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Calculus 1 : Rate

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Example Questions

Example Question #3825 : Calculus

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:

59078

81922

44074

22573

37631

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:

6:01

5:31

4:52

4:05

5:17

Correct answer:

4:05

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 1100 to 5300 between 3:30 and 3:15, we can solve for this constant of proportionality. Treat the minutes as decimals of an hour by dividing by 60:

$5300=1100e^{k(3.5-3.25)}$

$\frac{53}{11}=e^{0.25k}$

$0.25k=ln(\frac{53}{11})$

$k=\frac{ln(\frac{53}{11})}{0.25}=6.290$

Now that the constant of proportionality is known, we can use it to find the specified point in time:

$200000=5300e^{6.290(t-3.5)}$

$6.290(t-3.5)=ln(\frac{2000}{53})$

$t=3.5+\frac{ln(\frac{2000}{53})}{6.290}=4.077 \rightarrow 4:05$

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Example Question #3828 : Calculus

The rate of growth of the number of bacterial cells in a culture is proportional to the population. The population increased from 24 to 313 between 3:00 and 4:00. At what point in time apprxoimately will the population be 50,000?

Possible Answers:

5:59

6:14

6:39

6:52

5:25

Correct answer:

5:59

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 24 to 313 between 3:00 and 4:00, we can solve for this constant of proportionality:

$313=24e^{k(4-3)}$

$\frac{313}{24}=e^{k}$

$k=ln(\frac{313}{24})=2.568$

Now that the constant of proportionality is known, we can use it to find the specified point in time:

$50000=313e^{2.568(t-4)}$

$2.568(t-4)=ln(\frac{50000}{313})$

$t=4+\frac{ln(\frac{50000}{313})}{2.568} =5.976 \rightarrow 5:59$

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Example Question #3829 : Calculus

The rate of growth of the number of bacteriophages assaulting a culture is proportional to the population. The population increased from 11 to 165 between 1:00 and 3:00. At what point in time will the population be 800,000?

Possible Answers:

8:41

7:02

8:04

9:16

7:33

Correct answer:

9:16

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 11 to 165 between 1:00 and 3:00, we can solve for this constant of proportionality:

$165=11e^{k(3-1)}$

$15=e^{2k}$

2k=ln(15)

$k=\frac{ln(15)}{2}=1.354$

Now that the constant of proportionality is known, we can use it to find the specified point in time:

$800000=165e^{1.354(t-3)}$

$1.354(t-3)=ln(\frac{160000}{33})$

$t=3+\frac{ln(\frac{160000}{33})}{1.354}=9.268 \rightarrow 9:16$

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Example Question #105 : Constant Of Proportionality

The rate of growth of the number of red snapper in secluded waters is proportional to the population. The population increased from 12000 to 96000 between 2011 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.520

0.411

0.488

0.556

0.693

Correct answer:

0.520

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, and is the constant of proportionality.

Since the population increased from 12000 to 96000 between 2011 and 2015, we can solve for this constant of proportionality:

$96000=12000e^{k(2015-2011)}$

$8=e^{4k}$

4k=ln(8)

$k=\frac{ln(8)}{4}=0.520$

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Example Question #3831 : Calculus

The rate of growth of the number of trash-hungry pigeons is proportional to the population. The population increased from 1300 to 28,600 between 2010 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

15.455

1.482

0.618

3.412

0.993

Correct answer:

0.618

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, and is the constant of proportionality.

Since the population increased from 1300 to 28,600 between 2010 and 2015, we can solve for this constant of proportionality:

$28600=1300e^{k(2015-2010)}$

$22=e^{5k}$

5k=ln(22)

$k=\frac{ln(22)}{5}=0.618$

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Example Question #3832 : Calculus

The rate of growth of the number of pinethrushes in Beulica is proportional to the population. The population increased from 5400 to 37,800 between 2009 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.154

1.507

0.982

0.233

0.324

Correct answer:

0.324

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, and is the constant of proportionality.

Since the population increased from 5400 to 37,800 between 2009 and 2015, we can solve for this constant of proportionality:

$37800=5400e^{k(2015-2009)}$

$7=e^{6k}$

6k=ln(7)

$k=\frac{ln(7)}{6}=0.324$

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Example Question #111 : Constant Of Proportionality

The rate of growth of the number of pine martens in a section of Napa County is proportional to the population. The population increased by 38.9 percent between 2012 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.134

0.073

0.296

0.110

0.181

Correct answer:

0.110

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, and is the constant of proportionality.

Since the population increased by 38.9 percent between 2012 and 2015, we can solve for this constant of proportionality:

$(1+0.389)y_0=y_0e^{k(2015-2012)}$

$1.389=e^{3k}$

3k=ln(1.389)

$k=\frac{ln(1.389)}{3}=0.110$

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Example Question #111 : How To Find Constant Of Proportionality Of Rate

The rate of growth of the number of moths in the wardrobe leading to a land of talking animals is proportional to the population. The population increased by 17.5 percent between January and April. What is the constant of proportionality in years^-1?

Possible Answers:

0.054

0.719

0.645

0.100

0.882

Correct answer:

0.645

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, and is the constant of proportionality.

Since the population increased by 17.5 percent between January and April, we can solve for this constant of proportionality. Mind the units that we're looking for, though it may be helpful to treat the months as their number in the calendar:

$(1+0.175)y_0=y_0e^{k(\frac{4-1}{12})}$

$1.175=e^{\frac{1}{4}k}$

$\frac{1}{4}k=ln(1.175)$

$k=\frac{ln(1.175)}{\frac{1}{4}}=0.645$

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