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Calculus 1 : Rate of Change

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Example Questions

Example Question #2181 : Functions

A spherical balloon is being filled with air. What is the diameter of the sphere at the instance the rate of growth of the volume is 0.34 times the rate of growth of the circumference?

Possible Answers:

0.170

0.340

0.825

0.627

0.412

Correct answer:

0.825

Explanation:

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. Now given the problem information, the rate of growth of the volume is 0.34 times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(0.34)2\pi \frac{dr}{dt}$

$r^2=(0.34)\frac{1}{2}$

$r =\sqrt{\frac{0.34}{2}}=0.4123$

The diameter is therefore

d=0.825

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Example Question #301 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is the volume of the sphere at the instance the rate of growth of the surface area is 24 times the rate of growth of the circumference?

Possible Answers:

$288\pi$

$216\pi$

$108\pi$

$144\pi$

$316 \pi$

Correct answer:

$288\pi$

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

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 growth of the surface area is 24 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(24)2\pi \frac{dr}{dt}$

$r =\frac{24}{4}$

r=6

Now to solve for the volume:

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

$V=288\pi$

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Example Question #301 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is the diameter of the sphere at the instance the rate of growth of the surface area is 3.12 times the rate of growth of the circumference?

Possible Answers:

1.04

0.78

4.16

2.08

1.56

Correct answer:

1.56

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

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 growth of the surface area is 3.12 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(3.12)2\pi \frac{dr}{dt}$

$r =\frac{3.12}{4}$

r=0.78

The diameter is then

d=1.56

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Example Question #2181 : Functions

A spherical balloon is deflating, while maintaining its spherical shape. What is the circumference of the sphere at the instance the rate of shrinkage of the surface area is 11.12 times the rate of shrinkage of the circumference?

Possible Answers:

23.5

15.1

11.2

17.5

19.4

Correct answer:

17.5

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

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 11.12 times the rate of shrinkage of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(11.12)2\pi \frac{dr}{dt}$

$r =\frac{11.12}{4}$

r=2.78

The circumference is then

$C=2\pi r$

C=17.5

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Example Question #303 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the surface area of the sphere at the instance the rate of shrinkage of the surface area is 76 times the rate of shrinkage of the circumference?

Possible Answers:

$1444\pi$

$1458 \pi$

$722\pi$

$2888 \pi$

$729 \pi$

Correct answer:

$1444\pi$

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

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 76 times the rate of shrinkage of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(76)2\pi \frac{dr}{dt}$

$r =\frac{76}{4}$

r=19

Now all that remains is to find the surface area at this time

$A=4\pi r^2$

$A= 1444\pi$

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Example Question #304 : Rate Of Change

A spherical balloon is being filled with air. What is the radius of the sphere at the instance the rate of growth of the volume is 2.22 times the rate of growth of the surface area?

Possible Answers:

4.44

2.49

2.22

4.98

6.66

Correct answer:

4.44

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. So given our problem conditions, the rate of growth of the volume is 2.22 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(2.22)8\pi r \frac{dr}{dt}$

r =(2.22)2

r=4.44

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Example Question #305 : Rate Of Change

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 volume is 19.85 times the rate of shrinkage of the surface area?

Possible Answers:

79.4

67.4

39.7

158.8

33.7

Correct answer:

79.4

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. So given our problem conditions, the rate of shrinkage of the volume is 19.85 times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(19.85)8\pi r \frac{dr}{dt}$

r =(19.85)2

r=39.7

The diameter is then

d=79.4

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Example Question #306 : Rate Of Change

A spherical balloon is being filled with air. What is the volume of the sphere at the instance the rate of growth of the volume is 1.62 times the rate of growth of the circumference?

Possible Answers:

2.018

4.036

3.054

1.527

1.009

Correct answer:

3.054

Explanation:

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. Now given the problem information, the rate of growth of the volume is 1.62 times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(1.62)2\pi \frac{dr}{dt}$

$r^2=(1.62)\frac{1}{2}$

$r =\sqrt{\frac{1.62}{2}}$

r=.9

To find the volume:

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

$V=\frac{4}{3}\pi (0.9)^3=3.054$

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Example Question #307 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the surface area of the sphere at the instance the rate of shrinkage of the volume is 3.24 times the rate of shrinkage of the circumference?

Possible Answers:

5.09

6.21

12.41

20.36

24.82

Correct answer:

20.36

Explanation:

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. Now given the problem information, the rate of shrinkage of the volume is 3.24 times the rate of shrinkage of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(3.24)2\pi \frac{dr}{dt}$

$r^2=(3.24)\frac{1}{2}$

$r =\sqrt{\frac{3.24}{2}}$

$r=0.9\sqrt{2}$

Then, to find the surface area:

$A=4\pi r^2$

$A=4\pi (0.9\sqrt{2})^2=20.36$

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Example Question #308 : Rate Of Change

A spherical balloon is being filled with air. What is the circumference of the sphere at the instance the rate of growth of the surface area is 9.96 times the rate of growth of the circumference?

Possible Answers:

5.81

15.65

17.43

2.49

20.29

Correct answer:

15.65

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

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, , to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(9.96)2\pi \frac{dr}{dt}$

$r =\frac{9.96}{4}$

r=2.49

And finally to find the circumference:

$C=2\pi r$

$C=2\pi (2.49)=15.65$

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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 #2181 : Functions

Example Question #301 : How To Find Rate Of Change

Example Question #301 : How To Find Rate Of Change

Example Question #2181 : Functions

Example Question #303 : Rate Of Change

Example Question #304 : Rate Of Change

Example Question #305 : Rate Of Change

Example Question #306 : Rate Of Change

Example Question #307 : Rate Of Change

Example Question #308 : Rate Of Change

All Calculus 1 Resources

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