How to find constant of proportionality of rate - CLEP Calculus

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Question

The rate of decrease of the population of E. coli in response to a particular antibiotic is proportional to the population. The population decreased from 160,000 to 8,000 between 1:00 and 1:30. Determine the expected population at 1:45.

Answer

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:

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

Where 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 160,000 to 8,000 between 1:00 and 1:30, we can solve for this constant of proportionality. Treat the minutes as decimals after the hour by dividing by 60:

$8000=160000e^{k(1.5-1)}$

$\frac{1}{20}=e^{0.5k}$

$0.5k=ln(\frac{1}{20})$

$k=\frac{ln(\frac{1}{20})}{0.5}=-5.991$

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:

$P=8000e^{(1.75-1.5)(-5.991)} \approx 1789$

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Question

The rate of growth of the culture of bacteria on a dirty plate is proportional to the population. The population increased from 50 to 200 between 1:15 and 1:30. At what point in time approximately would the population be 700?

Answer

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 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 50 to 200 between 1:15 and 1:30, we can solve for this constant of proportionality. Treat the minutes as decimals after the hour by dividing by 60:

$200=50e^{k(1.5-1.25)}$

$4=e^{0.25k}$

$k=\frac{ln(4)}{0.25}=5.545$

Now that the constant of proportionality is known, we can use it to estimate our time point:

$700=200e^{5.545(t-1.5)}$

$5.545(t-1.5)=ln(\frac{7}{2})$

$t=1.5+\frac{ln(\frac{7}{2})}{5.545} = 1.726 \rightarrow 1:44$

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Question

Suppose a blood cell increases proportionally to the present amount. If there were blood cells to begin with, and blood cells are present after hours, what is the growth constant?

Answer

The population size after some time is given by:

where is the initial population.

At the start, there were 30 blood cells.

Substitute this value into the given formula.

After 2 hours, 45 blood cells were present. Write this in mathematical form.

Substitute this into , and solve for .

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

$ln\left(\frac{3}{2}\right)=ln(e^2^k)$

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

$k=\frac{1}{2}ln\left(\frac{3}{2}\right)$

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Question

The rate of decrease of the robomen population due to the transgalactic organic tai cyber conflict is proportional to the population. The population decreased from 6 billion to 450 million between 2063 and 2081. Determine the expected population in 2100.

Answer

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 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 6 billion to 450 million between 2063 and 2081, we can solve for this constant of proportionality:

$450000000=6000000000e^{k(2081-2063)}$

$\frac{3}{40}=e^{18k}$

$18k=ln(\frac{3}{40})$

$k=\frac{ln(\frac{3}{40})}{18}=-0.1439$

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:

$P=450000000e^{(2100-2081)(-0.1439)} \approx 29228600$

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Question

Find the direct constant of proportionality of from to .

Answer

To determine the direct constant of proportionality, we determine the rate of change from and for .

Rate of change is determined by

$r=\frac{\Delta y}{\Delta x}$ .

In our case, between and , the rate of change is

$r=\frac{2.5^2-1}{2.5-1}=\frac{5.25}{1.5}=3.5$ .

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Question

Given any linear function , determine the direct constant of proportionality

Answer

Direct constant of proportionality for any given function y, between any x values, is given by

$r=\frac{\Delta y}{\Delta x}$ , where is the direction constant of proportionality

In the case of a linear function

$\frac{\Delta y}{\Delta x}$ is the same thing as the slope.

Therefore, the constant of proportionality is

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Question

Find the direct constant of proportionality of from to .

Answer

Direct constant of proportionality is given by

$r=\frac{\Delta y}{\Delta x}$ .

Since and $\Delta x= 4-2=2$

$r= \frac{\Delta y}{\Delta x}=\frac{ln(34)-ln(32)}{2}= \frac{ln(2)}{2}$

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Question

Suppose a population of bacteria increases from to in . What is the constant of growth?

Answer

The equation for population growth is given by $P=Ae^{kt}$ . is the population, is the intial value, is time, and is the growth constant. We can plug in the values we know at time and solve for .

$100=Ae^{0k}=A$

Now that we solved for , we can plug in what we know for time and solve for .

$1000=100e^{30k}$

$e^{30k}=10$

$30k=\ln(10)$

$k=\frac{\ln(10)}{30}$

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Question

A population of deer grew from 50 to 200 in 7 years. What is the growth constant for this population?

Answer

The equation for population growth is given by $P=Ae^{kt}$ . P is the population, is the intial value, is time, and is the growth constant. We can plug in the values we know at time and solve for .

$50=Ae^{0k}=A$

Now that we have solved for we can solve for at

$200=50e^{7k}$

$e^{7k}=4$

$7k=\ln(4)$

$k=\frac{\ln(4)}{7}$

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Question

A population of mice has 200 mice. After 6 weeks, there are 1600 mice in the population. What is the constant of growth?

Answer

The equation for population growth is given by $P=Ae^{kt}$ . is the population, is the intial value, is time, and is the growth constant. We can plug in the values we know at time and solve for .

$200=Ae^{0k}=A$

Now that we have we can solve for at .

$1600=200e^{6k}$

$e^{6k}=8$

$6k=\ln(8)$

$k=\frac{\ln(8)}{6}$

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Question

Find the direct constant of proportionality of from to $x=\frac{\pi}{3}$ .

Answer

Direct constant of proportionality is given by

$k=\frac{\Delta y}{\Delta x}$ , where $\Delta y$ is the change in the position and $\Delta x$ is the change in the position.

Since , and we're going from to $x=\frac{\pi}{3}$

$\Delta y=3sec(\frac{\pi}{3})-3sec(0)=6-3=3$

$\Delta x=\frac{\pi}{3}-0=\frac{\pi}{3}$

$k=\frac{3}{\frac{\pi}{3}}=\frac{9}{\pi}$

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Question

The rate of growth of the duck population in Wingfield is proportional to the population. The population increased by 15 percent between 2001 and 2008. What is the constant of proportionality?

Answer

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 is an initial population value, and is the constant of proportionality.

Since the population increased by 15 percent between 2001 and 2008, we can solve for this constant of proportionality:

$(1+.15)y_0=y_0e^{k(2008-2001)}$

$1.15=e^{7k}$

$k=\frac{ln(1.15)}{7}=0.01997$

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Question

The rate of growth of the culture of E.coli on a piece of room temperature meat is proportional to the population. The population increased from 400 to 1600 between 1:15 and 1:45. Determine the expected population at 3:15.

Answer

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 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 400 to 1600 between 1:15 and 1:45, we can solve for this constant of proportionality. Treat the minutes as decimals after the hour by dividing by 60:

$1600=400e^{k(1.75-1.25)}$

$4=e^{0.5k}$

$k=\frac{ln(4)}{0.5}=2.773$

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:

$P=1600e^{(3.25-1.75)(2.773)} \approx 102463$

Which is why we use refrigerators.

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Question

The rate of growth of the bacteria in an agar dish is proportional to the population. The population increased by 150 percent between 1:15 and 2:30. What is the constant of proportionality?

Answer

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 is an initial population value, and is the constant of proportionality.

Since the population increased by 150 percent between 1:15 and 2:30, we can solve for this constant of proportionality:

$(1+1.50)y_0=y_0e^{k(2:30 - 1:15)}$

Dealing in minutes:

$2.50=e^{75k}$

$k=\frac{ln(2.50)}{75}=0.012$

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Question

The rate of growth of the Martian Transgalactic Constituency is proportional to the population. The population increased by 23 percent between 2530 and 2534 AD. What is the constant of proportionality?

Answer

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 is an initial population value, and is the constant of proportionality.

Since the population increased by 23 percent between 2530 and 2534 AD, we can solve for this constant of proportionality:

$(1+0.23)y_0=y_0e^{k(2534-2530)}$

$1.23=e^{4k}$

$k=\frac{ln(1.23)}{4}=0.05$

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Question

The rate of decrease of the dwindling wolf population of Zion National Park is proportional to the population. The population decreased by 7 percent between 2009 and 2011. What is the constant of proportionality?

Answer

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 is an initial population value, and is the constant of proportionality.

Since the population decreased by 7 percent between 2009 and 2011, we can solve for this constant of proportionality:

$(1-0.07)y_0=y_0e^{k(2011-2009)}$

$0.93=e^{2k}$

$k=\frac{ln(0.93)}{2}=-0.036$

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Question

The rate of decrease of the panda population is proportional to the population. The population decreased by 12 percent between 1990 and 2001. What is the constant of proportionality?

Answer

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 is an initial population value, and is the constant of proportionality.

Since the population decreased by 12 percent between 1990 and 2001, we can solve for this constant of proportionality:

$(1-0.12)y_0=y_0e^{k(2001-1990)}$

$0.88=e^{11k}$

$k=\frac{ln(0.88)}{11}=-0.012$

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Question

The rate of growth of the salmon population of Yuba is proportional to the population. The population increased by 21 percent over the course of seven years. What is the constant of proportionality?

Answer

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 is an initial population value, and is the constant of proportionality.

Since the population increased by 21 percent over the course of seven years, we can solve for this constant of proportionality:

$(1+0.21)y_0=y_0e^{7k}$

$1.21=e^{7k}$

$k=\frac{ln(1.21)}{7}=0.027$

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Question

The rate of growth of the robot army of Cloudweb is proportional to the population. The population increased by 89 percent between 2034 and 2037. What is the constant of proportionality for this terrifying army which threatens man's dominance?

Answer

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 is an initial population value, and is the constant of proportionality.

Since the population increased by 89 percent between 2034 and 2037, we can solve for this constant of proportionality:

$(1+0.89)y_0=y_0e^{k(2037-2034)}$

$1.89=e^{3k}$

$k=\frac{ln(1.89)}{3}=0.21$

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Question

The rate of decrease of the number of concert attendees to former teen heartthrob Justice Beaver is proportional to the population. The population decreased by 34 percent between 2013 and 2015. What is the constant of proportionality?

Answer

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 is an initial population value, and is the constant of proportionality.

Since the population decreased by 34 percent between 2013 and 2015, we can solve for this constant of proportionality:

$(1-0.34)y_0=y_0e^{k(2015-2013)}$

$0.66=e^{2k}$

$k=\frac{ln(0.66)}{2}=-0.21$

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