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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #508 : 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:

$\frac{9}{4}$

$\frac{9}{16}$

$\frac{3}{4}$

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

$36\sqrt{3}$

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:

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

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

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

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Example Question #509 : 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:

$2\sqrt{6}$

$4\sqrt{3}$

$8\sqrt{6}$

$8\sqrt{3}$

$4\sqrt{6}$

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 #510 : 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:

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

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

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

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

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

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

Example Question #591 : Rate

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 $\frac{\sqrt{2}}{2}$ ?

Possible Answers:

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

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

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

$\frac{\sqrt{6}}{8}$

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

Correct answer:

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

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:

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

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

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

Report an Error

Example Question #601 : Rate

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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

$\phi=\frac{\sqrt{3}(\frac{\sqrt{3}}{3})^2}{4}=\frac{\sqrt{3}}{12}$

Report an Error

Example Question #602 : Rate

A regular tetrahedron is burgeoning in size. What is the ratio of the rate of change of the surface area of the tetrahedron to the rate of change of its height when its sides have length 17?

Possible Answers:

$17\sqrt{6}$

$51\sqrt{6}$

$17\sqrt{3}$

$51\sqrt{2}$

$51\sqrt{3}$

Correct answer:

$51\sqrt{2}$

Explanation:

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

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

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

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

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

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

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:

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

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

$\phi=3(17)\sqrt{2}=51\sqrt{2}$

Report an Error

Example Question #603 : Rate

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

Possible Answers:

$\sqrt{2}$

$\sqrt{3}$

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

$\sqrt{6}$

Correct answer:

Explanation:

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

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

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

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

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

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

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:

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

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

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

Report an Error

Example Question #604 : Rate

A regular tetrahedron is burgeoning in size. What is the ratio of the rate of change of the surface area of the tetrahedron to the rate of change of its height when its sides have length 56?

Possible Answers:

$42\sqrt{3}$

$42\sqrt{6}$

$168\sqrt{6}$

$168\sqrt{3}$

$168\sqrt{2}$

Correct answer:

$168\sqrt{2}$

Explanation:

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

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

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

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

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

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

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:

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

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

$\phi=3(56)\sqrt{2}=168\sqrt{2}$

Report an Error

Example Question #605 : Rate

A regular tetrahedron is burgeoning in size. What is the ratio of the rate of change of the surface area of the tetrahedron to the rate of change of its height when its sides have length 25?

Possible Answers:

$75\sqrt{6}$

$15\sqrt{3}$

$15\sqrt{2}$

$75\sqrt{2}$

$75\sqrt{3}$

Correct answer:

$75\sqrt{2}$

Explanation:

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

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

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

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

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

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

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:

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

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

$\phi=3(25)\sqrt{2}=75\sqrt{2}$

Report an Error

Example Question #606 : Rate

Possible Answers:

$126\sqrt{3}$

$126\sqrt{2}$

$126\sqrt{6}$

$84\sqrt{3}$

$84\sqrt{2}$

Correct answer:

$84\sqrt{3}$

Explanation:

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

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

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

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

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

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

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:

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

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

$\phi=3(14\sqrt{6})\sqrt{2}=84\sqrt{3}$

Report an Error

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #508 : Rate Of Change

Example Question #509 : Rate Of Change

Example Question #510 : Rate Of Change

Example Question #591 : Rate

Example Question #601 : Rate

Example Question #602 : Rate

Example Question #603 : Rate

Example Question #604 : Rate

Example Question #605 : Rate

Example Question #606 : Rate

All Calculus 1 Resources

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