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Calculus 1 : Rate

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Example Questions

Example Question #45 : How To Find Constant Of Proportionality Of Rate

The rate of growth of the wolf population in Yosemite is proportional to the population. The population increased from 2500 to 4200 between 2012 and 2015. At what year approximately would the population reach at least 14000?

Possible Answers:

2024

2026

2020

2022

2018

Correct answer:

2022

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, represents a measure of elapsed time relative to this population value, and is the constant of proportionality.

Since the population increased from 2500 to 4200 between 2012 and 2015, we can solve for this constant of proportionality:

$4200=2500e^{k(2015-2012)}$

$\frac{42}{25}=e^{3k}$

$3k=ln(\frac{42}{25})$

$k=\frac{ln(\frac{42}{25})}{3}=0.173$

Now that the constant of proportionality is known, we can use it to estimate our time point:

$14000=4200e^{0.173(t-2015)}$

$0.173(t-2015)=ln(\frac{10}{3})$

$t=2015+\frac{ln(\frac{10}{3})}{0.173}=2021.9 \rightarrow 2022$

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Example Question #46 : How To Find Constant Of Proportionality Of Rate

The rate of growth of the yeast in a sour dough is proportional to the population. The population increased from 3100 to 11000 in 60 minutes. How many more minutes will it take approximately for the population to reach 25000?

Possible Answers:

Correct answer:

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, represents a measure of elapsed time relative to this population value, and is the constant of proportionality.

Since the population increased from 3100 to 11000 in 60 minutes, we can solve for this constant of proportionality:

$11000=3100e^{60k}$

$\frac{110}{31}=e^{60k}$

$60k=ln(\frac{110}{31})$

$k=\frac{ln(\frac{110}{31})}{60}=0.0211$

Now that the constant of proportionality is known, we can use it to estimate our time point:

$25000=11000e^{0.0211(t)}$

$0.0211(t)=ln(\frac{25}{11})$

$t=\frac{ln(\frac{25}{11})}{0.0211}=38.9 \rightarrow 39$

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Example Question #47 : How To Find Constant Of Proportionality Of Rate

The rate of growth of the population of salmon of Bodega Bay is proportional to the population. The population increased from 6750 to 9000 between January and March. At what month approximately would the population be 16000?

Possible Answers:

August

June

October

September

July

Correct answer:

July

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, represents a measure of elapsed time relative to this population value, and is the constant of proportionality.

Since the population increased from 2700 to 9000 between January and March, we can solve for this constant of proportionality. Use the number of the months as they fall in the calendar:

$9000=6750e^{k(3-1)}$

$\frac{4}{3}=e^{2k}$

$2k=ln(\frac{4}{3})$

$k=\frac{ln(\frac{4}{3})}{2}=0.1438$

Now that the constant of proportionality is known, we can use it to estimate our time point:

$16000=9000e^{0.1438(t-3)}$

$0.1438(t-3)=ln(\frac{16}{9})$

$t=3+\frac{ln(\frac{16}{9})}{0.1438}=7.001 \rightarrow July$

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Example Question #48 : How To Find Constant Of Proportionality Of Rate

The rate of growth of the population of hounds of Baskervilles is proportional to the population. The population increased from 13 to 32 between 2013 and 2015. At what year approximately would the population finally exceed 300?

Possible Answers:

2020

2028

2025

2018

2032

Correct answer:

2020

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, represents a measure of elapsed time relative to this population value, and is the constant of proportionality.

Since the population increased from 13 to 32 between 2013 and 2015, we can solve for this constant of proportionality:

$32=13e^{k(2015-2013)}$

$\frac{32}{13}=e^{2k}$

$2k=ln(\frac{32}{13})$

$k=\frac{ln(\frac{32}{13})}{2}=0.4504$

Now that the constant of proportionality is known, we can use it to estimate our time point:

$300=32e^{0.4504(t-2015)}$

$0.4504(t-2015)=ln(\frac{75}{8})$

$t=2015+\frac{ln(\frac{75}{8})}{0.4504}=2019.97 \rightarrow 2020$

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Example Question #41 : Constant Of Proportionality

The rate of growth of the population of kodiak bears in Juno is proportional to the population. The population increased from 1100 to 1300 between 2014 and 2015. At what year approximately would the population be 3000?

Possible Answers:

2024

2017

2020

2027

2019

Correct answer:

2020

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, represents a measure of elapsed time relative to this population value, and is the constant of proportionality.

Since the population increased from 1100 to 1300 between 2014 and 2015, we can solve for this constant of proportionality:

$1300=1100e^{k(2015-2014)}$

$\frac{13}{11}=e^{k}$

$k=ln(\frac{13}{11})=0.1671$

Now that the constant of proportionality is known, we can use it to estimate our time point:

$3000=1300e^{0.1671(t-2015)}$

$0.1671(t-2015)=ln(\frac{3000}{1300})$

$t=2015+\frac{ln(\frac{30}{13})}{0.1671} \approx 2020$

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Example Question #50 : How To Find Constant Of Proportionality Of Rate

The rate of growth of the population of yeast in some old grape juice is proportional to the population. The population increased from 430 to 1180 after 50 minutes. How many more minutes will it take for the population to exceed 5000?

Possible Answers:

Correct answer:

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, represents a measure of elapsed time relative to this population value, and is the constant of proportionality.

Since the population increased from 430 to 1180 after 50 minutes, we can solve for this constant of proportionality:

$1180=430e^{50k}$

$\frac{1180}{430}=e^{50k}$

$50k=ln(\frac{1180}{430})$

$k=\frac{ln(\frac{1180}{430})}{50}=0.0202$

Now that the constant of proportionality is known, we can use it to estimate our time point:

$5000=1180e^{0.0202(t)}$

$0.0202t=ln(\frac{250}{59})$

$t=\frac{ln(\frac{250}{59})}{0.0202}=71.48 \rightarrow 72$

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Example Question #2741 : Functions

The rate of growth of the bacteria on a piece of raw poultry is proportional to the population. The population increased from 1800 to 7200 between 4:15 and 5:00. At what point in time to the nearest minute will the population be 30000?

Possible Answers: