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Calculus 1 : Rate of Change

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Example Questions

Example Question #381 : How To Find Rate Of Change

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of the area of a single face when its sides have length 10?

Possible Answers:

$\displaystyle 15$

$\displaystyle 5$

$\displaystyle 10$

$\displaystyle 30$

$\displaystyle 20$

Correct answer:

$\displaystyle 15$

Explanation:

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

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

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

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

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

$\frac{dA}{dt}=2s\frac{ds}{dt}$ $\displaystyle \frac{dA}{dt}=2s\frac{ds}{dt}$

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

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

$\phi=\frac{3s}{2}$ $\displaystyle \phi=\frac{3s}{2}$

$\phi =\frac{3(10)}{2}=15$ $\displaystyle \phi =\frac{3(10)}{2}=15$

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Example Question #471 : Rate

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 diagonal when its sides have length $\sqrt[4]{54}$ $\displaystyle \sqrt[4]{54}$?

Possible Answers:

$3\sqrt{2}$ $\displaystyle 3\sqrt{2}$

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

$9\sqrt{2}$ $\displaystyle 9\sqrt{2}$

$6\sqrt{6}$ $\displaystyle 6\sqrt{6}$

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

Correct answer:

$9\sqrt{2}$ $\displaystyle 9\sqrt{2}$

Explanation:

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

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

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

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

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

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

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

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

$\phi=s^2\sqrt{3}$ $\displaystyle \phi=s^2\sqrt{3}$

$\phi =(\sqrt[4]{54})^2\sqrt{3}=9\sqrt{2}$ $\displaystyle \phi =(\sqrt[4]{54})^2\sqrt{3}=9\sqrt{2}$

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Example Question #2261 : Functions

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its diagonal when its sides have length $11\sqrt[4]{6}$ $\displaystyle 11\sqrt[4]{6}$?

Possible Answers:

$121\sqrt{2}$ $\displaystyle 121\sqrt{2}$

$363\sqrt{2}$ $\displaystyle 363\sqrt{2}$

$121\sqrt{3}$ $\displaystyle 121\sqrt{3}$

$121\sqrt{6}$ $\displaystyle 121\sqrt{6}$

$363\sqrt{3}$ $\displaystyle 363\sqrt{3}$

Correct answer:

$363\sqrt{2}$ $\displaystyle 363\sqrt{2}$

Explanation:

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

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

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

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

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

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

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

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

$\phi=s^2\sqrt{3}$ $\displaystyle \phi=s^2\sqrt{3}$

$\phi =(11\sqrt[4]{6})^2\sqrt{3}=363\sqrt{2}$ $\displaystyle \phi =(11\sqrt[4]{6})^2\sqrt{3}=363\sqrt{2}$

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Example Question #381 : How To Find Rate Of Change

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 sides when its sides have length 4?

Possible Answers:

$\displaystyle 48$

$\displaystyle 8$

$\displaystyle 16$

$\displaystyle 32$

$\displaystyle 64$

Correct answer:

$\displaystyle 48$

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 $\displaystyle V=s^3$

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

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

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

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

$\phi=3s^2$ $\displaystyle \phi=3s^2$

$\phi =3(4)^2=48$ $\displaystyle \phi =3(4)^2=48$

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Example Question #472 : Rate

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length 7?

Possible Answers:

$\displaystyle 21$

343 $\displaystyle 343$

147 $\displaystyle 147$

1026 $\displaystyle 1026$

$\displaystyle 49$

Correct answer:

147 $\displaystyle 147$

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 $\displaystyle V=s^3$

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

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

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

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

$\phi=3s^2$ $\displaystyle \phi=3s^2$

$\phi =3(7)^2=147$ $\displaystyle \phi =3(7)^2=147$

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Example Question #385 : How To Find 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 $\sqrt{48}$ $\displaystyle \sqrt{48}$?

Possible Answers:

$48\sqrt{2}$ $\displaystyle 48\sqrt{2}$

$\displaystyle 24$

$48\sqrt{3}$ $\displaystyle 48\sqrt{3}$

$\displaystyle 48$

$24\sqrt{3}$ $\displaystyle 24\sqrt{3}$

Correct answer:

$\displaystyle 48$

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

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

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

$\phi =4(\sqrt{48})\sqrt{3}=48$ $\displaystyle \phi =4(\sqrt{48})\sqrt{3}=48$

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Example Question #2271 : Functions

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 $12\sqrt{2}$ $\displaystyle 12\sqrt{2}$?

Possible Answers:

$24\sqrt{2}$ $\displaystyle 24\sqrt{2}$

$24\sqrt{3}$ $\displaystyle 24\sqrt{3}$

$48\sqrt{2}$ $\displaystyle 48\sqrt{2}$

$48\sqrt{6}$ $\displaystyle 48\sqrt{6}$

$48\sqrt{3}$ $\displaystyle 48\sqrt{3}$

Correct answer:

$48\sqrt{6}$ $\displaystyle 48\sqrt{6}$

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 $\displaystyle A=6s^2$

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

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

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

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

$\phi =4(12\sqrt{2})\sqrt{3}=48\sqrt{6}$ $\displaystyle \phi =4(12\sqrt{2})\sqrt{3}=48\sqrt{6}$

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Example Question #3301 : Calculus

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 sides when its sides have length 3?

Possible Answers:

$\displaystyle 36$

$\displaystyle 6$

$\displaystyle 9$

$\displaystyle 19$

$\displaystyle 3$

Correct answer:

$\displaystyle 36$

Explanation:

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

A=6s^2 $\displaystyle A=6s^2$

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

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

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

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

$\phi=12s$ $\displaystyle \phi=12s$

$\phi =12(3)=36$ $\displaystyle \phi =12(3)=36$

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Example Question #389 : 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 sides when its sides have length $\frac{1}{24}$ $\displaystyle \frac{1}{24}$?

Possible Answers:

$\frac{1}{3}$ $\displaystyle \frac{1}{3}$

$\frac{1}{6}$ $\displaystyle \frac{1}{6}$

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

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

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

Correct answer:

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

Explanation:

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

A=6s^2 $\displaystyle A=6s^2$

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

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

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

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

$\phi=12s$ $\displaystyle \phi=12s$

$\phi =12(\frac{1}{24})=\frac{1}{2}$ $\displaystyle \phi =12(\frac{1}{24})=\frac{1}{2}$

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Example Question #390 : Rate Of Change

Find the slope of the following function at x=2 $\displaystyle x=2$.

$f(x)= x^2 + \frac{1}{6}x^6$ $\displaystyle f(x)= x^2 + \frac{1}{6}x^6$

Possible Answers:

$\displaystyle 12$

$\displaystyle 36$

$\displaystyle 43$

$\displaystyle 87$

$\displaystyle 23$

Correct answer:

$\displaystyle 36$

Explanation:

In order to find the slope of a function, you must first differentiate that function.

In this case, the derivative of the given function is: $\displaystyle f'(x) = 2x + x^5$

Then, plug x=2 $\displaystyle x=2$ into the function to get the slope: $\displaystyle f'(2) = 2(2) + (2)^5$

Therefore, the slope is: $\displaystyle 36$

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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #381 : How To Find Rate Of Change

Example Question #471 : Rate

Example Question #2261 : Functions

Example Question #381 : How To Find Rate Of Change

Example Question #472 : Rate

Example Question #385 : How To Find Rate Of Change

Example Question #2271 : Functions

Example Question #3301 : Calculus

Example Question #389 : Rate Of Change

Example Question #390 : Rate Of Change

All Calculus 1 Resources

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