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Calculus 1 : How to find rate of change

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Example Questions

Example Question #351 : 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 diagonal when its sides have length 3.33?

Possible Answers:

21.0

16.5

10.8

13.4

19.2

Correct answer:

19.2

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

$d=s\sqrt{3}$

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

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

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

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

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

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

$\phi =(3.33)^2\sqrt{3}=19.2$

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Example Question #352 : 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 6.5?

Possible Answers:

$6.5\sqrt{3}$

$26\sqrt{3}$

$13\sqrt{3}$

Correct answer:

$26\sqrt{3}$

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

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

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

$\phi =4(6.5)\sqrt{3}=26\sqrt{3}$

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Example Question #353 : 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 surface area to the rate of loss of its diagonal when its sides have length $\frac{\sqrt{3}}{3}$ ?

Possible Answers:

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

$4\sqrt{3}$

$\frac{4}{3}$

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

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

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

$\phi =4(\frac{\sqrt{3}}{3})\sqrt{3}=4$

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Example Question #354 : How To Find Rate Of Change

A regular tetrahedron is growing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length 7.9?

Possible Answers:

1.02

0.67

0.90

0.81

1.44

Correct answer:

0.81

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{7.9\sqrt{6}}{24}=0.81$

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Example Question #355 : How To Find Rate Of Change

A regular tetrahedron is growing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length $23\sqrt{3}$ ?

Possible Answers:

$\frac{23\sqrt{3}}{2}$

$\frac{23\sqrt{3}}{8}$

$\frac{23\sqrt{2}}{6}$

$\frac{23\sqrt{2}}{8}$

$\frac{23\sqrt{6}}{4}$

Correct answer:

$\frac{23\sqrt{2}}{8}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{(23\sqrt{3})\sqrt{6}}{24}=\frac{23\sqrt{2}}{8}$

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Example Question #356 : How To Find Rate Of Change

A regular tetrahedron is shrinking in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its surface area when its sides have length $18\sqrt{2}$ ?

Possible Answers:

$\sqrt{2}$

$\frac{3\sqrt{2}}{4}$

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

$3\sqrt{3}$

$\sqrt{3}$

Correct answer:

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

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{(18\sqrt{2})\sqrt{6}}{24}=\frac{3\sqrt{3}}{2}$

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Example Question #357 : How To Find Rate Of Change

Possible Answers:

$\frac{21\sqrt{6}}{2}$

$\frac{21\sqrt{2}}{2}$

$\frac{21}{4}$

$\frac{21\sqrt{3}}{2}$

$\frac{21}{2}$

Correct answer:

$\frac{21}{4}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and surface area in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$A=\sqrt{3}s^2$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

$\phi=\frac{s\sqrt{6}}{24}$

$\phi=\frac{(21\sqrt{6})\sqrt{6}}{24}=\frac{21}{4}$

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Example Question #357 : How To Find Rate Of Change

A regular tetrahedron is increaseing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length $19\sqrt{3}$ ?

Possible Answers:

$\frac{1083\sqrt{2}}{8}$

$\frac{1083\sqrt{3}}{2}$

$\frac{1083\sqrt{2}}{4}$

$361\sqrt{2}$

$\frac{1083\sqrt{3}}{4}$

Correct answer:

$\frac{1083\sqrt{3}}{4}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

$\phi=\frac{\sqrt{3}(19\sqrt{3})^2}{4}=\frac{1083\sqrt{3}}{4}$

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Example Question #358 : How To Find Rate Of Change

A regular tetrahedron is increaseing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length $\sqrt[3]{6}$ ?

Possible Answers:

$\frac{3\sqrt[3]{3}}{2\sqrt[6]{2}}$

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

$\frac{3\sqrt[6]{3}}{2\sqrt[3]{2}}$

$\frac{3\sqrt[6]{3}}{2\sqrt[6]{2}}$

$\frac{3\sqrt[4]{3}}{2\sqrt[3]{2}}$

Correct answer:

$\frac{3\sqrt[6]{3}}{2\sqrt[3]{2}}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

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

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Example Question #358 : How To Find Rate Of Change

A regular tetrahedron is diminishing in size. What is the ratio of the rate of change of the volume of the tetrahedron to the rate of change of its height when its sides have length $\sqrt[4]{8}$ ?

Possible Answers:

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

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

$\frac{\sqrt{6}}{4}$

$\frac{\sqrt{6}}{2}$

$\frac{\sqrt{6}}{6}$

Correct answer:

$\frac{\sqrt{6}}{2}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its volume and height in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

$h=\frac{\sqrt{6}}{3}s$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\phi)\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\phi=\frac{\sqrt{3}s^2}{4}$

$\phi=\frac{\sqrt{3}(\sqrt[4]{8})^2}{4}=\frac{\sqrt{6}}{2}$

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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 #351 : How To Find Rate Of Change

Example Question #352 : How To Find Rate Of Change

Example Question #353 : How To Find Rate Of Change

Example Question #354 : How To Find Rate Of Change

Example Question #355 : How To Find Rate Of Change

Example Question #356 : How To Find Rate Of Change

Example Question #357 : How To Find Rate Of Change

Example Question #357 : How To Find Rate Of Change

Example Question #358 : How To Find Rate Of Change

Example Question #358 : How To Find Rate Of Change

All Calculus 1 Resources

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