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Calculus 1 : Rate

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Example Questions

Example Question #1021 : Rate

The rate of change of the number of leopard geckos is proportional to the population. The population increased by 45.5 percent between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.110

0.281

0.529

0.375

0.639

Correct answer:

0.375

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 45.5 percent between 2014 and 2015, we can solve for this constant of proportionality:

$(1+0.455)y_0=y_0e^{k(2015-2014)}$

$1.455=e^{k}$

k=ln(1.455)=0.375

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Example Question #2811 : Functions

The rate of decrease of the number of rupees due to plundering green-clad adventurers is proportional to the population. The population decreased from 999 to 25 between 3:30 and 5:42. What is the constant of proportionality in hours^-1?

Possible Answers:

-1.903

-0.259

-1.676

-2.222

-0.454

Correct answer:

-1.676

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 decreased from 999 to 25 between 3:30 and 5:42, we can solve for this constant of proportionality. Mind the units requested; divide the minutes by 60 to give decimals after the hour:

$25=999e^{k(5.7-3.5)}$

$\frac{25}{999}=e^{2.2k}$

$2.2k=ln(\frac{25}{999})$

$k=\frac{ln(\frac{25}{999})}{2.2}=-1.676$

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Example Question #1021 : Rate

The rate of loss of the number of jewelers' gold rings due to thieving hedgehogs is proportional to the population. The population decreased from 1000 to 50 between 3:30 and 3:45. What is the constant of proportionality in minutes^-1?

Possible Answers:

-9.909

-5.117

-11.983

-3.098

-0.200

Correct answer:

-0.200

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 decreased from 1000 to 50 between 3:30 and 3:45, we can solve for this constant of proportionality:

$50=1000e^{k(3:45-3:30)}$

$\frac{1}{20}=e^{15k}$

$15k=ln(\frac{1}{20})$

$k=\frac{ln(\frac{1}{20})}{15}=-0.200$

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Example Question #1024 : Rate

The rate of change of the number of math problems is proportional to the population. The population increased from 10 to 2000 between 2014 and 2015. Determine the expected population in 2017.

Possible Answers:

$7.995 \cdot 10^8$

$7.995 \cdot 10^4$

$7.995 \cdot 10^5$

$7.995 \cdot 10^7$

$7.995 \cdot 10^6$

Correct answer:

$7.995 \cdot 10^7$

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 10 to 2000 between 2014 and 2015, we can solve for this constant of proportionality:

$2000=10e^{k(2015-2014)}$

$200=e^{k}$

k=ln(200)=5.298

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:

$P=2000e^{(5.298)(2017-2015)} \approx 7.995 \cdot 10^7$

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Example Question #2812 : Functions

The rate of change of the number of bacteria on an old dish rag is proportional to the population. The population increased from 13 to 169 between 3:15 and 3:30. Determine the expected population at 4:00.

Possible Answers:

12783

28564

36418

84380

4174

Correct answer:

28564

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 13 to 169 between 3:15 and 3:30, we can solve for this constant of proportionality. Treating the minutes as decimals by dividing by 60:

$169=13e^{k(3.5-3.25)}$

$13=e^{0.25k}$

0.25k=ln(13)

$k=\frac{ln(13)}{0.25}=10.260$

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:

$P=169e^{(10.260)(4-3.5)} \approx 28564$

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Example Question #1022 : Rate

The rate of change of the number of quiblobs due to the influence of morplibs is proportional to the population. The population increased from 18 to 324 between 2908 and 2909. Determine the expected population in 2921.

Possible Answers:

$8.21\cdot10^{19}$

$8.21\cdot10^{15}$

$3.73\cdot10^{13}$

$3.73\cdot10^{9}$

$3.73\cdot10^{17$

Correct answer:

$3.73\cdot10^{17$

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 18 to 324 between 2908 and 2909, we can solve for this constant of proportionality:

$324=18e^{k(2909-2908)}$

$18=e^{k}$

k=ln(18)=2.890

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:

$P=324e^{(2.890)(2921-2909)} \approx 3.73\cdot10^{17}$

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Example Question #131 : How To Find Constant Of Proportionality Of Rate

The rate of chamge of the number of vape shops is inexplicably proportional to the population. The population increased from 18 to 90 between 2014 and 2015. Which year will the population pass 56,000?

Possible Answers:

2021

2022

2019

2020

2018

Correct answer:

2019

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 18 to 90 between 2014 and 2015, we can solve for this constant of proportionality:

$90=18e^{k(2015-2014)}$

$5=e^{k}$

k=ln(5)=1.609

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

$56000=90e^{1.609(t-2015)}$

$1.609(t-2015)=ln(\frac{5600}{9})$

$t=2015+\frac{ln(\frac{5600}{9})}{1.609} =2018.998 \rightarrow 2019$

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Example Question #131 : How To Find Constant Of Proportionality Of Rate

The rate of chamge of the number of yeast cells in a rising pizza dough is proportional to the population. The population increased from 80 to 560 between 3:15 and 3:27. At what point in time will the population be 84,000?

Possible Answers: