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Calculus 1 : Rate

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Example Questions

Example Question #3795 : Calculus

The rate of growth of the number of cottonwood trees in Reno is proportional to the population. The population increased from 1600 to 1900 between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.172

0.190

0.223

0.144

0.275

Correct answer:

0.172

Explanation:

We're told that the rate of growth 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 1600 to 1900 between 2014 and 2015, we can solve for this constant of proportionality:

$1900=1600e^{k(2015-2014)}$

$\frac{19}{16}=e^{k}$

$k=ln(\frac{19}{16})=0.172$

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Example Question #2761 : Functions

The rate of growth of the number of gray squirrels is proportional to the population. The population increased from 23000 to 52000 between 2011 and 2014. What is the constant of proportionality in years^-1?

Possible Answers:

0.388

0.228

0.272

0.039

0.251

Correct answer:

0.272

Explanation:

We're told that the rate of growth 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 23000 to 52000 between 2011 and 2014, we can solve for this constant of proportionality:

$52000=23000e^{k(2014-2011)}$

$\frac{52}{23}=e^{3k}$

$3k=ln(\frac{52}{23})$

$k=\frac{ln(\frac{52}{23})}{3}=0.272$

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Example Question #3797 : Calculus

The rate of decrease of the number of African elephants is proportional to the population. The population decreased from 2700 to 1900 between 2013 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-0.132

-0.099

-0.176

-0.209

-0.241

Correct answer:

-0.176

Explanation:

We're told that the rate of decrease 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 from 2700 to 1900 between 2013 and 2015, we can solve for this constant of proportionality:

$1900=2700e^{k(2015-2013)}$

$\frac{19}{27}=e^{2k}$

$2k=ln(\frac{19}{27})$

$k=\frac{ln(\frac{19}{27})}{2}=-0.176$

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Example Question #3798 : Calculus

The rate of decrease of typhoid sufferers, due to the advent of modern medicine, is proportional to the population. The population decreased from 250,000 to 80,000 between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-1.139

-0.775

-2.913

-0.988

-0.533

Correct answer:

-1.139

Explanation:

We're told that the rate of decrease 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 from 250,000 to 80,000 between 2014 and 2015, we can solve for this constant of proportionality:

$80000=250000e^{k(2015-2014)}$

$\frac{8}{25}=e^{k}$

$k=ln(\frac{8}{25})=-1.139$

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Example Question #3799 : Calculus

The rate of decrease of the number of Amur Leopards is proportional to the population. The population decreased from 113 to 27 between 1999 and 2007. What is the constant of proportionality in years^-1?

Possible Answers:

-0.648

0.331

-3.511

-1.432

-0.179

Correct answer:

-0.179

Explanation:

We're told that the rate of decrease 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 from 113 to 27 between 1999 and 2007, we can solve for this constant of proportionality:

$27=113e^{k(2007-1999)}$

$\frac{27}{113}=e^{8k}$

$8k=ln(\frac{27}{113})$

$k=\frac{ln(\frac{27}{113})}{8}=-0.179$

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Example Question #3800 : Calculus

The rate of growth of the number of cholera causing bacteria in a pound is proportional to the population. The population increased from 100 to 750 between 3:20 and 4:00. What is the constant of proportionality in minutes^-1?

Possible Answers:

0.109

0.081

0.050

0.038

0.062

Correct answer:

0.050

Explanation:

We're told that the rate of growth 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 100 to 750 between 3:20 and 4:00, we can solve for this constant of proportionality:

$750=100e^{k(4:00-3:20)}$

$\frac{15}{2}=e^{40k}$

$40k=ln(\frac{15}{2})$

$k=\frac{ln(\frac{15}{2})}{40}=0.050$

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Example Question #81 : Constant Of Proportionality

The rate of growth of the number of E. coli in a room-temperature stew is proportional to the population. The population increased from 1000 to 3500 between 3:15 and 3:30. What is the constant of proportionality in hours^-1?

Possible Answers:

2.119

3.806

0.835

0.612

5.011

Correct answer:

5.011

Explanation:

We're told that the rate of growth 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 1000 to 3500 between 3:15 and 3:30, we can solve for this constant of proportionality. Mind the units; treat the minutes as decimals of an hour by dividing by 60:

$3500=1000e^{k(3.5-3.25)}$

$\frac{7}{2}=e^{0.25k}$

$0.25k=ln(\frac{7}{2})$

$k=\frac{ln(\frac{7}{2})}{0.25}=5.011$

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Example Question #82 : Constant Of Proportionality

The rate of growth of the numer of maned wolves in a select portion of South American grassland is proportional to the population. The population increased from 1800 to 2300 between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.020

0.189

0.053

0.245

0.117

Correct answer:

0.245

Explanation:

We're told that the rate of growth 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 1800 to 2300 between 2014 and 2015, we can solve for this constant of proportionality:

$2300=1800e^{k(2015-2014)}$

$\frac{23}{18}=e^{k}$

$k=ln(\frac{23}{18})=0.245$

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Example Question #83 : Constant Of Proportionality

The rate of decrease of the number of anacondas in an encroached portion of river is proportional to the population. The population decreased from 1808 to 1117 between 2013 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-0.595

-1.175

-0.241

1.175

0.241

Correct answer:

-0.241

Explanation:

We're told that the rate of decrease 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 from 1808 to 1117 between 2013 and 2015, we can solve for this constant of proportionality:

$1117=1808e^{k(2015-2013)}$

$\frac{1117}{1808}=e^{2k}$

$2k=ln(\frac{1117}{1808})$

$k=\frac{ln(\frac{1117}{1808})}{2}=-0.241$

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Example Question #84 : Constant Of Proportionality

The rate of decrease of the number of Tazmanian devils near a growing housing development is proportional to the population. The population decreased from 1800 to 713 between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-1.117

-1.365

-1.669

-0.926

-1.008

Correct answer:

-0.926

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 from 1800 to 713 between 2014 and 2015, we can solve for this constant of proportionality:

$713=1800e^{k(2015-2014)}$

$\frac{713}{1800}=e^{k}$

$k=ln(\frac{713}{1800})=-0.926$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3795 : Calculus

Example Question #2761 : Functions

Example Question #3797 : Calculus

Example Question #3798 : Calculus

Example Question #3799 : Calculus

Example Question #3800 : Calculus

Example Question #81 : Constant Of Proportionality

Example Question #82 : Constant Of Proportionality

Example Question #83 : Constant Of Proportionality

Example Question #84 : Constant Of Proportionality

All Calculus 1 Resources

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