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Calculus 1 : Rate of Change

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Example Questions

Example Question #3331 : Calculus

A spherical balloon is deflating, while maintaining its spherical shape. What is the diameter of the sphere at the instance the rate of shrinkage of the surface area is 0.08 times the rate of shrinkage of the circumference?

Possible Answers:

0.02 $\displaystyle 0.02$

0.04 $\displaystyle 0.04$

0.01 $\displaystyle 0.01$

0.16 $\displaystyle 0.16$

0.08 $\displaystyle 0.08$

Correct answer:

0.04 $\displaystyle 0.04$

Explanation:

Start by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

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

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

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now we can use the relation given in the problem statement, the rate of shrinkage of the surface area is 0.08 times the rate of shrinkage of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(0.08)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(0.08)2\pi \frac{dr}{dt}$

$r =\frac{0.08}{4}=0.02$ $\displaystyle r =\frac{0.08}{4}=0.02$

The diameter is then:

d=2r $\displaystyle d=2r$

d=0.04 $\displaystyle d=0.04$

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Example Question #421 : How To Find Rate Of Change

Find the equation for the rate of change of the volume of a cylindrical pitcher with respect to radius. The height of the cylinder is three times the size of the radius.

Possible Answers:

$\frac{dV}{dr}=9\pi r^{2}$ $\displaystyle \frac{dV}{dr}=9\pi r^{2}$

$\frac{dV}{dr}=9\pi r^{2}h$ $\displaystyle \frac{dV}{dr}=9\pi r^{2}h$

$\frac{dV}{dr}=3\pi r^{2}$ $\displaystyle \frac{dV}{dr}=3\pi r^{2}$

$\frac{dV}{dr}=3\pi r^{2}h$ $\displaystyle \frac{dV}{dr}=3\pi r^{2}h$

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

Correct answer:

$\frac{dV}{dr}=9\pi r^{2}$ $\displaystyle \frac{dV}{dr}=9\pi r^{2}$

Explanation:

The volume of a cylinder is $\pi r^{2}h$ $\displaystyle \pi r^{2}h$. Before we take the derivative of the volume equation, we must first make all of our variable in terms of radius. Therefore, $V=\pi r^{2}(3r)=3\pi r^{3}$ $\displaystyle V=\pi r^{2}(3r)=3\pi r^{3}$.

Now, we take the derivative with respect to radius: $dV=9 \pi r^{2}dr.$ $\displaystyle dV=9 \pi r^{2}dr.$

Rearrange to get: $\frac{dV}{dr}=9 \pi r^{2}.$ $\displaystyle \frac{dV}{dr}=9 \pi r^{2}.$

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Example Question #3335 : 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 surface area when the radius is 92?

Possible Answers:

$\displaystyle 92$

$\displaystyle 23$

184 $\displaystyle 184$

61.3 $\displaystyle 61.3$

$\displaystyle 46$

Correct answer:

$\displaystyle 46$

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:

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

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

$\phi =\frac{92}{2}=46$ $\displaystyle \phi =\frac{92}{2}=46$

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Example Question #2303 : 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 140?

Possible Answers:

$\displaystyle 70$

$\displaystyle 20$

$\displaystyle 56$

$\displaystyle 28$

140 $\displaystyle 140$

Correct answer:

$\displaystyle 70$

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:

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

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

$\phi =\frac{140}{2}=70$ $\displaystyle \phi =\frac{140}{2}=70$

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Example Question #3332 : 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 surface area when the radius is 162?

Possible Answers:

216 $\displaystyle 216$

$\displaystyle 81$

$\displaystyle 18$

$\displaystyle 54$

108 $\displaystyle 108$

Correct answer:

$\displaystyle 81$

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:

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

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

$\phi =\frac{162}{2}=81$ $\displaystyle \phi =\frac{162}{2}=81$

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Example Question #3331 : Calculus

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 910?

Possible Answers:

130 $\displaystyle 130$

455 $\displaystyle 455$

260 $\displaystyle 260$

$\displaystyle 91$

520 $\displaystyle 520$

Correct answer:

455 $\displaystyle 455$

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:

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

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

$\phi =\frac{910}{2}=455$ $\displaystyle \phi =\frac{910}{2}=455$

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Example Question #421 : How To Find Rate Of Change

Possible Answers:

106 $\displaystyle 106$

79.5 $\displaystyle 79.5$

26.5 $\displaystyle 26.5$

$\displaystyle 53$

318 $\displaystyle 318$

Correct answer:

79.5 $\displaystyle 79.5$

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:

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

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

$\phi =\frac{159}{2}=79.5$ $\displaystyle \phi =\frac{159}{2}=79.5$

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Example Question #421 : Rate Of Change

Possible Answers:

310 $\displaystyle 310$

248 $\displaystyle 248$

124 $\displaystyle 124$

$\displaystyle 31$

496 $\displaystyle 496$

Correct answer:

496 $\displaystyle 496$

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:

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

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

$\phi =\frac{992}{2}=496$ $\displaystyle \phi =\frac{992}{2}=496$

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Example Question #422 : 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 radius when the radius is 1?

Possible Answers:

$2\pi^2$ $\displaystyle 2\pi^2$

$\displaystyle 4$

$4\pi^2$ $\displaystyle 4\pi^2$

$2\pi$ $\displaystyle 2\pi$

$4\pi$ $\displaystyle 4\pi$

Correct answer:

$4\pi$ $\displaystyle 4\pi$

Explanation:

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

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

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

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

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 radius, divide:

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

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

$\phi =4\pi(1)^2=4\pi$ $\displaystyle \phi =4\pi(1)^2=4\pi$

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Example Question #423 : 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 radius when the radius is 0.25?

Possible Answers:

$\pi$ $\displaystyle \pi$

$\frac{\pi}{8}$ $\displaystyle \frac{\pi}{8}$

$\frac{\pi}{2}$ $\displaystyle \frac{\pi}{2}$

$\frac{\pi}{4}$ $\displaystyle \frac{\pi}{4}$

$\frac{\pi}{16}$ $\displaystyle \frac{\pi}{16}$

Correct answer:

$\frac{\pi}{4}$ $\displaystyle \frac{\pi}{4}$

Explanation:

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

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

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

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

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 radius, divide:

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

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

$\phi =4\pi(0.25)^2=\frac{\pi}{4}$ $\displaystyle \phi =4\pi(0.25)^2=\frac{\pi}{4}$

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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 #3331 : Calculus

Example Question #421 : How To Find Rate Of Change

Example Question #3335 : Calculus

Example Question #2303 : Functions

Example Question #3332 : Calculus

Example Question #3331 : Calculus

Example Question #421 : How To Find Rate Of Change

Example Question #421 : Rate Of Change

Example Question #422 : How To Find Rate Of Change

Example Question #423 : Rate Of Change

All Calculus 1 Resources

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