Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

varsity tutors app store varsity tutors android store

Example Questions

Example Question #282 : 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 surface area when its sides have length 13.6?

Possible Answers:

\displaystyle 3.4

\displaystyle 10.2

\displaystyle 15.3

\displaystyle 5.1

\displaystyle 6.8

Correct answer:

\displaystyle 3.4

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:

\displaystyle V=s^3

\displaystyle 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:

\displaystyle \frac{dV}{dt}=3s^2\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 volume and surface area:

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

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

\displaystyle \phi =3.4

Example Question #2163 : 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 80.4?

Possible Answers:

\displaystyle 40.2

\displaystyle 60.3

\displaystyle 5.1

\displaystyle 30.2

\displaystyle 20.1

Correct answer:

\displaystyle 20.1

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:

\displaystyle V=s^3

\displaystyle 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:

\displaystyle \frac{dV}{dt}=3s^2\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 volume and surface area:

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

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

\displaystyle \phi =20.1

Example Question #371 : 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 surface area when its sides have length 0.08?

Possible Answers:

\displaystyle 0.06

\displaystyle 0.16

\displaystyle 0.12

\displaystyle 0.02

\displaystyle 0.01

Correct answer:

\displaystyle 0.02

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:

\displaystyle V=s^3

\displaystyle 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:

\displaystyle \frac{dV}{dt}=3s^2\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 volume and surface area:

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

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

\displaystyle \phi =0.02

Example Question #371 : 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 surface area when its sides have length 5.4?

Possible Answers:

\displaystyle 1.35

\displaystyle 0.8

\displaystyle 2.55

\displaystyle 1.8

\displaystyle 3.85

Correct answer:

\displaystyle 1.35

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:

\displaystyle V=s^3

\displaystyle 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:

\displaystyle \frac{dV}{dt}=3s^2\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 volume and surface area:

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

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

\displaystyle \phi =1.35

Example Question #371 : 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\displaystyle \sqrt[4]{27}?

Possible Answers:

\displaystyle 3\sqrt{3}

\displaystyle 3

\displaystyle 9

\displaystyle 27

\displaystyle 9\sqrt{3}

Correct answer:

\displaystyle 9

Explanation:

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

\displaystyle V=s^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:

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

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

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

\displaystyle \phi =9

Example Question #287 : 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 diagonal when its sides have length \displaystyle 2\sqrt[4]{3}?

Possible Answers:

\displaystyle 3

\displaystyle 12

\displaystyle 4\sqrt{3}

\displaystyle 2\sqrt{3}

\displaystyle 6

Correct answer:

\displaystyle 12

Explanation:

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

\displaystyle V=s^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:

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

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

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

\displaystyle \phi =6

Example Question #282 : 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 its diagonal when its sides have length \displaystyle 8\sqrt{2}?

Possible Answers:

\displaystyle 96\sqrt{3}

\displaystyle 128\sqrt{3}

\displaystyle 32\sqrt{3}

\displaystyle 192\sqrt{3}

\displaystyle 64\sqrt{3}

Correct answer:

\displaystyle 128\sqrt{3}

Explanation:

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

\displaystyle V=s^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:

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

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

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

\displaystyle \phi =128\sqrt{3}

Example Question #281 : 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 its diagonal when its sides have length \displaystyle 5\sqrt[4]{27}?

Possible Answers:

\displaystyle 25\sqrt{3}

\displaystyle 45\sqrt{3}

\displaystyle 75

\displaystyle 225

\displaystyle 45

Correct answer:

\displaystyle 225

Explanation:

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

\displaystyle V=s^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:

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

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

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

\displaystyle \phi =225

Example Question #2171 : Functions

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 \displaystyle 2\sqrt{27}?

Possible Answers:

\displaystyle 90

\displaystyle 180

\displaystyle 72

\displaystyle 18

\displaystyle 36

Correct answer:

\displaystyle 72

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:

\displaystyle A=6s^2

\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:

\displaystyle \frac{dA}{dt}=12s\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:

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

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

\displaystyle \phi =72

Example Question #3202 : 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 diagonal when its sides have length \displaystyle 3\sqrt{5}?

Possible Answers:

\displaystyle 3\sqrt{15}

\displaystyle 60\sqrt{3}

\displaystyle 6\sqrt{5}

\displaystyle 12\sqrt{5}

\displaystyle 12\sqrt{15}

Correct answer:

\displaystyle 12\sqrt{15}

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:

\displaystyle A=6s^2

\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:

\displaystyle \frac{dA}{dt}=12s\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:

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

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

\displaystyle \phi =12\sqrt{15}

Learning Tools by Varsity Tutors