sqrt{3}=9$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #318 : Rate Of Change

Example Question #319 : Rate Of Change

Example Question #320 : Rate Of Change

Example Question #321 : Rate Of Change

Example Question #322 : Rate Of Change

Example Question #323 : Rate Of Change

Example Question #324 : Rate Of Change

Example Question #325 : Rate Of Change

Example Question #326 : Rate Of Change

Example Question #327 : Rate Of Change

All Calculus 1 Resources

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