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Calculus 1 : Rate

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Example Questions

Example Question #2781 : Functions

The rate of growth of the number of bacterial cells in a dog bowl is proportional to the population. The population increased from 11300 to 28900 between 3:00 and 3:45. Determine the expected population at 5:15.

Possible Answers:

112503

210302

159665

79984

189018

Correct answer:

189018

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

Since the population increased from 11300 to 28900 between 3:00 and 3:45, we can solve for this constant of proportionality. Treat the minutes as decimals by dividing them by 60:

$28900=11300e^{k(3.75-3)}$

$\frac{289}{113}=e^{0.75k}$

$0.75k=ln(\frac{289}{113})$

$k=\frac{ln(\frac{289}{113})}{0.75}=1.252$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=28900e^{(1.252)(5.25-3.75)} \approx 189018$

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Example Question #2782 : Functions

The rate of growth of the number of albatrosses plaguing wayward ships is proportional to the population. The population increased from 1123 to 1839 between 1798 and 1802. Determine the expected population in 1834.

Possible Answers:

94181

107224

86433

90312

111545

Correct answer:

94181

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

Since the population increased from 1123 to 1839 between 1798 and 1802, we can solve for this constant of proportionality:

$1839=1123e^{k(1802-1798)}$

$\frac{1839}{1123}=e^{4k}$

$4k=ln(\frac{1839}{1123})$

$k=\frac{ln(\frac{1839}{1123})}{4}=0.123$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=1839e^{(0.123)(1834-1802)} \approx 94181$

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Example Question #91 : How To Find Constant Of Proportionality Of Rate

The rate of decrease of the number of living yeast cells as the bread is put into the oven is proportional to the population. The population decreased from 4,500,000,000 to 1,500,000 between 3:15 and 3:45. Determine the expected population at 4:15.

Possible Answers:

500

432

887

Correct answer:

500

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

Since the population decreased from 4,500,000,000 to 1,500,000 between 3:15 and 3:45, we can solve for this constant of proportionality. Convert minutes to decimals by dividing by 60:

$1500000=4500000000e^{k(3.75-3.25)}$

$\frac{1}{3000}=e^{0.5k}$

$0.5k=ln(\frac{1}{3000})$

$k=\frac{ln(\frac{1}{3000})}{0.5}=-16.013$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=1500000e^{(-16.013)(4.25-3.75)} \approx 500$

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Example Question #2783 : Functions

The rate of decrease of the number of albino squirrels is proportional to the population. The population decreased from 413 to 136 between 2010 and 2015. Determine the expected population in 2018.

Possible Answers:

112

265

Correct answer:

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

Since the population decreased from 413 to 136 between 2010 and 2015, we can solve for this constant of proportionality:

$136=413e^{k(2015-2010)}$

$\frac{136}{413}=e^{5k}$

$5k=ln(\frac{136}{413})$

$k=\frac{ln(\frac{136}{413})}{5}=-0.222$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=136e^{(-0.222)(2018-2015)} \approx 70$

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Example Question #2784 : Functions

The rate of decrease of the number of pollen particles in the air of a room with a vacuum on is proportional to the population. The population decreased from 12345 to 6789 between 3:00 and 4:00. Determine the expected population at 7:00.

Possible Answers:

1234

1588

816

1129

1002

Correct answer:

1129

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

Since the population decreased from 12345 to 6789 between 3:00 and 4:00, we can solve for this constant of proportionality:

$6789=12345e^{k(4-3)}$

$\frac{6789}{12345}=e^{k}$

$k=ln(\frac{6789}{12345})=-0.598$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=6789e^{(-0.598)(7-4)} \approx 1129$

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Example Question #91 : How To Find Constant Of Proportionality Of Rate

The rate of decrease of the number of tunnel snakes is proportional to the population. The population decreased from 2445 to 813 between 2267 and 2269. Determine the expected population in 2277.

Possible Answers:

213

Correct answer:

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

Since the population decreased from 2445 to 813 between 2267 and 2269, we can solve for this constant of proportionality:

$813=2445e^{k(2269-2267)}$

$\frac{271}{815}=e^{2k}$

$2k=ln(\frac{271}{815})$

$k=\frac{ln(\frac{271}{815})}{2}=-0.551$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=813e^{(-0.551)(2277-2269)} \approx 10$

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Example Question #101 : Constant Of Proportionality

The rate of decrease of the number of baceterium due to an introduction of soap is proportional to the population. The population decreased from 15800 to 2400 between 3:00 and 5:00. Determine the expected population at 5:30.

Possible Answers:

1499

832

1212

1047

1809

Correct answer:

1499

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

Since the population decreased from 15800 to 2400 between 3:00 and 5:00, we can solve for this constant of proportionality:

$2400=15800e^{k(5-3)}$

$\frac{12}{79}=e^{2k}$

$2k=ln(\frac{12}{79})$

$k=\frac{ln(\frac{12}{79})}{2}=-0.942$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points. Convert minutes to decimals by dividing by 60:

$P=2400e^{(-0.942)(5.5-5)} \approx 1499$

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Example Question #102 : Constant Of Proportionality

The rate of decrease of the number of black lotuses is proportional to the population. The population decreased from 1100 to 85 between 1993 and 2013. Determine the expected population in 2015.

Possible Answers:

Correct answer:

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

Since the population decreased from 1100 to 85 between 1993 and 2013, we can solve for this constant of proportionality:

$85=1100e^{k(2013-1993)}$

$\frac{17}{220}=e^{10k}$

$10k=ln(\frac{17}{220})$

$k=\frac{ln(\frac{17}{220})}{10}=-0.256$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=85e^{(-0.256)(2015-2013)} \approx 51$

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Example Question #103 : Constant Of Proportionality

The rate of growth of the number of tree frogs is proportional to the population. The population increased from 43800 to 87215 between 1990 and 1995. Determine the expected population in 2015.

Possible Answers:

2001865

1377983

1208617

1836714

987162

Correct answer:

1377983

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

Since the population increased from 43800 to 87215 between 1990 and 1995, we can solve for this constant of proportionality:

$87215=43800e^{k(1995-1990)}$

$\frac{87215}{43800}=e^{5k}$

$5k=ln(\frac{87215}{43800})$

$k=\frac{ln(\frac{87215}{43800})}{5}=0.138$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=87215e^{(0.138)(2015-1995)} \approx 1377983$

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Example Question #104 : Constant Of Proportionality

The rate of growth of the number of glowing mushrooms in a flooded cave is proportional to the population. The population increased from 110 to 310 between January and July. Determine the expected population in October.

Possible Answers:

521

1312

419

677

806

Correct answer:

521

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

Since the population increased from 110 to 310 between January and July, we can solve for this constant of proportionality. Use the number of the months as they're ordered in the calendar:

$310=110e^{k(7-1)}$

$\frac{31}{11}=e^{6k}$

$6k=ln(\frac{31}{11})$

$k=\frac{ln(\frac{31}{11})}{6}=0.173$

Now that the constant of proportionality is known, we can use it to find an expected population value relative to an initial population value due to the difference in time points:

$P=310e^{(0.173)(10-7)} \approx 521$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2781 : Functions

Example Question #2782 : Functions

Example Question #91 : How To Find Constant Of Proportionality Of Rate

Example Question #2783 : Functions

Example Question #2784 : Functions

Example Question #91 : How To Find Constant Of Proportionality Of Rate

Example Question #101 : Constant Of Proportionality

Example Question #102 : Constant Of Proportionality

Example Question #103 : Constant Of Proportionality

Example Question #104 : Constant Of Proportionality

All Calculus 1 Resources

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