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Calculus 1 : Rate

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Example Questions

Example Question #31 : How To Find Constant Of Proportionality Of Rate

The rate of growth of the population of sentient self-replicating toy soldiers is proportional to the population. The population increased from 12 to 43 between Monday and Wednesday. Determine the expected population on Saturday.

Possible Answers:

271

253

292

188

207

Correct answer:

292

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 to 43 between Monday and Wednesday, we can solve for this constant of proportionality. Assign the days numbers based on the occurence in the week:

$43=12e^{k(4-2)}$

$\frac{43}{12}=e^{2k}$

$2k=ln(\frac{43}{12})$

$k=\frac{ln(\frac{43}{12})}{2}=0.638$

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:

$P=43e^{(7-4)(0.638)} \approx 292$

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Example Question #2723 : Functions

The rate of growth of the prokaryotes of Lake Tahoe is proportional to the population. The population increased from 193678 to 254183 between 2014 and 2015. Determine the expected population in 2021.

Possible Answers:

1398602

1200068

897775

1299915

1136241

Correct answer:

1299915

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 193678 to 254183 between 2014 and 2015, we can solve for this constant of proportionality:

$254183=193678e^{k(2015-2014)}$

$\frac{254183}{193678}=e^{k}$

$k=ln(\frac{254183}{193678})=0.272$

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:

$P=254183e^{(2021-2015)(0.272)} \approx 1299915$

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Example Question #931 : Rate

The rate of growth of the alien spores flooding the atmostphere is proportional to the population. The population increased from 19000 to 58000 between 2013 and 2014. Determine the expected population in 2017.

Possible Answers:

1213964

1649855

2136621

2549818

1887903

Correct answer:

1649855

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 19000 to 58000 between 2013 and 2014, we can solve for this constant of proportionality:

$58000=19000e^{k(2014-2013)}$

$\frac{58}{19}=e^{k}$

$k=ln(\frac{58}{19})=1.116$

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:

$P=58000e^{(2017-2014)(1.116)} \approx 1649855$

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Example Question #932 : Rate

The rate of growth of the algae in Bill's unkempt swimming pool is proportional to the population. The population increased from 17902 to 39551 between March and May. Determine the expected population in July.

Possible Answers:

87321

154479

192788

102256

136894

Correct answer:

87321

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 17902 to 39551 between March and May, we can solve for this constant of proportionality. Assign the months their number in the calendar to determine time elapsed:

$39551=17902e^{k(5-3)}$

$\frac{39551}{17902}=e^{2k}$

$2k=ln(\frac{39551}{17902})$

$k=\frac{ln(\frac{39551}{17902})}{2}=0.396$

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:

$P=39551e^{(7-5)(0.396)} \approx 87321$

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Example Question #933 : Rate

The rate of decrease of the population of Bengal tigers has been proportional to the population. The population decreased from 45000 to 1800 between 1900 and 1972. Determine the constant of proportionality for this decrease.

Possible Answers:

0.0188

-0.1808

0.0039

-0.0065

-0.0447

Correct answer:

-0.0447

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 decreased from 45000 to 1800 between 1900 and 1972, we can solve for this constant of proportionality:

$1800=45000e^{k(1972-1900)}$

$\frac{1800}{45000}=e^{72k}$

$72k=ln(\frac{1800}{45000})$

$k=\frac{ln(\frac{1800}{45000})}{72}=-0.0447$

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Example Question #934 : Rate

The rate of growth of the population of robomen is proportional to the population. The population increased from 100 units to 1500 between 2008 and 2015. Determine the expected population in 2055.

Possible Answers:

7924462944

77812342

6012348

583094

8941326112

Correct answer:

7924462944

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 100 units to 1500 between 2008 and 2015, we can solve for this constant of proportionality:

$1500=100e^{k(2015-2008)}$

$15=e^{7k}$

7k=ln(15)

$k=\frac{ln(15)}{7}=0.387$

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:

$P=1500e^{(2055-2015)(0.387)} \approx 7924462944$

Better unplug that production facility.

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Example Question #41 : How To Find Constant Of Proportionality Of Rate

The rate of decrease of the robomen population due to the transgalactic organic tai cyber conflict is proportional to the population. The population decreased from 6 billion to 450 million between 2063 and 2081. Determine the expected population in 2100.

Possible Answers:

45231

29228600

8057651

37995623

112804

Correct answer:

29228600

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 decreased from 6 billion to 450 million between 2063 and 2081, we can solve for this constant of proportionality:

$450000000=6000000000e^{k(2081-2063)}$

$\frac{3}{40}=e^{18k}$

$18k=ln(\frac{3}{40})$

$k=\frac{ln(\frac{3}{40})}{18}=-0.1439$

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:

$P=450000000e^{(2100-2081)(-0.1439)} \approx 29228600$

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Example Question #3761 : Calculus

The rate of growth of the culture of E.coli on a piece of room temperature meat is proportional to the population. The population increased from 400 to 1600 between 1:15 and 1:45. Determine the expected population at 3:15.

Possible Answers:

102463

8954

87112

11366

28796

Correct answer:

102463

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 400 to 1600 between 1:15 and 1:45, we can solve for this constant of proportionality. Treat the minutes as decimals after the hour by dividing by 60:

$1600=400e^{k(1.75-1.25)}$

$4=e^{0.5k}$

0.5k=ln(4)

$k=\frac{ln(4)}{0.5}=2.773$

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:

$P=1600e^{(3.25-1.75)(2.773)} \approx 102463$

Which is why we use refrigerators.

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Example Question #42 : How To Find Constant Of Proportionality Of Rate

The rate of decrease of the population of E. coli in response to a particular antibiotic is proportional to the population. The population decreased from 160,000 to 8,000 between 1:00 and 1:30. Determine the expected population at 1:45.

Possible Answers:

812

932

4896

1789

Correct answer:

1789

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 decreased from 160,000 to 8,000 between 1:00 and 1:30, we can solve for this constant of proportionality. Treat the minutes as decimals after the hour by dividing by 60:

$8000=160000e^{k(1.5-1)}$

$\frac{1}{20}=e^{0.5k}$

$0.5k=ln(\frac{1}{20})$

$k=\frac{ln(\frac{1}{20})}{0.5}=-5.991$

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:

$P=8000e^{(1.75-1.5)(-5.991)} \approx 1789$

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Example Question #2731 : Functions

The rate of growth of the culture of bacteria on a dirty plate is proportional to the population. The population increased from 50 to 200 between 1:15 and 1:30. At what point in time approximately would the population be 700?

Possible Answers: