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Calculus 1 : How to find rate of change

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Example Questions

Example Question #461 : Rate Of Change

Find the slope at x=0 given the following function: f(x)= (x^4 + ln(x) - 3)^2

Possible Answers:

254.5

890.1

45.8

24.8

Answer not listed

Correct answer:

890.1

Explanation:

In order to find the slope of a certain point, you first find the derivative of the following function: f(x)= (x^4 + ln(x) - 3)^2

The derivative is: $f'(x)= 2(x^4 + ln(x) - 3) (4x^{3} + \frac{1}{x} )$

Then, plug x= 2 into the derivative: $f'(2)= 2({\color{Blue} 2}^4 + ln({\color{Blue} 2}) - 3) (4({\color{Blue} 2})^{3} + \frac{1}{{\color{Blue} 2}} )$

Therefore, the slope is: 890.1

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Example Question #462 : Rate Of Change

Find the slope at x=0 given the following function: $f(x)= \sin(2x^4 - e^{2x})$

Possible Answers:

1.476

23.7

-.004

Answer not listed

3.41

Correct answer:

Answer not listed

Explanation:

In order to find the slope of a certain point, you first find the derivative of the following function: $f(x)= \sin(2x^4 - e^{2x})$

The derivative is: $f'(x)= \cos(2x^4 - e^{2x}) (8x^{3} - 2e^{2x})$

Then, plug x= 0 into the derivative: $f'(0)= \cos(2({\color{Blue} 0})^4 - e^{2({\color{Blue} 0})}) (8({\color{Blue} 0})^{3} - 2e^{2({\color{Blue} 0})})$

Therefore, the slope is: -1.08

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Example Question #463 : Rate Of Change

Find the slope at x=1 given the following function: $f(x)= e^{4x^5 + ln(x)}$

Possible Answers:

23.597

9.245

513.5

1146.6

Answer not listed

Correct answer:

1146.6

Explanation:

In order to find the slope of a certain point, you first find the derivative of the following function: $f(x)= e^{4x^5 + ln(x)}$

The derivative is: $f'(x)= e^{4x^5 + ln(x)} (20x^{4} + \frac{1}{x})$

Then, plug x= 1 into the derivative: $f'(x)= e^{4({\color{Blue} 1})^5 + ln({\color{Blue} 1})} (20({\color{Blue} 1})^{4} + \frac{1}{{\color{Blue} 1}})$

Therefore, the slope is: 1146.6

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Example Question #464 : Rate Of Change

Find the slope at x=0 given the following function: $f(x)= -\sin( 2x+ e^{2x})$

Possible Answers:

4.94

9.24

-7.81

3.96

Answer not listed

Correct answer:

3.96

Explanation:

In order to find the slope of a certain point, you first find the derivative of the following function: $f(x)= -\sin( 2x+ e^{2x})$

The derivative is: $f'(x)= -\cos( 2x+ e^{2x}) (2 + 2e^{2x})$

Then, plug x= 0 into the derivative: $f'(0)= -\cos( 2({\color{Blue} 0})+ e^{2({\color{Blue} 0})}) (2 + 2e^{2({\color{Blue} 0})})$

Therefore, the slope is: 3.96

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Example Question #465 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's surface area to the rate of growth of its diagonal when its sides have length $7\sqrt{2}$ ?

Possible Answers:

$28\sqrt{2}$

$7\sqrt{6}$

$14\sqrt{2}$

$14\sqrt{3}$

$28\sqrt{6}$

Correct answer:

$28\sqrt{6}$

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

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

$\phi=4s\sqrt{3}$

$\phi =4(7\sqrt{2})\sqrt{3}=28\sqrt{6}$

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Example Question #466 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's surface area to the rate of growth of its diagonal when its sides have length 11?

Possible Answers:

$44\sqrt{3}$

$11\sqrt{3}$

$11\sqrt{2}$

$44\sqrt{6}$

$22\sqrt{6}$

Correct answer:

$44\sqrt{3}$

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

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

$\phi=4s\sqrt{3}$

$\phi =4(11)\sqrt{3}=44\sqrt{3}$

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Example Question #467 : Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's surface area to the rate of growth of its diagonal when its sides have length $5\sqrt{6}$ ?

Possible Answers:

$30\sqrt{3}$

$20\sqrt{6}$

$30\sqrt{2}$

$60\sqrt{3}$

$60\sqrt{2}$

Correct answer:

$60\sqrt{2}$

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

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

$\phi=4s\sqrt{3}$

$\phi =4(5\sqrt{6})\sqrt{3}=60\sqrt{2}$

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Example Question #468 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its diagonal when its sides have length $7\sqrt{3}$ ?

Possible Answers:

$84\sqrt{3}$

$28\sqrt{3}$

$28\sqrt{6}$

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

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

$\phi=4s\sqrt{3}$

$\phi =4(7\sqrt{3})\sqrt{3}=84$

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Example Question #469 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its diagonal when its sides have length $\frac{\sqrt{3}}{2}$ ?

Possible Answers:

$\frac{3}{2}$

$2\sqrt{3}$

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

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

$\phi=4s\sqrt{3}$

$\phi =4(\frac{\sqrt{3}}{2})\sqrt{3}=6$

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Example Question #470 : Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's surface area to the rate of loss of its diagonal when its sides have length 103?

Possible Answers:

$412\sqrt{6}$

$412\sqrt{3}$

$412\sqrt{2}$

$206\sqrt{3}$

$206\sqrt{6}$

Correct answer:

$412\sqrt{3}$

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

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

$\phi=4s\sqrt{3}$

$\phi =4(103)\sqrt{3}=412\sqrt{3}$

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Calculus 1 : How to find rate of change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #461 : Rate Of Change

Example Question #462 : Rate Of Change

Example Question #463 : Rate Of Change

Example Question #464 : Rate Of Change

Example Question #465 : Rate Of Change

Example Question #466 : Rate Of Change

Example Question #467 : Rate Of Change

Example Question #468 : Rate Of Change

Example Question #469 : Rate Of Change

Example Question #470 : Rate Of Change

All Calculus 1 Resources

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