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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #623 : Rate

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 15 times the rate of growth of its diagonal?

Possible Answers:

$15\sqrt{3}$

$\frac{5\sqrt{3}}{4}$

$\frac{5\sqrt{3}}{2}$

$\frac{15\sqrt{3}}{4}$

$\frac{4\sqrt{3}}{5}$

Correct answer:

$\frac{5\sqrt{3}}{4}$

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 15 times the rate of growth of its diagonal:

$12s\frac{ds}{dt}=(15)\sqrt{3}\frac{ds}{dt}$

$s=(15)\frac{\sqrt{3}}{12}=\frac{5\sqrt{3}}{4}$

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Example Question #3451 : 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 surface area is equal to 4 times the rate of growth of its diagonal?

Possible Answers:

$\sqrt{3}$

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, 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 times the rate of growth of its diagonal

$12s\frac{ds}{dt}=(4)\sqrt{3}\frac{ds}{dt}$

$s=(4)\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{3}$

The surface area is then:

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

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Example Question #2422 : Functions

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 surface area is equal to 24 times the rate of growth of its diagonal?

Possible Answers:

$12\sqrt{3}$

$8\sqrt{3}$

$24\sqrt{3}$

Correct answer:

$24\sqrt{3}$

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, 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 24 times the rate of growth of its diagonal

$12s\frac{ds}{dt}=(24)\sqrt{3}\frac{ds}{dt}$

$s=(24)\frac{\sqrt{3}}{12}=2\sqrt{3}$

Then to find the volume:

V=s^3

$V=(2\sqrt{3})^3=24\sqrt{3}$

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Example Question #541 : 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 72 times the rate of growth of its diagonal?

Possible Answers:

$2\sqrt{3}$

$6\sqrt{3}$

$24\sqrt{3}$

$36\sqrt{3}$

$12\sqrt{3}$

Correct answer:

$6\sqrt{3}$

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, 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 72 times the rate of growth of its diagonal:

$12s\frac{ds}{dt}=(72)\sqrt{3}\frac{ds}{dt}$

$s=(72)\frac{\sqrt{3}}{12}=6\sqrt{3}$

Report an Error

Example Question #542 : Rate Of Change

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 surface area is equal to $\frac{1}{3}$ times the rate of shrinkage of its diagonal?

Possible Answers:

$\frac{\sqrt{3}}{12}$

$\frac{1}{36}$

$\frac{\sqrt{3}}{18}$

$\frac{\sqrt{3}}{36}$

$\frac{1}{12}$

Correct answer:

$\frac{\sqrt{3}}{36}$

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 $\frac{1}{3}$ times the rate of shrinkage of its diagonal:

$12s\frac{ds}{dt}=(\frac{1}{3})\sqrt{3}\frac{ds}{dt}$

$s=(\frac{1}{3})\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{36}$

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Example Question #543 : 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 surface area is equal to $\frac{1}{5}$ times the rate of shrinkage of its diagonal?

Possible Answers:

$\frac{\sqrt{3}}{20}$

$\frac{\sqrt{3}}{10}$

$\frac{\sqrt{3}}{60}$

$\frac{1}{20}$

$\frac{1}{60}$

Correct answer:

$\frac{1}{20}$

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 $\frac{1}{5}$ times the rate of shrinkage of its diagonal:

$12s\frac{ds}{dt}=(\frac{1}{5})\sqrt{3}\frac{ds}{dt}$

$s=(\frac{1}{5})\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{60}$

The length of the diagonal is then:

$d=(\frac{\sqrt{3}}{60})\sqrt{3}=\frac{1}{20}$

Report an Error

Example Question #544 : 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 surface area is equal to 12 times the rate of growth of its diagonal?

Possible Answers:

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 growth of the cube's surface area is equal to 12 times the rate of growth of its diagonal:

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

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

The area of a face is then:

a=s^2

$a=(\sqrt{3})^2=3$

Report an Error

Example Question #2421 : Functions

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 $\frac{3}{4}$ times the rate of shrinkage of its diagonal?

Possible Answers:

$\frac{3}{256}$

$\frac{9}{256}$

$\frac{3}{128}$

$\frac{9}{64}$

$\frac{9}{128}$

Correct answer:

$\frac{3}{256}$

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 $\frac{3}{4}$ times the rate of shrinkage of its diagonal:

$12s\frac{ds}{dt}=(\frac{3}{4})\sqrt{3}\frac{ds}{dt}$

$s=(\frac{3}{4})\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{16}$

An area of a face is then:

a=s^2

$a=(\frac{\sqrt{3}}{16})^2=\frac{3}{256}$

Report an Error

Example Question #546 : Rate Of Change

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 surface area is equal to $\frac{1}{2}$ times the rate of shrinkage of its diagonal?

Possible Answers:

$\frac{1}{8}$

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{16}$

$\frac{1}{32}$

Correct answer:

$\frac{1}{32}$

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 $\frac{1}{2}$ times the rate of shrinkage of its diagonal:

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

$s=(\frac{1}{2})\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{24}$

The surface area is then:

$A=6(\frac{\sqrt{3}}{24})^2=\frac{1}{32}$

Report an Error

Example Question #631 : Rate

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 surface area is equal to $\frac{15}{16}$ times the rate of shrinkage of its diagonal?

Possible Answers:

$\frac{125\sqrt{3}}{262144}$

$\frac{25\sqrt{3}}{64}$

$\frac{375\sqrt{3}}{262144}$

$\frac{125\sqrt{3}}{64}$

$\frac{375\sqrt{3}}{64}$

Correct answer:

$\frac{375\sqrt{3}}{262144}$

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 $\frac{15}{16}$ times the rate of shrinkage of its diagonal:

$12s\frac{ds}{dt}=(\frac{15}{16})\sqrt{3}\frac{ds}{dt}$

$s=(\frac{15}{16})\frac{\sqrt{3}}{12}=\frac{5\sqrt{3}}{64}$

The volume is then:

V=s^3

$V=(\frac{5\sqrt{3}}{64})^3=\frac{375\sqrt{3}}{262144}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #623 : Rate

Example Question #3451 : Calculus

Example Question #2422 : Functions

Example Question #541 : Rate Of Change

Example Question #542 : Rate Of Change

Example Question #543 : Rate Of Change

Example Question #544 : Rate Of Change

Example Question #2421 : Functions

Example Question #546 : Rate Of Change

Example Question #631 : Rate

All Calculus 1 Resources

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