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Calculus 1 : Rate of Change

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Example Questions

Example Question #361 : 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 $\sqrt[4]{39}$ ?

Possible Answers:

$\frac{3\sqrt{13}}{4}$

$\frac{13\sqrt{3}}{4}$

$13\sqrt{3}$

$\frac{3\sqrt{13}}{2}$

$\frac{13\sqrt{3}}{2}$

Correct answer:

$\frac{3\sqrt{13}}{4}$

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:

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

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

$\phi=\frac{\sqrt{3}(\sqrt[4]{39})^2}{4}=\frac{3\sqrt{13}}{4}$

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Example Question #362 : How To Find Rate Of Change

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 $5\sqrt{6}$ ?

Possible Answers:

$15\sqrt{6}$

$15\sqrt{2}$

$15\sqrt{3}$

$30\sqrt{2}$

$30\sqrt{3}$

Correct answer:

$30\sqrt{3}$

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:

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

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

$\phi=3(5\sqrt{6})\sqrt{2}=30\sqrt{3}$

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Example Question #363 : How To Find Rate Of Change

Possible Answers:

$22\sqrt{6}$

$11\sqrt{6}$

$22\sqrt{3}$

$33\sqrt{2}$

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:

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

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

$\phi=3(11\sqrt{2})\sqrt{2}=66$

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Example Question #364 : 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 $\frac{\sqrt{6}}{8}$ ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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

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

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Example Question #365 : How To Find Rate Of Change

Possible Answers:

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

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

$\sqrt{6}$

$2\sqrt{3}$

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

Correct answer:

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

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:

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

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

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

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Example Question #366 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the surface area when the radius is 108?

Possible Answers:

Correct answer:

Explanation:

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$A=4\pi r^2$

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

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{108}{2}=54$

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Example Question #367 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the surface area when the radius is 216?

Possible Answers:

108

162

Correct answer:

108

Explanation:

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$A=4\pi r^2$

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

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{216}{2}=108$

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Example Question #368 : How To Find Rate Of Change

A spherical balloon is deflating, although it maintains its spherical shape. What is ratio of the rate of loss of the volume of the sphere to the rate of loss of the surface area when the radius is 38?

Possible Answers:

Correct answer:

Explanation:

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$A=4\pi r^2$

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

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{38}{2}=19$

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Example Question #369 : How To Find Rate Of Change

Possible Answers:

160

128

Correct answer:

Explanation:

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$A=4\pi r^2$

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

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and surface area, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)8\pi r \frac{dr}{dt}$

$\phi =\frac{r}{2}$

$\phi =\frac{192}{2}=96$

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Example Question #370 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the volume of the sphere to the rate of growth of the circumference when the radius is 1.5?

Possible Answers:

3.5

4.5

2.3

1.5

Correct answer:

4.5

Explanation:

Let's begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

$V=\frac{4}{3}\pi r^3$

$C=2\pi r^$

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

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the volume and circumference, divide:

$4\pi r^2 \frac{dr}{dt}=(\phi)2\pi \frac{dr}{dt}$

$\phi =2r^2$

$\phi =2(1.5)^2=4.5$

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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #361 : How To Find Rate Of Change

Example Question #362 : How To Find Rate Of Change

Example Question #363 : How To Find Rate Of Change

Example Question #364 : How To Find Rate Of Change

Example Question #365 : How To Find Rate Of Change

Example Question #366 : How To Find Rate Of Change

Example Question #367 : How To Find Rate Of Change

Example Question #368 : How To Find Rate Of Change

Example Question #369 : How To Find Rate Of Change

Example Question #370 : How To Find Rate Of Change

All Calculus 1 Resources

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