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Calculus 1 : Calculus

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Example Questions

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:

6:17

6:59

5:46

5:23

6:35

Correct answer:

5:46

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 1800 to 7200 between 4:15 and 5:00, we can solve for this constant of proportionality. Convert the minutes to decimals after the hour by dividing by 60:

$7200=1800e^{k(5-4.25)}$

$4=e^{0.75k}$

0.75k=ln(4)

$k=\frac{ln(4)}{0.75}=1.8484$

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

$30000=7200e^{1.8484(t-5)}$

$1.8484(t-5)=ln(\frac{25}{6})$

$t=5+\frac{ln(\frac{25}{6})}{1.8484}=5.772 \rightarrow 5:46$

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Example Question #51 : How To Find Constant Of Proportionality Of Rate

The rate of growth of the population of wildflowers in a Denver meadow is proportional to the population. The population increased from 1600 to 2000 between 2010 and 2015. At what year approximately will the population reach 6000?

Possible Answers:

2040

2020

2025

2035

2030

Correct answer:

2040

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 1600 to 2000 between 2010 and 2015, we can solve for this constant of proportionality:

$2000=1600e^{k(2015-2010)}$

$\frac{5}{4}=e^{5k}$

$5k=ln(\frac{5}{4})$

$k=\frac{ln(\frac{5}{4})}{5}=0.0446$

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

$6000=2000e^{0.0446(t-2015)}$

0.0446(t-2015)=ln(3)

$t=2015+\frac{ln(3)}{0.0446}-2039.6 \rightarrow2040$

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Example Question #2743 : Functions

The rate of growth of the population of jellyfish in the Gulf of Mexico is proportional to the population. The population increased from 12,000 to 32,000 between 2012 and 2015. At what year approximately will the population exceed 600,000?

Possible Answers:

2021

2027

2030

2018

2024

Correct answer:

2024

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 12,000 to 32,000 between 2012 and 2015, we can solve for this constant of proportionality:

$32000=12000e^{k(2015-2012)}$

$\frac{8}{3}=e^{3k}$

$3k=ln(\frac{8}{3})$

$k=\frac{ln(\frac{8}{3})}{3}=0.327$

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

$600000=32000e^{0.327(t-2015)}$

$0.327(t-2015)=ln(\frac{75}{4})$

$t=2015+\frac{ln(\frac{75}{4})}{0.327}=2023.9 \rightarrow 2024$

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Example Question #2744 : Functions

The rate of decrease of the number of Staphylococcus in the presence of a potent antibiotic is proportional to the population. The population decreases from 15000 to 4000 over the course of twenty minutes. How many more minutes will it take for the population to be reduced to 10 or less?

Possible Answers:

147

113

162

Correct answer:

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

Since the population decreases from 15000 to 4000 over the course of twenty minutes, we can solve for this constant of proportionality:

$4000=15000e^{20k}$

$\frac{4}{15}=e^{20k}$

$20k=ln(\frac{4}{15})$

$k=\frac{ln(\frac{4}{15})}{20}=-0.0661$

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

$10=4000e^{-0.0661t}$

$-0.0661t=ln(\frac{1}{400})$

$t=\frac{ln(\frac{1}{400})}{-0.0661}=90.6 \rightarrow 91$

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Example Question #3771 : Calculus

The rate of decrease of the bacteria in the presence of a UV sterilization lamp is proportional to the population. The population on a surgical instrument decreases from 2900 to 500 in the course of an hour. How many more minutes will it take to bring the count under 25?

Possible Answers:

265

103

134

Correct answer:

103

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 decreases from 2900 to 500 in the course of an hour (or 60 minutes), we can solve for this constant of proportionality:

$500=2900e^{60k}$

$\frac{5}{29}=e^{60k}$

$60k=ln(\frac{5}{29})$

$k=\frac{ln(\frac{5}{29})}{60}=-0.0293$

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

$25=500e^{-0.0293t}$

$-0.0293t=ln(\frac{1}{20})$

$t=\frac{ln(\frac{1}{20})}{-0.0293}=102.2 \rightarrow103$

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Example Question #3771 : Calculus

The rate of change of the yeast in a lump of unbaked sour dough is proportional to the population. The population increased by 5 percent over the course of 15 minutes. What is the constant of proportionality in minutes^-1?

Possible Answers:

0.0111

0.0033

0.0004

0.0756

0.0283

Correct answer:

0.0033

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 by 5 percent over the course of 15 minutes, we can solve for this constant of proportionality:

$(1+0.05)y_0=y_0e^{15k}$

$1.05=e^{15k}$

15k=ln(1.05)

$k=\frac{ln(1.05)}{15}=0.0033$

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Example Question #953 : Rate

The rate of change of the culture size of baceteria on a dirty plate is proportional to the population. The population increased by 32 percent over the course of 35 minutes. What is the constant of proportionality in minutes^-1?

Possible Answers:

0.0183

0.0079

0.0203

0.0018

0.099

Correct answer:

0.0079

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 by 32 percent over the course of 35 minutes, we can solve for this constant of proportionality:

$(1+0.32)y_0=y_0e^{35k}$

$1.32=e^{35k}$

35k=ln(1.32)

$k=\frac{ln(1.32)}{35}=0.0079$

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Example Question #954 : Rate

The rate of change of the population of Crystal Cove salmon is proportional to the population. The population increased by 23 percent over the course of two years. What is the constant of proportionality in years^-1?

Possible Answers:

0.122

0.233

0.157

0.104

0.188

Correct answer:

0.104

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 by 23 percent over the course of two years, we can solve for this constant of proportionality:

$(1+0.23)y_0=y_0e^{2k}$

$1.23=e^{2k}$

2k=ln(1.23)

$k=\frac{ln(1.23)}{2}=0.104$

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Example Question #955 : Rate

The rate of change of the Bjorg is proportional to the population. The population increased by 189 percent over the course of 1 year. What is the constant of proportionality in years^-1?

Possible Answers:

0.166

0.753

0.424

1.061

0.899

Correct answer:

1.061

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 by 189 percent over the course of 1 year, we can solve for this constant of proportionality:

$(1+1.89)y_0=y_0e^{k(1)}$

$2.89=e^{k}$

k=ln(2.89)=1.061

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Example Question #956 : Rate

The rate of change of the E. coli number on a piece of raw meat is proportional to the population. The population increased by 6300 percent over the course of 2 hours. What is the constant of proportionality in minutes^-1?

Possible Answers:

0.035

1.388

2.079

0.989

0.152

Correct answer:

0.035

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 , we can solve for this constant of proportionality. Remember what units the problem asks for:

$(1+63)y_0=y_0e^{k(2*60)}$

$64=e^{120k}$

120k=ln(64)

$k=\frac{ln(64)}{120}$

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