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Calculus 1 : Rate

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Example Questions

Example Question #165 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of vampires (not bats) is proportional to the population. The population decreased from 8000 to 250 between November 1st and 5th. What is the constant of proportionality in days^-1?

Possible Answers:

-0.115

-0.866

-0.622

-1.095

-0.467

Correct answer:

-0.866

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 8000 to 250 between November 1st and 5th, we can solve for this constant of proportionality:

$250=8000e^{k(5-1)}$

$\frac{1}{32}=e^{4k}$

$4k=ln(\frac{1}{32})$

$k=\frac{ln(\frac{1}{32})}{4}=-0.866$

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Example Question #166 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of werewolves is proportional to the population. The population decreased from 600 to 30 between November 1st and 15th. What is the constant of proportionality in days^-1?

Possible Answers:

-0.273

-0.214

-0.121

-0.199

-0.236

Correct answer:

-0.214

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 600 to 30 between November 1st and 15th, we can solve for this constant of proportionality:

$30=600e^{k(15-1)}$

$\frac{1}{20}=e^{14k}$

$14k=ln(\frac{1}{20})$

$k=\frac{ln(\frac{1}{20})}{14}=-0.214$

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Example Question #167 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of friendly ghosts is proportional to the population. The population decreased from 10000 to 100 between November 28th and 30th due to the pressures of holiday spending. What is the constant of proportionality in days^-1?

Possible Answers:

-1.266

-0.369

-2.303

-1.874

-2.954

Correct answer:

-2.303

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 10000 to 100 between November 28th and 30th, we can solve for this constant of proportionality:

$100=10000e^{k(30-28)}$

$\frac{1}{100}=e^{2k}$

$2k=ln(\frac{1}{100})$

$k=\frac{ln(\frac{1}{100})}{2}=-2.303$

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Example Question #168 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of spooky, scary skeletons is proportional to the population. The population decreased from 1800 to 30 between November 1st and 2nd. What is the constant of proportionality in days^-1?

Possible Answers:

-4.094

-2.888

-1.028

-3.015

-0.702

Correct answer:

-4.094

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 30 between November 1st and 2nd, we can solve for this constant of proportionality:

$30=1800e^{k(2-1)}$

$\frac{1}{60}=e^{k}$

$k=ln(\frac{1}{60})=-4.094$

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Example Question #169 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of swamp things is proportional to the population. The population decreased from 1900 to 38 between November 2nd and 6th. What is the constant of proportionality in days^-1?

Possible Answers:

-1.447

-0.775

-0.809

-0.978

-1.731

Correct answer:

-0.978

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 1900 to 38 between November 2nd and 6th, we can solve for this constant of proportionality:

$38=1900e^{k(6-2)}$

$\frac{1}{50}=e^{4k}$

$4k=ln(\frac{1}{50})$

$k=\frac{ln(\frac{1}{50})}{4}=-0.978$

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Example Question #2861 : Functions

The rate of change of the number of sycamores is proportional to the population. The population increased from 1400 to 12600 between 2010 and 2012. What is the constant of proportionality in years^-1?

Possible Answers:

1.10

1.99

1.50

4.55

2.34

Correct answer:

1.10

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 1400 to 12600 between 2010 and 2012, we can solve for this constant of proportionality:

$12600=1400e^{k(2012-2010)}$

$9=e^{2k}$

2k=ln(9)

$k=\frac{ln(9)}{2}=1.10$

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Example Question #171 : Constant Of Proportionality

The rate of change of the gila mobsters in the lizard mafia is proportional to the population. The population increased from 2300 to 13200 between 2013 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

1.69

0.33

0.90

1.10

1.42

Correct answer:

0.90

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 2300 to 13200 between 2013 and 2015, we can solve for this constant of proportionality:

$13200=2300e^{k(2015-2013)}$

$6=e^{2k}$

2k=ln(6)

$k=\frac{ln(6)}{2}=0.90$

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Example Question #3891 : Calculus

The rate of change of the number of gila monsters is proportional to the population. The population increased from 1800 to 9000 between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

1.61

2.09

1.21

0.93

1.35

Correct answer:

1.61

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 1800 to 9000 between 2014 and 2015, we can solve for this constant of proportionality:

$9000=1800e^{k(2015-2014)}$

$5=e^{k}$

k=ln(5)=1.61

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Example Question #172 : Constant Of Proportionality

The rate of change of the number of leaves on trees in fall is proportional to the population. The population decreased from 120000 to 1000 between October 18th and 22nd. What is the constant of proportionality in days^-1?

Possible Answers:

-2.22

-6.17

-1.20

-3.40

-3.97

Correct answer:

-1.20

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 120000 to 1000 between October 18th and 22nd, we can solve for this constant of proportionality:

$1000=120000e^{k(22-18)}$

$\frac{1}{120}=e^{4k}$

$4k=ln(\frac{1}{120})$

$k=\frac{ln(\frac{1}{120})}{4}=-1.20$

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Example Question #173 : Constant Of Proportionality

The rate of change of the number of chromatic dragons is proportional to the population. The population increased from 100 to 1700 between March 3rd and 8th. What is the constant of proportionality in days^-1?

Possible Answers:

0.89

1.45

0.21

0.57

1.01

Correct answer:

0.57

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 100 to 1700 between March 3rd and 8th, we can solve for this constant of proportionality:

$1700=100e^{k(8-3)}$

$17=e^{5k}$

5k=ln(17)

$k=\frac{ln(17)}{5}=0.57$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #165 : How To Find Constant Of Proportionality Of Rate

Example Question #166 : How To Find Constant Of Proportionality Of Rate

Example Question #167 : How To Find Constant Of Proportionality Of Rate

Example Question #168 : How To Find Constant Of Proportionality Of Rate

Example Question #169 : How To Find Constant Of Proportionality Of Rate

Example Question #2861 : Functions

Example Question #171 : Constant Of Proportionality

Example Question #3891 : Calculus

Example Question #172 : Constant Of Proportionality

Example Question #173 : Constant Of Proportionality

All Calculus 1 Resources

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