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Calculus 1 : Rate

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Example Questions

Example Question #2801 : Functions

The rate of growth of the number of E. coli cells in a forgotten slab of hamburger is proportional to the population. The population increased by 300 percent between 3:00 and 5:00. What is the constant of proportionality in minutes^-1?

Possible Answers:

0.012

2.740

0.476

0.693

1.189

Correct answer:

0.012

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased by 300 percent between 3:00 and 5:00, we can solve for this constant of proportionality. Be mindful of the units requested:

$(1+3)y_0=y_0e^{k(60(5-3))}$

$4=e^{120k}$

120k=ln(4)

$k=\frac{ln(4)}{120}=0.012$

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Example Question #3836 : Calculus

The rate of growth of the number of electric eels in Baldwin's Lake is proportional to the population. The population increased by 19.6 percent between 2012 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.362

1.008

0.006

0.060

1.761

Correct answer:

0.060

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased by 19.6 percent between 2012 and 2015, we can solve for this constant of proportionality:

$(1+0.196)y_0=y_0e^{k(2015-2012)}$

$1.196=e^{3k}$

3k=ln(1.196)

$k=\frac{ln(1.196)}{3}=0.060$

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Example Question #1011 : Rate

The rate of growth of the number of mongooses in Dale's Wild Woodland Safari is proportional to the population. The population increased by 59.2 percent between January of 2014 and February of 2015. What is the constant of proportionality in months^-1?

Possible Answers:

0.036

0.717

0.333

0.099

0.122

Correct answer:

0.036

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased by 59.2 percent between January of 2014 and February of 2015, a timeframe of 13 months, we can solve for this constant of proportionality:

$(1+0.592)y_0=y_0e^{k(13)}$

$1.592=e^{13k}$

13k=ln(1.592)

$k=\frac{ln(1.592)}{13}=0.036$

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Example Question #3838 : Calculus

The rate of growth of the number of man-of-war jellyfish due to climate change is proportional to the population. The population increased by 125.4 percent between 2013 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.881

0.120

0.239

0.406

1.625

Correct answer:

0.406

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased by 125.4 percent between 2013 and 2015, we can solve for this constant of proportionality:

$(1+1.254)y_0=y_0e^{k(2015-2013)}$

$2.254=e^{2k}$

2k=ln(2.254)

$k=\frac{ln(2.254)}{2}=0.406$

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Example Question #3839 : Calculus

The rate of change of the beluga population in an overhunted location is proportional to the population. The population decreased by 92.3 percent between 2013 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-0.131

-1.282

-0.040

-0.852

-0.357

Correct answer:

-1.282

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population decreased by 92.3 percent between 2013 and 2015, we can solve for this constant of proportionality:

$(1-0.923)y_0=y_0e^{k(2015-2013)}$

$0.077=e^{2k}$

2k=ln(0.077)

$k=\frac{ln(0.077)}{2}=-1.282$

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Example Question #111 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of E. coli baceteria after exposure to a disinfectant is proportional to the population. The population decreased by 93.1 percent between 3:15 and 4:00. What is the constant of proportionality in hours^-1?

Possible Answers:

-0.002

-3.565

-1.304

-0.059

-0.095

Correct answer:

-3.565

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population decreased by 93.1 percent between 3:15 and 4:00, we can solve for this constant of proportionality. Given the units requested in the problem statement, convert the minutes to the decimals of an hour by dividing by 60:

$(1-0.931)y_0=y_0e^{k(4-3.25)}$

$0.069=e^{0.75k}$

0.75k=ln(0.069)

$k=\frac{ln(0.069)}{0.75}=-3.565$

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Example Question #121 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of yeast cells in a loaf of bread as it's put into the oven is proportional to the population. The population decreased by 99.7 percent between 5:00 and 6:00. What is the constant of proportionality in hours^-1?

Possible Answers:

-0.003

-0.097

-2.993

-5.809

-8.145

Correct answer:

-5.809

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population decreased by 99.7 percent between 5:00 and 6:00, we can solve for this constant of proportionality:

$(1-0.997)y_0=y_0e^{k(6-5)}$

$0.003=e^{k}$

k=ln(0.003)=-5.809

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Example Question #122 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of cyborgs on Planet X-038 is proportional to the population. The population increased by 513 percent between 2039 and 2043. What is the constant of proportionality in years^-1?

Possible Answers:

0.453

1.281

0.427

0.318

1.927

Correct answer:

0.453

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased by 513 percent between 2039 and 2043, we can solve for this constant of proportionality:

$(1+5.13)y_0=y_0e^{k(2043-2039)}$

$6.13=e^{4k}$

4k=ln(6.13)

$k=\frac{ln(6.13)}{4}=0.453$

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Example Question #123 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of nanobots in an advancing horror of gray goo is proportional to the population. The population increased by 113800 percent between 2025 and 2028. What is the constant of proportionality in years^-1?

Possible Answers:

2.346

3.567

5.939

7.101

4.046

Correct answer:

2.346

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased by 113800 percent between 2025 and 2028, we can solve for this constant of proportionality:

$(1+1138)y_0=y_0e^{k(2028-2025)}$

$1139=e^{3k}$

3k=ln(1139)

$k=\frac{ln(1139)}{3}=2.346$

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Example Question #124 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of bacteriophages in a room where nobody covers their face after sneezing is proportional to the population. The population increased from 1,600 to 320,000 between 4:15 and 5:00. What is the constant of proportionality in minutes^-1?

Possible Answers:

5.586

1.492

0.510

7.064

0.118

Correct answer:

0.118

Explanation:

We're told that the rate of change of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased from 1,600 to 320,000 between 4:15 and 5:00, we can solve for this constant of proportionality:

$320000=1600e^{k(5:00-4:15)}$

$200=e^{45k}$

45k=ln(200)

$k=\frac{ln(200)}{45}=0.118$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2801 : Functions

Example Question #3836 : Calculus

Example Question #1011 : Rate

Example Question #3838 : Calculus

Example Question #3839 : Calculus

Example Question #111 : How To Find Constant Of Proportionality Of Rate

Example Question #121 : How To Find Constant Of Proportionality Of Rate

Example Question #122 : How To Find Constant Of Proportionality Of Rate

Example Question #123 : How To Find Constant Of Proportionality Of Rate

Example Question #124 : How To Find Constant Of Proportionality Of Rate

All Calculus 1 Resources

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