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Calculus 1 : How to find constant of proportionality of rate

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Example Questions

Example Question #71 : How To Find Constant Of Proportionality Of Rate

The rate of decrease of the number of grey wolves is proportional to the population. The population decreased by 33 percent between 2008 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-0.103 $\displaystyle -0.103$

-0.333 $\displaystyle -0.333$

-0.019 $\displaystyle -0.019$

-0.057 $\displaystyle -0.057$

-0.084 $\displaystyle -0.084$

Correct answer:

-0.057 $\displaystyle -0.057$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population decreased by 33 percent between 2008 and 2015, we can solve for this constant of proportionality:

$(1-0.33)y_0=y_0e^{k(2015-2008)}$ $\displaystyle (1-0.33)y_0=y_0e^{k(2015-2008)}$

$0.67=e^{7k}$ $\displaystyle 0.67=e^{7k}$

$\displaystyle 7k=ln(0.67)$

$k=\frac{ln(0.67)}{7}=-0.057$ $\displaystyle k=\frac{ln(0.67)}{7}=-0.057$

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Example Question #962 : Rate

The rate of decrease of the Siberian tiger population of northern Asia is proportional to the population. The population decreased by 18 percent between 2009 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-0.088 $\displaystyle -0.088$

-0.044 $\displaystyle -0.044$

-0.033 $\displaystyle -0.033$

-0.055 $\displaystyle -0.055$

-0.066 $\displaystyle -0.066$

Correct answer:

-0.033 $\displaystyle -0.033$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population decreased by 18 percent between 2009 and 2015, we can solve for this constant of proportionality:

$(1-0.18)y_0=y_0e^{k(2015-2009)}$ $\displaystyle (1-0.18)y_0=y_0e^{k(2015-2009)}$

$0.82=e^{6k}$ $\displaystyle 0.82=e^{6k}$

$\displaystyle 6k=ln(0.82)$

$k=\frac{ln(0.82)}{6}=-0.033$ $\displaystyle k=\frac{ln(0.82)}{6}=-0.033$

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Example Question #2762 : Functions

The rate of decrease of the number of puffins is proportional to the population. The population decreased by 27 percent between 2003 and 2007. What is the constant of proportionality in years^-1?

Possible Answers:

-0.225 $\displaystyle -0.225$

-1.701 $\displaystyle -1.701$

-0.028 $\displaystyle -0.028$

-0.079 $\displaystyle -0.079$

-0.011 $\displaystyle -0.011$

Correct answer:

-0.079 $\displaystyle -0.079$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population decreased by 27 percent between 2003 and 2007, we can solve for this constant of proportionality:

$(1-0.27)y_0=y_0e^{k(2007-2003)}$ $\displaystyle (1-0.27)y_0=y_0e^{k(2007-2003)}$

$0.73=e^{4k}$ $\displaystyle 0.73=e^{4k}$

$\displaystyle 4k=ln(0.73)$

$k=\frac{ln(0.73)}{4}=-0.079$ $\displaystyle k=\frac{ln(0.73)}{4}=-0.079$

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Example Question #72 : How To Find Constant Of Proportionality Of Rate

The rate of growth of the numbers of sunflowers in Nevada is proportional to the population. The population increased from 11000 to 17000 between 2012 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.122 $\displaystyle 0.122$

0.167 $\displaystyle 0.167$

0.112 $\displaystyle 0.112$

0.145 $\displaystyle 0.145$

0.098 $\displaystyle 0.098$

Correct answer:

0.145 $\displaystyle 0.145$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population increased from 11000 to 17000 between 2012 and 2015, we can solve for this constant of proportionality:

$17000=11000e^{k(2015-2012)}$ $\displaystyle 17000=11000e^{k(2015-2012)}$

$\frac{17}{11}=e^{3k}$ $\displaystyle \frac{17}{11}=e^{3k}$

$3k=ln(\frac{17}{11})$ $\displaystyle 3k=ln(\frac{17}{11})$

$k=\frac{ln(\frac{17}{11})}{3}=0.145$ $\displaystyle k=\frac{ln(\frac{17}{11})}{3}=0.145$

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Example Question #71 : Constant Of Proportionality

The rate of growth of the number of cottonwood trees in Reno is proportional to the population. The population increased from 1600 to 1900 between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

0.172 $\displaystyle 0.172$

0.144 $\displaystyle 0.144$

0.223 $\displaystyle 0.223$

0.190 $\displaystyle 0.190$

0.275 $\displaystyle 0.275$

Correct answer:

0.172 $\displaystyle 0.172$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population increased from 1600 to 1900 between 2014 and 2015, we can solve for this constant of proportionality:

$1900=1600e^{k(2015-2014)}$ $\displaystyle 1900=1600e^{k(2015-2014)}$

$\frac{19}{16}=e^{k}$ $\displaystyle \frac{19}{16}=e^{k}$

$k=ln(\frac{19}{16})=0.172$ $\displaystyle k=ln(\frac{19}{16})=0.172$

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Example Question #2761 : Functions

The rate of growth of the number of gray squirrels is proportional to the population. The population increased from 23000 to 52000 between 2011 and 2014. What is the constant of proportionality in years^-1?

Possible Answers:

0.388 $\displaystyle 0.388$

0.228 $\displaystyle 0.228$

0.272 $\displaystyle 0.272$

0.039 $\displaystyle 0.039$

0.251 $\displaystyle 0.251$

Correct answer:

0.272 $\displaystyle 0.272$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population increased from 23000 to 52000 between 2011 and 2014, we can solve for this constant of proportionality:

$52000=23000e^{k(2014-2011)}$ $\displaystyle 52000=23000e^{k(2014-2011)}$

$\frac{52}{23}=e^{3k}$ $\displaystyle \frac{52}{23}=e^{3k}$

$3k=ln(\frac{52}{23})$ $\displaystyle 3k=ln(\frac{52}{23})$

$k=\frac{ln(\frac{52}{23})}{3}=0.272$ $\displaystyle k=\frac{ln(\frac{52}{23})}{3}=0.272$

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Example Question #3791 : Calculus

The rate of decrease of the number of African elephants is proportional to the population. The population decreased from 2700 to 1900 between 2013 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-0.132 $\displaystyle -0.132$

-0.099 $\displaystyle -0.099$

-0.241 $\displaystyle -0.241$

-0.209 $\displaystyle -0.209$

-0.176 $\displaystyle -0.176$

Correct answer:

-0.176 $\displaystyle -0.176$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population decreased from 2700 to 1900 between 2013 and 2015, we can solve for this constant of proportionality:

$1900=2700e^{k(2015-2013)}$ $\displaystyle 1900=2700e^{k(2015-2013)}$

$\frac{19}{27}=e^{2k}$ $\displaystyle \frac{19}{27}=e^{2k}$

$2k=ln(\frac{19}{27})$ $\displaystyle 2k=ln(\frac{19}{27})$

$k=\frac{ln(\frac{19}{27})}{2}=-0.176$ $\displaystyle k=\frac{ln(\frac{19}{27})}{2}=-0.176$

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Example Question #3798 : Calculus

The rate of decrease of typhoid sufferers, due to the advent of modern medicine, is proportional to the population. The population decreased from 250,000 to 80,000 between 2014 and 2015. What is the constant of proportionality in years^-1?

Possible Answers:

-1.139 $\displaystyle -1.139$

-0.775 $\displaystyle -0.775$

-2.913 $\displaystyle -2.913$

-0.988 $\displaystyle -0.988$

-0.533 $\displaystyle -0.533$

Correct answer:

-1.139 $\displaystyle -1.139$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population decreased from 250,000 to 80,000 between 2014 and 2015, we can solve for this constant of proportionality:

$80000=250000e^{k(2015-2014)}$ $\displaystyle 80000=250000e^{k(2015-2014)}$

$\frac{8}{25}=e^{k}$ $\displaystyle \frac{8}{25}=e^{k}$

$k=ln(\frac{8}{25})=-1.139$ $\displaystyle k=ln(\frac{8}{25})=-1.139$

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Example Question #72 : How To Find Constant Of Proportionality Of Rate

The rate of decrease of the number of Amur Leopards is proportional to the population. The population decreased from 113 to 27 between 1999 and 2007. What is the constant of proportionality in years^-1?

Possible Answers:

-0.648 $\displaystyle -0.648$

-1.432 $\displaystyle -1.432$

-3.511 $\displaystyle -3.511$

0.331 $\displaystyle 0.331$

-0.179 $\displaystyle -0.179$

Correct answer:

-0.179 $\displaystyle -0.179$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population decreased from 113 to 27 between 1999 and 2007, we can solve for this constant of proportionality:

$27=113e^{k(2007-1999)}$ $\displaystyle 27=113e^{k(2007-1999)}$

$\frac{27}{113}=e^{8k}$ $\displaystyle \frac{27}{113}=e^{8k}$

$8k=ln(\frac{27}{113})$ $\displaystyle 8k=ln(\frac{27}{113})$

$k=\frac{ln(\frac{27}{113})}{8}=-0.179$ $\displaystyle k=\frac{ln(\frac{27}{113})}{8}=-0.179$

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Example Question #3791 : Calculus

The rate of growth of the number of cholera causing bacteria in a pound is proportional to the population. The population increased from 100 to 750 between 3:20 and 4:00. What is the constant of proportionality in minutes^-1?

Possible Answers:

0.109 $\displaystyle 0.109$

0.050 $\displaystyle 0.050$

0.038 $\displaystyle 0.038$

0.081 $\displaystyle 0.081$

0.062 $\displaystyle 0.062$

Correct answer:

0.050 $\displaystyle 0.050$

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:

$p(t)=p_0e^{kt}$ $\displaystyle p(t)=p_0e^{kt}$

Where p_0 $\displaystyle p_0$ is an initial population value, and $\displaystyle k$ is the constant of proportionality.

Since the population increased from 100 to 750 between 3:20 and 4:00, we can solve for this constant of proportionality:

$750=100e^{k(4:00-3:20)}$ $\displaystyle 750=100e^{k(4:00-3:20)}$

$\frac{15}{2}=e^{40k}$ $\displaystyle \frac{15}{2}=e^{40k}$

$40k=ln(\frac{15}{2})$ $\displaystyle 40k=ln(\frac{15}{2})$

$k=\frac{ln(\frac{15}{2})}{40}=0.050$ $\displaystyle k=\frac{ln(\frac{15}{2})}{40}=0.050$

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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 #71 : How To Find Constant Of Proportionality Of Rate

Example Question #962 : Rate

Example Question #2762 : Functions

Example Question #72 : How To Find Constant Of Proportionality Of Rate

Example Question #71 : Constant Of Proportionality

Example Question #2761 : Functions

Example Question #3791 : Calculus

Example Question #3798 : Calculus

Example Question #72 : How To Find Constant Of Proportionality Of Rate

Example Question #3791 : Calculus

All Calculus 1 Resources

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