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Calculus 1 : How to find constant of proportionality of rate

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Example Question #61 : Constant Of Proportionality

The rate of change of the staph bacterial number in a petri dish is proportional to the population. The population increased by 300 percent over the course of 40 minutes. What is the constant of proportionality in hours^-1?

Possible Answers:

0.034

0.582

8.225

0.007

2.079

Correct answer:

2.079

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, and is the constant of proportionality.

Since the population increased by 300 percent over the course of 40 minutes, we can solve for this constant of proportionality. Mind the units required:

$(1+3)y_0=y_0e^{k(\frac{40}{60})}$

$4=e^{\frac{2}{3}k}$

$\frac{2}{3}k=ln(4)$

$k=\frac{ln(4)}{\frac{2}{3}}=2.079$

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Example Question #62 : Constant Of Proportionality

The rate of change of the salmonella prescence in a contaminated egg is proportional to the population. The population increased by 0.33 percent over the course of 89 seconds. What is the constant of proportionality in hours^-1?

Possible Answers:

4.83107

0.00003

0.13326

0.02824

0.00522

Correct answer:

0.13326

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, and is the constant of proportionality.

Since the population increased by 0.33 percent over the course of 89 seconds, we can solve for this constant of proportionality. Mind the unites:

$(1+0.0033)y_0=y_0e^{k(\frac{89}{3600})}$

$1.0033=e^{\frac{89}{3600}k}$

$\frac{89}{3600}k=ln(1.0033)$

$k=\frac{ln(1.0033)}{\frac{89}{3600}}=0.1333$

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Example Question #63 : Constant Of Proportionality

The rate of change of the number of black bears in an untouched stretch of California forest is proportional to the population. The population increased by 57 percent between January of 2013 and January of 2015. What is the constant of proportionality in months^-1?

Possible Answers:

10.826

0.628

0.019

0.451

5.233

Correct answer:

0.019

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, and is the constant of proportionality.

Since the population increased by 57 percent between January of 2013 and January of 2015, we can solve for this constant of proportionality. Be careful of the units used, however:

$(1+0.57)y_0=y_0e^{k(12(2015-2013))}$

$1.57=e^{24k}$

24k=ln(1.57)

$k=\frac{ln(1.57)}{24}=0.019$

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Example Question #64 : Constant Of Proportionality

The rate of change of the number cholera causing bacteria in a stagnant pool is proportional to the population. The population increased by 83 percent between 3:15 and 4:30. What is the constant of proportionality in hours^-1?

Possible Answers:

0.405

0.029

8.113

0.008

0.483

Correct answer:

0.483

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, and is the constant of proportionality.

Since the population increased by 83 percent between 3:15 and 4:30, we can solve for this constant of proportionality. Since we're asked for the units of inverse hours, convert the minutes to decimals by dividing by 60:

$(1+0.83)y_0=y_0e^{k(4.5-3.25)}$

$1.83=e^{1.25k}$

1.25k=ln(1.83)

$k=\frac{ln(1.83)}{1.25}=0.483$

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Example Question #61 : Constant Of Proportionality

The rate of change of the trout population in Lake Tahoe is proportional to the population. The population increased by 42 percent between January and April. What is the constant of proportionality in months^-1?

Possible Answers:

0.117

0.351

0.088

-0.748

-1.036

Correct answer:

0.117

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, and is the constant of proportionality.

Since the population increased by 42 percent between January and April, we can solve for this constant of proportionality. Treat the months as their number in the calendar:

$(1+0.42)y_0=y_0e^{k(4-1)}$

$1.42=e^{3k}$

3k=ln(1.42)

$k=\frac{ln(1.42)}{3}=0.117$

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Example Question #962 : Rate

The rate of change of the the number of E. coli in room-temperature hamburger is proportional to the population. The population increased by 100 percent between 4:10 and 4:30. What is the constant of proportionality in seconds^-1?

Possible Answers:

0.0116

0.6932

831.7766

0.0006

0.0347

Correct answer:

0.0006

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, and is the constant of proportionality.

Since the population increased by 100 percent between 4:10 and 4:30, we can solve for this constant of proportionality. Note that units we're asked to use:

$(1+1)y_0=y_0e^{k(20*60)}$

$2=e^{1200k}$

1200k=ln(2)

$k=\frac{ln(2)}{1200}=0.0006$

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Example Question #963 : Rate

The rate of change of the number of pike in Lake Mead is proportional to the population. The population increased by 48 percent between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.392

0.056

0.893

0.128

0.204

Correct answer:

0.392

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, and is the constant of proportionality.

Since the population increased by 48 percent between 2014 and 2015, we can solve for this constant of proportionality:

$(1+0.48)y_0=y_0e^{k(2015-2014)}$

$1.48=e^{k}$

k=ln(1.48)=0.392

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Example Question #964 : Rate

The rate of growth of the population of wild foxes in Britain is proportional to the population. The population increased by 113 percent between 2011 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.092

0.189

0.240

0.213

0.283

Correct answer:

0.189

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, and is the constant of proportionality.

Since the population increased by 113 percent between 2011 and 2015, we can solve for this constant of proportionality:

$(1+1.13)y_0=y_0e^{k(2015-2011)}$

$2.13=e^{4k}$

4k=ln(2.13)

$k=\frac{ln(2.13)}{4}=0.189$

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Example Question #965 : Rate

The rate of growth of the number of swallows in Saksegawa is proportional to the population. The population increased by 8 percent between 2013 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.049

0.027

0.060

0.071

0.038

Correct answer:

0.038

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, and is the constant of proportionality.

Since the population increased by 8 percent between 2013 and 2015, we can solve for this constant of proportionality:

$(1+0.08)y_0=y_0e^{k(2015-2013)}$

$1.08=e^{2k}$

2k=ln(1.08)

$k=\frac{ln(1.08)}{2}=0.038$

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Example Question #966 : Rate

The rate of growth of the number of belugas in the pacific is proportional to the population. The population increased by 148 percent between 2010 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.202

0.182

0.138

0.121

0.165

Correct answer:

0.182

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, and is the constant of proportionality.

Since the population increased by 148 percent between 2010 and 2015, we can solve for this constant of proportionality:

$(1+1.48)y_0=y_0e^{k(2015-2010)}$

$2.48=e^{5k}$

5k=ln(2.48)

$k=\frac{ln(2.48)}{5}=0.182$

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Calculus 1 : How to find constant of proportionality of rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #61 : Constant Of Proportionality

Example Question #62 : Constant Of Proportionality

Example Question #63 : Constant Of Proportionality

Example Question #64 : Constant Of Proportionality

Example Question #61 : Constant Of Proportionality

Example Question #962 : Rate

Example Question #963 : Rate

Example Question #964 : Rate

Example Question #965 : Rate

Example Question #966 : Rate

All Calculus 1 Resources

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