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Calculus 1 : How to find constant of proportionality of rate

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Example Questions

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:

2022

2020

2018

2021

2019

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:

3:49

4:22

3:58

4:06

3:23

Correct answer:

3:58

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 80 to 560 between 3:15 and 3:27, we can solve for this constant of proportionality. Convert the minutes to decimals by dividing by 60:

$560=80e^{k(3.45-3.25)}$

$7=e^{0.2k}$

0.2k=ln(7)

$k=\frac{ln(7)}{0.2}=9.730$

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

$84000=560e^{9.730(t-3.45)}$

9.730(t-3.45)=ln(150)

$t=3.45+\frac{ln(150)}{9.730}=3.965 \rightarrow 3:58$

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Example Question #132 : How To Find Constant Of Proportionality Of Rate

The rate of chamge of the number of fungal spores in the presence of an herbicide is proportional to the population. The population decreased from 9600 to 1600 between 3:21 and 3:36. At what point in time will the population be 20?

Possible Answers:

4:48

5:01

4:32

3:59

4:13

Correct answer:

4:13

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 9600 to 1600 between 3:21 and 3:36, we can solve for this constant of proportionality. Convert minutes to decimals by dividing by 60:

$1600=9600e^{k(3.60-3.35)}$

$\frac{1}{6}=e^{0.25k}$

$0.25k=ln(\frac{1}{6})$

$k=\frac{ln(\frac{1}{6})}{0.25}=-7.167$

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

$20=1600e^{-7.167(t-3.60)}$

$-7.167(t-3.60)=ln(\frac{1}{80})$

$t=3.60+\frac{ln(\frac{1}{80})}{-7.167}=4.211 \rightarrow 4:13$

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Example Question #131 : Constant Of Proportionality

The rate of change of the number of direwolves is proportional to the population. The population increased by 259 percent between 1605 and 1607. What is the constant of proportionality in years^-1?

Possible Answers:

0.43

0.86

0.64

0.32

0.59

Correct answer:

0.64

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 259 percent between 1605 and 1607, we can solve for this constant of proportionality:

$(1+2.59)y_0=y_0e^{k(1607-1605)}$

$3.59=e^{2k}$

2k=ln(3.59)

$k=\frac{ln(3.59)}{2}=0.64$

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Example Question #134 : How To Find Constant Of Proportionality Of Rate

The rate of change of the number of samoyeds is proportional to the population. The population increased by 269 percent between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

2.05

1.25

1.09

1.31

1.73

Correct answer:

1.31

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

$(1+2.69)y_0=y_0e^{k(2015-2014)}$

$3.69=e^{k}$

k=ln(3.69)=1.31

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Example Question #1031 : Rate

The rate of change of the number of bacteria in a coffee mug is proportional to the population. The population increased by 518 percent between 3:15 and 3:25. What is the constant of proportionality in minutes^-1?

Possible Answers:

2.108

0.115

0.182

0.599

0.164

Correct answer:

0.182

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 518 percent between 3:15 and 3:25, we can solve for this constant of proportionality:

$(1+5.18)y_0=y_0e^{k(3:25-3:15)}$

$6.18=e^{10k}$

10k=ln(6.18)

$k=\frac{ln(6.18)}{10}=0.182$

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Example Question #1032 : Rate

The rate of change of the number of emus is proportional to the population. The population decreased by 54 percent between 2013 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-1.47

-1.31

-1.84

-0.31

-0.39

Correct answer:

-0.39

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 by 54 percent between 2013 and 2015, we can solve for this constant of proportionality:

$(1-0.54)y_0=y_0e^{k(2015-2013)}$

$0.46=e^{2k}$

2k=ln(0.46)

$k=\frac{ln(0.46)}{2}=-0.39$

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Example Question #1033 : Rate

The rate of change of the number of E. coli cells in the presence of a disinfectant spray is proportional to the population. The population decreased by 99.8 percent between 3:10 and 3:15. What is the constant of proportionality in minutes^-1?

Possible Answers:

-3.91

-7.82

-6.08

-4.55

-1.24

Correct answer:

-1.24

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 by 99.8 percent between 3:10 and 3:15, we can solve for this constant of proportionality:

$(1-0.998)y_0=y_0e^{k(3:15-3:10)}$

$0.002=e^{5k}$

5k=ln(0.002)

$k=\frac{ln(0.002)}{5}=-1.24$

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Example Question #1034 : Rate

The rate of change of the number of sabre cats is proportional to the population. The population decreased by 84.1 percent between 1001 and 1005. What is the constant of proportionality in years^-1?

Possible Answers:

-1.56

-3.23

-1.84

-0.46

-0.04

Correct answer:

-0.46

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 by 84.1 percent between 1001 and 1005, we can solve for this constant of proportionality:

$(1-0.841)y_0=y_0e^{k(1005-1001)}$

$0.159=e^{4k}$

4k=ln(0.159)

$k=\frac{ln(0.159)}{4}=-0.46$

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Example Question #1035 : Rate

The rate of change of the number of body snatchers is proportional to the population. The population increased from 11 to 1331 between 8 am and 9am. What is the constant of proportionality in hours^-1?

Possible Answers:

9.59

13.31

7.19

2.45

4.80

Correct answer:

4.80

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 11 to 1331 between 8 am and 9am, we can solve for this constant of proportionality:

$1331=11e^{k(9-8)}$

$121=e^{k}$

k=ln(121)=4.80

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