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Calculus 1 : How to find rate of change

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Example Questions

Example Question #2381 : Functions

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 $24\sqrt{6}$ ?

Possible Answers:

$\sqrt{6}$

$12\sqrt{6}$

Correct answer:

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:

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

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

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

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Example Question #2382 : Functions

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 $12\sqrt{6}$ ?

Possible Answers:

$12\sqrt{3}$

$24\sqrt{6}$

Correct answer:

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:

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

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

$\phi=\frac{(12\sqrt{6})\sqrt{6}}{24}=3$

Report an Error

Example Question #2383 : Functions

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 14?

Possible Answers:

$\frac{7\sqrt{6}}{12}$

$\frac{7\sqrt{3}}{4}$

$\frac{7\sqrt{3}}{12}$

$\frac{7\sqrt{6}}{6}$

$\frac{7\sqrt{2}}{3}$

Correct answer:

$\frac{7\sqrt{6}}{12}$

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:

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

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

$\phi=\frac{(14)\sqrt{6}}{24}=\frac{7\sqrt{6}}{12}$

Report an Error

Example Question #2384 : Functions

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 $48\sqrt{2}$ ?

Possible Answers:

$2\sqrt{6}$

$2\sqrt{3}$

$4\sqrt{3}$

$8\sqrt{3}$

$8\sqrt{6}$

Correct answer:

$4\sqrt{3}$

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:

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

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

$\phi=\frac{(48\sqrt{2})\sqrt{6}}{24}=4\sqrt{3}$

Report an Error

Example Question #2385 : Functions

Possible Answers:

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

$9\sqrt{2}$

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

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

Correct answer:

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:

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

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

$\phi=\frac{(36\sqrt{6})\sqrt{6}}{24}=9$

Report an Error

Example Question #2386 : Functions

Possible Answers:

$12\sqrt{2}$

$8\sqrt{2}$

$12\sqrt{3}$

$8\sqrt{3}$

$4\sqrt{6}$

Correct answer:

$12\sqrt{2}$

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:

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

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

$\phi=\frac{(96\sqrt{3})\sqrt{6}}{24}=12\sqrt{2}$

Report an Error

Example Question #501 : 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 6?

Possible Answers:

$18\sqrt{6}$

$18\sqrt{3}$

$9\sqrt{3}$

$9\sqrt{6}$

$3\sqrt{3}$

Correct answer:

$9\sqrt{3}$

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:

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

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

$\phi=\frac{\sqrt{3}(6)^2}{4}=9\sqrt{3}$

Report an Error

Example Question #502 : 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 $\frac{\sqrt{3}}{2}$ ?

Possible Answers:

$\frac{9\sqrt{3}}{16}$

$\frac{9}{16}$

$36\sqrt{3}$

$\frac{9}{4}$

$\frac{3}{4}$

Correct answer:

$\frac{9\sqrt{3}}{16}$

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{\sqrt{3}}{2})^2}{4}=\frac{9\sqrt{3}}{16}$

Report an Error

Example Question #503 : 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 $4\sqrt{2}$ ?

Possible Answers:

$4\sqrt{3}$

$2\sqrt{6}$

$4\sqrt{6}$

$8\sqrt{6}$

$8\sqrt{3}$

Correct answer:

$8\sqrt{3}$

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:

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

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

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

Report an Error

Example Question #504 : 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 $\frac{1}{2}$ ?

Possible Answers:

$\frac{\sqrt{3}}{16}$

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

$\frac{\sqrt{3}}{32}$

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

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

Correct answer:

$\frac{\sqrt{3}}{16}$

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{1}{2})^2}{4}=\frac{\sqrt{3}}{16}$

Report an Error

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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 #2381 : Functions

Example Question #2382 : Functions

Example Question #2383 : Functions

Example Question #2384 : Functions

Example Question #2385 : Functions

Example Question #2386 : Functions

Example Question #501 : How To Find Rate Of Change

Example Question #502 : How To Find Rate Of Change

Example Question #503 : How To Find Rate Of Change

Example Question #504 : How To Find Rate Of Change

All Calculus 1 Resources

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