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Calculus 1 : Rate

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Example Questions

Example Question #2394 : Functions

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 $\frac{\sqrt{3}}{3}$ $\displaystyle \frac{\sqrt{3}}{3}$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

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

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Example Question #2395 : Functions

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:

$51\sqrt{2}$ $\displaystyle 51\sqrt{2}$

$51\sqrt{3}$ $\displaystyle 51\sqrt{3}$

$51\sqrt{6}$ $\displaystyle 51\sqrt{6}$

$17\sqrt{3}$ $\displaystyle 17\sqrt{3}$

$17\sqrt{6}$ $\displaystyle 17\sqrt{6}$

Correct answer:

$51\sqrt{2}$ $\displaystyle 51\sqrt{2}$

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #2396 : Functions

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

Possible Answers:

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

$\displaystyle 3$

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

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

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

Correct answer:

$\displaystyle 3$

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #2397 : Functions

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:

$42\sqrt{3}$ $\displaystyle 42\sqrt{3}$

$42\sqrt{6}$ $\displaystyle 42\sqrt{6}$

$168\sqrt{6}$ $\displaystyle 168\sqrt{6}$

$168\sqrt{2}$ $\displaystyle 168\sqrt{2}$

$168\sqrt{3}$ $\displaystyle 168\sqrt{3}$

Correct answer:

$168\sqrt{2}$ $\displaystyle 168\sqrt{2}$

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #2398 : Functions

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:

$75\sqrt{3}$ $\displaystyle 75\sqrt{3}$

$15\sqrt{2}$ $\displaystyle 15\sqrt{2}$

$75\sqrt{6}$ $\displaystyle 75\sqrt{6}$

$15\sqrt{3}$ $\displaystyle 15\sqrt{3}$

$75\sqrt{2}$ $\displaystyle 75\sqrt{2}$

Correct answer:

$75\sqrt{2}$ $\displaystyle 75\sqrt{2}$

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #601 : Rate

Possible Answers:

$126\sqrt{6}$ $\displaystyle 126\sqrt{6}$

$84\sqrt{2}$ $\displaystyle 84\sqrt{2}$

$126\sqrt{3}$ $\displaystyle 126\sqrt{3}$

$84\sqrt{3}$ $\displaystyle 84\sqrt{3}$

$126\sqrt{2}$ $\displaystyle 126\sqrt{2}$

Correct answer:

$84\sqrt{3}$ $\displaystyle 84\sqrt{3}$

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #511 : How To Find Rate Of Change

Possible Answers:

$27\sqrt{2}$ $\displaystyle 27\sqrt{2}$

$27\sqrt{3}$ $\displaystyle 27\sqrt{3}$

$27\sqrt{6}$ $\displaystyle 27\sqrt{6}$

$18\sqrt{3}$ $\displaystyle 18\sqrt{3}$

$18\sqrt{6}$ $\displaystyle 18\sqrt{6}$

Correct answer:

$27\sqrt{6}$ $\displaystyle 27\sqrt{6}$

Explanation:

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

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

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

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

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

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

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

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

$\phi=3(9\sqrt{3})\sqrt{2}=27\sqrt{6}$ $\displaystyle \phi=3(9\sqrt{3})\sqrt{2}=27\sqrt{6}$

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Example Question #512 : How To Find Rate Of Change

A regular tetrahedron is decreasing 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 55?

Possible Answers:

$33\sqrt{3}$ $\displaystyle 33\sqrt{3}$

$165\sqrt{3}$ $\displaystyle 165\sqrt{3}$

$165\sqrt{2}$ $\displaystyle 165\sqrt{2}$

$33\sqrt{6}$ $\displaystyle 33\sqrt{6}$

$33\sqrt{2}$ $\displaystyle 33\sqrt{2}$

Correct answer:

$165\sqrt{2}$ $\displaystyle 165\sqrt{2}$

Explanation:

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

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

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

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

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

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

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

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

$\phi=3(55)\sqrt{2}=165\sqrt{2}$ $\displaystyle \phi=3(55)\sqrt{2}=165\sqrt{2}$

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Example Question #513 : How To Find Rate Of Change

A regular tetrahedron is decreasing 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 $33\sqrt{2}$ $\displaystyle 33\sqrt{2}$ ?

Possible Answers:

$99\sqrt{2}$ $\displaystyle 99\sqrt{2}$

$198\sqrt{3}$ $\displaystyle 198\sqrt{3}$

198 $\displaystyle 198$

$99\sqrt{3}$ $\displaystyle 99\sqrt{3}$

$198\sqrt{2}$ $\displaystyle 198\sqrt{2}$

Correct answer:

198 $\displaystyle 198$

Explanation:

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

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

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

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

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

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

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

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

$\phi=3(33\sqrt{2})\sqrt{2}=198$ $\displaystyle \phi=3(33\sqrt{2})\sqrt{2}=198$

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Example Question #521 : How To Find Rate Of Change

A regular tetrahedron is decreasing 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 30?

Possible Answers:

$15\sqrt{2}$ $\displaystyle 15\sqrt{2}$

$30\sqrt{2}$ $\displaystyle 30\sqrt{2}$

$90\sqrt{2}$ $\displaystyle 90\sqrt{2}$

$60\sqrt{2}$ $\displaystyle 60\sqrt{2}$

$120\sqrt{2}$ $\displaystyle 120\sqrt{2}$

Correct answer:

$90\sqrt{2}$ $\displaystyle 90\sqrt{2}$

Explanation:

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

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

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

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

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

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

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

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

$\phi=3(30)\sqrt{2}=90\sqrt{2}$ $\displaystyle \phi=3(30)\sqrt{2}=90\sqrt{2}$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2394 : Functions

Example Question #2395 : Functions

Example Question #2396 : Functions

Example Question #2397 : Functions

Example Question #2398 : Functions

Example Question #601 : Rate

Example Question #511 : How To Find Rate Of Change

Example Question #512 : How To Find Rate Of Change

Example Question #513 : How To Find Rate Of Change

Example Question #521 : How To Find Rate Of Change

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