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Calculus 1 : How to find constant of proportionality of rate

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Example Questions

Example Question #3831 : Calculus

The rate of growth of the number of trash-hungry pigeons is proportional to the population. The population increased from 1300 to 28,600 between 2010 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

15.455

1.482

0.618

3.412

0.993

Correct answer:

0.618

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 1300 to 28,600 between 2010 and 2015, we can solve for this constant of proportionality:

$28600=1300e^{k(2015-2010)}$

$22=e^{5k}$

5k=ln(22)

$k=\frac{ln(22)}{5}=0.618$

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Example Question #3832 : Calculus

The rate of growth of the number of pinethrushes in Beulica is proportional to the population. The population increased from 5400 to 37,800 between 2009 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.154

1.507

0.982

0.233

0.324

Correct answer:

0.324

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 5400 to 37,800 between 2009 and 2015, we can solve for this constant of proportionality:

$37800=5400e^{k(2015-2009)}$

$7=e^{6k}$

6k=ln(7)

$k=\frac{ln(7)}{6}=0.324$

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Example Question #3833 : Calculus

The rate of growth of the number of pine martens in a section of Napa County is proportional to the population. The population increased by 38.9 percent between 2012 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.110

0.134

0.073

0.181

0.296

Correct answer:

0.110

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 38.9 percent between 2012 and 2015, we can solve for this constant of proportionality:

$(1+0.389)y_0=y_0e^{k(2015-2012)}$

$1.389=e^{3k}$

3k=ln(1.389)

$k=\frac{ln(1.389)}{3}=0.110$

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Example Question #3834 : Calculus

The rate of growth of the number of moths in the wardrobe leading to a land of talking animals is proportional to the population. The population increased by 17.5 percent between January and April. What is the constant of proportionality in years^-1?

Possible Answers:

0.645

0.719

0.882

0.100

0.054

Correct answer:

0.645

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 17.5 percent between January and April, we can solve for this constant of proportionality. Mind the units that we're looking for, though it may be helpful to treat the months as their number in the calendar:

$(1+0.175)y_0=y_0e^{k(\frac{4-1}{12})}$

$1.175=e^{\frac{1}{4}k}$

$\frac{1}{4}k=ln(1.175)$

$k=\frac{ln(1.175)}{\frac{1}{4}}=0.645$

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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 #3837 : Calculus

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

0.099

0.717

0.333

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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Calculus 1 : How to find constant of proportionality of rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3831 : Calculus

Example Question #3832 : Calculus

Example Question #3833 : Calculus

Example Question #3834 : Calculus

Example Question #2801 : Functions

Example Question #3836 : Calculus

Example Question #3837 : Calculus

Example Question #3838 : Calculus

Example Question #3839 : Calculus

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

All Calculus 1 Resources

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