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Calculus 1 : How to find rate of change

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Example Questions

Example Question #3534 : Calculus

Find the slope at x=7 given the following function:

$f(x)= e^{ln(x)}$

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative.

The function simplifies to: $f(x)= e^{ln(x)} =x$

In this case the derivative is: f'(x)= 1

Then plug x=7 into the derivative function: f'(7)= 1

Therefore, the slope is:

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Example Question #3535 : Calculus

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 112 times the rate of growth of its surface area?

Possible Answers:

448

112

224

Correct answer:

448

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 112 times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(112)12s\frac{ds}{dt}$

s=(112)4

s=448

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Example Question #3536 : Calculus

A cube is growing in size. What is the length of the diagonal of the cube at the time that the rate of growth of the cube's volume is equal to 127 times the rate of growth of its surface area?

Possible Answers:

254

$508\sqrt{3}$

$127\sqrt{3}$

$254\sqrt{3}$

508

Correct answer:

$508\sqrt{3}$

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 127 times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(127)12s\frac{ds}{dt}$

s=(127)4=508

The diagonal is then:

$d=s\sqrt{3}$

$d=508\sqrt{3}$

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Example Question #3537 : Calculus

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 95 times the rate of growth of its surface area?

Possible Answers:

1732800

1299600

288800

866400

144400

Correct answer:

866400

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 95 times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(95)12s\frac{ds}{dt}$

s=(95)4=380

The surface area is then:

A=6s^2

A=6(380)^2=866400

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Example Question #3538 : Calculus

A cube is diminishing in size. What is the length of the sides of the cube at the time that the rate of shrinkage of the cube's volume is equal to $\frac{9}{10}$ times the rate of shrinkage of its surface area?

Possible Answers:

$\frac{9}{5}$

$\frac{3}{5}$

$\frac{36}{5}$

$\frac{18}{5}$

$\frac{72}{5}$

Correct answer:

$\frac{18}{5}$

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 $\frac{9}{10}$ times the rate of shrinkage of its surface area:

$3s^2\frac{ds}{dt}=(\frac{9}{10})12s\frac{ds}{dt}$

$s=(\frac{9}{10})4=\frac{18}{5}$

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Example Question #2505 : Functions

A cube is diminishing in size. What is volume of the cube at the time that the rate of shrinkage of the cube's volume is equal to $\frac{14}{15}$ times the rate of shrinkage of its surface area?

Possible Answers:

$\frac{97321}{625}$

$\frac{3136}{225}$

$\frac{175616}{3375}$

$\frac{108}{25}$

$\frac{9834496}{50625}$

Correct answer:

$\frac{175616}{3375}$

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 $\frac{14}{15}$ times the rate of shrinkage of its surface area

$3s^2\frac{ds}{dt}=(\frac{14}{15})12s\frac{ds}{dt}$

$s=(\frac{14}{15})4=\frac{56}{15}$

The volume is then:

V=s^3

$V=(\frac{56}{15})^3=\frac{175616}{3375}$

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Example Question #2511 : Functions

A cube is diminishing in size. What is the surface area of the cube at the time that the rate of shrinkage of the cube's volume is equal to $\frac{5}{6}$ times the rate of shrinkage of its surface area?

Possible Answers:

$\frac{400}{3}$

$\frac{400}{9}$

$\frac{100}{3}$

$\frac{200}{9}$

$\frac{200}{3}$

Correct answer:

$\frac{200}{3}$

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 $\frac{5}{6}$ times the rate of shrinkage of its surface area

$3s^2\frac{ds}{dt}=(\frac{5}{6})12s\frac{ds}{dt}$

$s=(\frac{5}{6})4=\frac{10}{3}$

The surface area at this time is then:

A=6s^2

$A=6(\frac{10}{3})^2=\frac{200}{3}$

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Example Question #2512 : Functions

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 1164?

Possible Answers:

291

485

388

194

Correct answer:

291

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:

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

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

$\phi =\frac{1164}{4}=291$

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Example Question #2513 : Functions

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 620?

Possible Answers:

310

186

155

930

Correct answer:

155

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:

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

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

$\phi =\frac{620}{4}=155$

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Example Question #2514 : Functions

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 1656?

Possible Answers:

6624

207

552

414

Correct answer:

414

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:

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

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

$\phi =\frac{1656}{4}=414$

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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 #3534 : Calculus

Example Question #3535 : Calculus

Example Question #3536 : Calculus

Example Question #3537 : Calculus

Example Question #3538 : Calculus

Example Question #2505 : Functions

Example Question #2511 : Functions

Example Question #2512 : Functions

Example Question #2513 : Functions

Example Question #2514 : Functions

All Calculus 1 Resources

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