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Calculus 1 : How to find rate of change

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Example Question #481 : How To Find 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 $\frac{1}{96}$ times the rate of growth of its surface area?

Possible Answers:

$\frac{1}{16}$

$\frac{1}{72}$

$\frac{1}{24}$

$\frac{1}{96}$

$\frac{1}{256}$

Correct answer:

$\frac{1}{96}$

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 $\frac{1}{96}$ times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(\frac{1}{96})12s\frac{ds}{dt}$

$s=(\frac{1}{96})4=\frac{1}{24}$

Then to find the surface area:

A=6s^2

$A=6(\frac{1}{24})^2=\frac{1}{96}$

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Example Question #482 : How To Find Rate Of Change

A cube is growing in size. What is the area of a face of the cube at the time that the rate of growth of the cube's volume is equal to $\frac{1}{16}$ times the rate of growth of its surface area?

Possible Answers:

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{1}{8}$

Correct answer:

$\frac{1}{16}$

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 $\frac{1}{16}$ times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(\frac{1}{16})12s\frac{ds}{dt}$

$s=(\frac{1}{16})4=\frac{1}{4}$

The area of a face of the cube is given by:

a=s^2

$a=(\frac{1}{4})^2=\frac{1}{16}$

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Example Question #483 : How To Find Rate Of Change

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 9 times the rate of growth of its sides?

Possible Answers:

$\frac{\sqrt{3}}{3}$

$3\sqrt{3}$

$9\sqrt{3}$

Correct answer:

Explanation:

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

V=s^3

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

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

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 9 times the rate of growth of its sides:

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

$s^2=\frac{(9)}{3}$

$s=\sqrt{3}$

The diagonal is given by

$d=s\sqrt{3}$

d=3

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Example Question #484 : How To Find 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 15 times the rate of growth of its sides?

Possible Answers:

$5\sqrt{6}$

$6\sqrt{5}$

Correct answer:

Explanation:

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

V=s^3

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

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

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 15 times the rate of growth of its sides:

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

$s^2=\frac{(15)}{3}$

$s=\sqrt{5}$

The surface area is then:

A=6s^2

$A=6(\sqrt{5})^2=30$

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Example Question #485 : How To Find 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 12 times the rate of growth of its sides?

Possible Answers:

$\frac{4\sqrt{3}}{3}$

$2\sqrt{3}$

$\frac{2\sqrt{3}}{3}$

Correct answer:

Explanation:

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

V=s^3

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

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

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 12 times the rate of growth of its sides:

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

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

$s=\sqrt{4}=2$

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Example Question #486 : How To Find Rate Of Change

A cube is growing in size. What is the area of a face of the cube at the time that the rate of growth of the cube's volume is equal to 3 times the rate of growth of its sides?

Possible Answers:

Correct answer:

Explanation:

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

V=s^3

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

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

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 3 times the rate of growth of its sides:

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

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

$s=\sqrt{1}=1$

The area of a face of the cube is then:

a=s^2

a=1

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Example Question #487 : How To Find Rate Of Change

A cube is growing in size. What is the volume of the cube at the time that the rate of growth of the cube's volume is equal to 75 times the rate of growth of its sides?

Possible Answers:

125

$\frac{25}{9}$

$\frac{125}{27}$

Correct answer:

125

Explanation:

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

V=s^3

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

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

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 75 times the rate of growth of its sides:

$3s^2\frac{ds}{dt}=(75)\frac{ds}{dt}$

$s^2=\frac{(75)}{3}$

$s=\sqrt{25}=5$

Then to find the volume:

V=s^3

V=5^3=125

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Example Question #3401 : 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 11 times the rate of growth of its sides?

Possible Answers:

$\sqrt{11}$

$\frac{\sqrt{11}}{3}$

$3\sqrt{11}$

$\sqrt{\frac{11}{3}}$

Correct answer:

$\sqrt{11}$

Explanation:

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

V=s^3

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

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

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 11 times the rate of growth of its sides:

$3s^2\frac{ds}{dt}=(11)\frac{ds}{dt}$

$s^2=\frac{(11)}{3}$

$s=\sqrt{\frac{11}{3}}$

The diagonal is then found as follows:

$d=s\sqrt{3}$

$d=(\sqrt{\frac{11}{3}})\sqrt{3}=\sqrt{11}$

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Example Question #3402 : Calculus

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the radius when the radius is $4\pi$ ?

Possible Answers:

$16\pi^2$

$64\pi$

$64\pi^2$

$64\pi^3$

$4\pi^2$

Correct answer:

$64\pi^3$

Explanation:

Let's begin by writing the equation for the volume of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

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

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and radius, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)\frac{dr}{dt}$

$\phi =4\pi r^2$

$\phi =4\pi(4\pi)^2=64\pi^3$

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Example Question #3403 : Calculus

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the radius when the radius is ?

Possible Answers:

$90\pi$

$60\pi$

$225\pi$

$450\pi$

$900\pi$

Correct answer:

$900\pi$

Explanation:

Let's begin by writing the equation for the volume of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

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

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and radius, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)\frac{dr}{dt}$

$\phi =4\pi r^2$

$\phi =4\pi(15)^2=900\pi$

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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 #481 : How To Find Rate Of Change

Example Question #482 : How To Find Rate Of Change

Example Question #483 : How To Find Rate Of Change

Example Question #484 : How To Find Rate Of Change

Example Question #485 : How To Find Rate Of Change

Example Question #486 : How To Find Rate Of Change

Example Question #487 : How To Find Rate Of Change

Example Question #3401 : Calculus

Example Question #3402 : Calculus

Example Question #3403 : Calculus

All Calculus 1 Resources

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