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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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Example Questions

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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Example Question #101 : Constant Of Proportionality

The rate of growth of the number of jackals is proportional to the population. The population increased from 1500 to 5800 between 2013 and 2015. Determine the expected population 2018.

Possible Answers:

22573

44074

81922

37631

59078

Correct answer:

44074

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 1500 to 5800 between 2013 and 2015, we can solve for this constant of proportionality:

$5800=1500e^{k(2015-2013)}$

$\frac{58}{15}=e^{2k}$

$2k=ln(\frac{58}{15})$

$k=\frac{ln(\frac{58}{15})}{2}=0.676$

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:

$P=5800e^{(0.676)(2018-2015)} \approx 44074$

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Example Question #3826 : Calculus

The rate of growth of the number of space vipers in the Milky Way is proportional to the population. The population increased from 1200 to 58000 between 2189 and 2213. Determine the expected population 2225.

Possible Answers:

293504

403285

513405

362778

459032

Correct answer:

403285

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 1200 to 58000 between 2189 and 2213, we can solve for this constant of proportionality:

$58000=1200e^{k(2213-2189)}$

$\frac{145}{3}=e^{24k}$

$24k=ln(\frac{145}{3})$

$k=\frac{ln(\frac{145}{3})}{24}=0.1616$

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:

$P=58000e^{(0.1616)(2225-2213)} \approx 403285$

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Example Question #3827 : Calculus

The rate of growth of the number of yeast cells in rising dough is proportional to the population. The population increased from 1100 to 5300 between 3:30 and 3:15. At what point in time approximately will the population be 200,000?

Possible Answers: