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Calculus 1 : Rate

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Example Questions

Example Question #2421 : Functions

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

Possible Answers:

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

$\frac{1}{36}$ $\displaystyle \frac{1}{36}$

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

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

$\frac{1}{12}$ $\displaystyle \frac{1}{12}$

Correct answer:

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

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 $\displaystyle A=6s^2$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

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Example Question #632 : Rate

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

Possible Answers:

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

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

$\frac{1}{60}$ $\displaystyle \frac{1}{60}$

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

$\frac{1}{20}$ $\displaystyle \frac{1}{20}$

Correct answer:

$\frac{1}{20}$ $\displaystyle \frac{1}{20}$

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 $\displaystyle A=6s^2$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

The length of the diagonal is then:

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

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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:

$\displaystyle 12$

$\displaystyle 3$

$\displaystyle 15$

$\displaystyle 6$

$\displaystyle 18$

Correct answer:

$\displaystyle 3$

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 $\displaystyle A=6s^2$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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:

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

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

The area of a face is then:

a=s^2 $\displaystyle a=s^2$

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

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Example Question #2422 : 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 $\frac{3}{4}$ $\displaystyle \frac{3}{4}$ times the rate of shrinkage of its diagonal?

Possible Answers:

$\frac{3}{128}$ $\displaystyle \frac{3}{128}$

$\frac{9}{256}$ $\displaystyle \frac{9}{256}$

$\frac{9}{128}$ $\displaystyle \frac{9}{128}$

$\frac{3}{256}$ $\displaystyle \frac{3}{256}$

$\frac{9}{64}$ $\displaystyle \frac{9}{64}$

Correct answer:

$\frac{3}{256}$ $\displaystyle \frac{3}{256}$

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 $\displaystyle A=6s^2$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

An area of a face is then:

a=s^2 $\displaystyle a=s^2$

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

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Example Question #2423 : 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 surface area is equal to $\frac{1}{2}$ $\displaystyle \frac{1}{2}$ times the rate of shrinkage of its diagonal?

Possible Answers:

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

$\frac{1}{32}$ $\displaystyle \frac{1}{32}$

$\frac{1}{8}$ $\displaystyle \frac{1}{8}$

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

$\frac{1}{16}$ $\displaystyle \frac{1}{16}$

Correct answer:

$\frac{1}{32}$ $\displaystyle \frac{1}{32}$

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 $\displaystyle A=6s^2$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

The surface area is then:

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

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Example Question #2431 : Functions

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\displaystyle A=6s^2$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

The volume is then:

V=s^3 $\displaystyle V=s^3$

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

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Example Question #2432 : Functions

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 $\frac{26}{3}$ $\displaystyle \frac{26}{3}$ times the rate of shrinkage of its diagonal?

Possible Answers:

$\frac{13\sqrt{3}}{6}$ $\displaystyle \frac{13\sqrt{3}}{6}$

$\frac{13\sqrt{3}}{9}$ $\displaystyle \frac{13\sqrt{3}}{9}$

$\frac{26\sqrt{3}}{13}$ $\displaystyle \frac{26\sqrt{3}}{13}$

$\frac{13\sqrt{3}}{18}$ $\displaystyle \frac{13\sqrt{3}}{18}$

$\frac{18\sqrt{3}}{13}$ $\displaystyle \frac{18\sqrt{3}}{13}$

Correct answer:

$\frac{13\sqrt{3}}{18}$ $\displaystyle \frac{13\sqrt{3}}{18}$

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 $\displaystyle A=6s^2$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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 $\frac{26}{3}$ $\displaystyle \frac{26}{3}$ times the rate of shrinkage of its diagonal:

$12s\frac{ds}{dt}=(\frac{26}{3})\sqrt{3}\frac{ds}{dt}$ $\displaystyle 12s\frac{ds}{dt}=(\frac{26}{3})\sqrt{3}\frac{ds}{dt}$

$s=(\frac{26}{3})\frac{\sqrt{3}}{12}=\frac{13\sqrt{3}}{18}$ $\displaystyle s=(\frac{26}{3})\frac{\sqrt{3}}{12}=\frac{13\sqrt{3}}{18}$

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Example Question #2433 : Functions

Find the slope at x=1 $\displaystyle x=1$ given the following function:

$f(x) = 2x^4 - \cos x$ $\displaystyle f(x) = 2x^4 - \cos x$

Possible Answers:

5.9 $\displaystyle 5.9$

2.4 $\displaystyle 2.4$

6.3 $\displaystyle 6.3$

Answer not listed

8.8 $\displaystyle 8.8$

Correct answer:

8.8 $\displaystyle 8.8$

Explanation:

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

In this case the derivative is: $f'(x) = 8x^{3} + \sin x$ $\displaystyle f'(x) = 8x^{3} + \sin x$

Then plug x=1 $\displaystyle x=1$ into the derivative function: $f'(1) = 8({\color{Blue} 1})^{3} + \sin ({\color{Blue} 1})$ $\displaystyle f'(1) = 8({\color{Blue} 1})^{3} + \sin ({\color{Blue} 1})$

Therefore, the slope is: 8.8 $\displaystyle 8.8$

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Example Question #2434 : Functions

Find the slope at x=1 $\displaystyle x=1$ given the following function:

$\displaystyle f(x) = 23+ e$

Possible Answers:

$\infty$ $\displaystyle \infty$

$\displaystyle 0$

$\displaystyle -1$

$\displaystyle 1$

Answer not listed

Correct answer:

$\displaystyle 0$

Explanation:

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

In this case the derivative is: $\displaystyle f'(x) = 0$

Then plug x=1 $\displaystyle x=1$ into the derivative function: $\displaystyle f'(1) = 0$

Therefore, the slope is: $\displaystyle 0$

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Example Question #631 : Rate

Find the slope at x=2 $\displaystyle x=2$ given the following function:

$\displaystyle f(x) = ln(7x +2)$

Possible Answers:

$\frac{3}{8}$ $\displaystyle \frac{3}{8}$

$\frac{2}{5}$ $\displaystyle \frac{2}{5}$

$\frac{7}{16}$ $\displaystyle \frac{7}{16}$

$\frac{9}{11}$ $\displaystyle \frac{9}{11}$

Answer not listed

Correct answer:

$\frac{7}{16}$ $\displaystyle \frac{7}{16}$

Explanation:

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

In this case the derivative is: $f'(x) = \frac{7}{7x+2}$ $\displaystyle f'(x) = \frac{7}{7x+2}$

Then plug x=2 $\displaystyle x=2$ into the derivative function: $f'(2) = \frac{7}{7({\color{Blue} 2})+2}$ $\displaystyle f'(2) = \frac{7}{7({\color{Blue} 2})+2}$

Therefore, the slope is: $\frac{7}{16}$ $\displaystyle \frac{7}{16}$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2421 : Functions

Example Question #632 : Rate

Example Question #544 : Rate Of Change

Example Question #2422 : Functions

Example Question #2423 : Functions

Example Question #2431 : Functions

Example Question #2432 : Functions

Example Question #2433 : Functions

Example Question #2434 : Functions

Example Question #631 : Rate

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