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Calculus 1 : Rate of Change

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Example Questions

Example Question #450 : 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 $\sqrt[4]{39}$ $\displaystyle \sqrt[4]{39}$?

Possible Answers:

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

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

$13\sqrt{3}$ $\displaystyle 13\sqrt{3}$

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

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

Correct answer:

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

Explanation:

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

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

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

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:

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

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

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

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Example Question #3271 : Calculus

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 $5\sqrt{6}$ $\displaystyle 5\sqrt{6}$?

Possible Answers:

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

$30\sqrt{3}$ $\displaystyle 30\sqrt{3}$

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

$15\sqrt{6}$ $\displaystyle 15\sqrt{6}$

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

Correct answer:

$30\sqrt{3}$ $\displaystyle 30\sqrt{3}$

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #3272 : Calculus

Possible Answers:

$\displaystyle 66$

$11\sqrt{6}$ $\displaystyle 11\sqrt{6}$

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

$22\sqrt{3}$ $\displaystyle 22\sqrt{3}$

$22\sqrt{6}$ $\displaystyle 22\sqrt{6}$

Correct answer:

$\displaystyle 66$

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:

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

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

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

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

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

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Example Question #361 : 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 $\frac{\sqrt{6}}{8}$ $\displaystyle \frac{\sqrt{6}}{8}$?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #3273 : Calculus

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

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Example Question #2246 : Functions

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:

$\displaystyle 36$

$\displaystyle 72$

$\displaystyle 96$

$\displaystyle 12$

$\displaystyle 54$

Correct answer:

$\displaystyle 54$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

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:

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

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

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

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Example Question #361 : 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:

$\displaystyle 72$

$\displaystyle 54$

108 $\displaystyle 108$

162 $\displaystyle 162$

$\displaystyle 36$

Correct answer:

108 $\displaystyle 108$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

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:

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

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

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

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Example Question #362 : 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:

$\displaystyle 46$

$\displaystyle 76$

$\displaystyle 92$

$\displaystyle 19$

$\displaystyle 23$

Correct answer:

$\displaystyle 19$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

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:

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

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

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

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Example Question #363 : How To Find Rate Of Change

Possible Answers:

128 $\displaystyle 128$

160 $\displaystyle 160$

$\displaystyle 64$

$\displaystyle 96$

$\displaystyle 32$

Correct answer:

$\displaystyle 96$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

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:

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

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

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

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Example Question #3281 : Calculus

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 $\displaystyle 3.5$

4.5 $\displaystyle 4.5$

$\displaystyle 3$

1.5 $\displaystyle 1.5$

2.3 $\displaystyle 2.3$

Correct answer:

4.5 $\displaystyle 4.5$

Explanation:

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

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

$C=2\pi r^$ $\displaystyle C=2\pi r^$

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

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

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

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:

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

$\phi =2r^2$ $\displaystyle \phi =2r^2$

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

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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 #450 : Rate

Example Question #3271 : Calculus

Example Question #3272 : Calculus

Example Question #361 : Rate Of Change

Example Question #3273 : Calculus

Example Question #2246 : Functions

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 #3281 : Calculus

All Calculus 1 Resources

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