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Calculus 1 : Rate

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Example Question #185 : Constant Of Proportionality

The population of a town grows exponentially from 200K to 800K in 3yrs . What is the population growth constant?

Possible Answers:

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

k=3

k=4

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

Correct answer:

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

Explanation:

Exponential growth is modeled by the equation $A=Pe^{kt}$

where is the final amount, is the inital amount, is the growth constant and is time.

In this problem, A=800K , P=200K and t=3 . Substituting these variables into the growth equation the solving for gives us

$800,000=200,000e^{3k}$

$\frac{800,000}{200,000}=e^{3k}$

$4=e^{3k}$

$ln(4)=ln(e^{3k})$

ln(4)=3k

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

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Example Question #186 : Constant Of Proportionality

Cobalt-60 has a half-life of 5.27yrs . What is the decay constant of Cobalt-60?

Possible Answers:

k=-0.5

$k=\frac{ln(0.5)}{60}$

$k=\frac{ln(60)}{5.27}$

$k=\frac{ln(0.5)}{5.27}$

Correct answer:

$k=\frac{ln(0.5)}{5.27}$

Explanation:

The half-life of an isotope is the time it takes for half the isotope to disappear. Isotopes decay exponentially.

Exponential decay is also modeled by the equation $A=Pe^{kt}$

where is the final amount, is the inital amount, is the growth constant and is time.

Since half the isotope has disappeared, the final amount is half the inital amount , or $\frac{A}{P}=0.5$ .

In this problem, t=5.27yrs .

Substituting these variables into the exponential equation and solving for gives us

$A=Pe^{kt}\Rightarrow \frac{A}{P}=e^{kt}$

$0.5=e^{5.27k}$

$ln(0.5)=ln(e^{5.27k})$

ln(0.5)=5.27k

$k=\frac{ln(0.5)}{5.27}$

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Example Question #187 : Constant Of Proportionality

The number of cats double every 64dys . How many cats will there be after 1 yr if there are cats initially?

Possible Answers:

$A\approx 730$

$A\approx 64$

$A\approx 104$

$A\approx 365$

Correct answer:

$A\approx 104$

Explanation:

Exponential growth is modeled by the equation $A=Pe^{kt}$

where is the final amount, is the inital amount, is the growth constant and is time.

After 64dys , the number of cats has doubled, or the final amount is double the inital amount , or $\frac{A}{P}=2$ .

In this problem, t=64 .

Substituting these variables into the exponential equation and solving for gives us

$A=Pe^{kt}\Rightarrow \frac{A}{P}=e^{kt}$

$2=e^{64k}$

$ln(2)=ln(e^{64k})$

ln(2)=64k

$k=\frac{ln(2)}{64}$

To find the number cats after year, P=2 , $k=\frac{ln(2)}{64}$ , and t=365

$A=Pe^{kt}=2e^{\frac{ln(2)}{64}*365}\approx 104$

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Example Question #188 : Constant Of Proportionality

The number of students enrolled in college has increased by $37\%$ every year since 2003 . If 460K students enrolled in 2003 , how many student enrolled in 2013 ?

Possible Answers:

y=10.3K

y=460M

y=2

y=10.7M

Correct answer:

y=10.7M

Explanation:

The exponential growth is modeled by the equation $y=a(1+r)^{t}$

where is the final amount, is the inital amount, is the growth rate and is time.

In this problem, a=460K , r=0.37 , and t=2013-2003=10 . Substituting these values into the equation gives us

$y=460,000(1+0.37)^{10}=10.7M$

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Example Question #189 : Constant Of Proportionality

The number of CD players owned has decreased by $15\%$ annually since 2008 . If people owned CD players in 2008 , how many people owned CD players in 2014 ?

Possible Answers:

y=1.13M

y=3K

y=390K

y=11M

Correct answer:

y=1.13M

Explanation:

The exponential growth is modeled by the equation $y=a(1+r)^{t}$

where is the final amount, is the inital amount, is the growth rate and is time.

In this problem, a=3M and t=2014-2008=6 . r=-0.15 because the rate is decreasing. Substituting these values into the equation gives us

$y=3,000,000(1-0.15)^{6}=1.13M$

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Example Question #190 : Constant Of Proportionality

You deposit $\$1500$ into your savings account. After 8 yrs , your account has $\$6500$ in it. What is the interest rate of this account if the account was untouched during the 8 yrs ?

Possible Answers:

$r=20.12\%$

$r=2.38\%$

$r=40.76\%$

$r=0.85\%$

Correct answer:

$r=20.12\%$

Explanation:

The exponential growth is modeled by the equation $y=a(1+r)^{t}$

where is the final amount, is the inital amount, is the growth rate and is time.

In this problem, $a=\$1500$ , $y=\$6500$ and t=8 yrs . Substituting these variables into the growth equation and solving for r gives us

$6500=1500(1+r)^{8}$

$\frac{13}{3}=(1+r)^{8}$

$\sqrt[8]{\frac{13}{3}}=\sqrt[8]{(1+r)^{8}}$

$\sqrt[8]{\frac{13}{3}}=1+r$

1+r=1.2012

$r=0.2012=20.12\%$

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Example Question #191 : How To Find Constant Of Proportionality Of Rate

Find the direct constant of proportionality of f(x)=e^t-cos(2t) from $t:(0,\pi)$ .

Possible Answers:

$\frac{e^\pi+1}{\pi}$

$\frac{e^\pi-1}{\pi}$

$\frac{-e^\pi+1}{\pi}$

$\frac{e^\pi}{\pi}$

Correct answer:

$\frac{e^\pi-1}{\pi}$

Explanation:

To find direct constant of proportionality , you need to find the slope of the line between the two points at t=0 and $t=\pi$ . To do so we use the formula:

f(x)=e^t-cos(2t)

$k=\frac{f(\pi)-f(0)}{\pi -0}=\frac{(e^\pi-cos(2*\pi))-(e^0-cos(0))}{\pi}=\frac{e^\pi-1}{\pi}$

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Example Question #1081 : Rate

The rate of growth of the population of Reindeer in Norway is proportional to the population. The population increased from 9876 to 10381 between 2013 and 2015. What is the expected population in 2030?

Possible Answers:

15150

53880

42581

49362

38914

Correct answer:

15150

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 from 9876 to 10381 between 2013 and 2015, we can solve for this constant of proportionality:

$10381=9876e^{k(2015-2013)}$

$\frac{10381}{9876}=e^{2k}$

$2k=ln(\frac{10381}{9876})$

$k=\frac{ln(\frac{10381}{9876})}{2}=0.0252$

Using this, we can calculate the expected value from 2015 to 2030:

$P=10381e^{(2030-2015)(0.0252)} \approx 15150$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #185 : Constant Of Proportionality

Example Question #186 : Constant Of Proportionality

Example Question #187 : Constant Of Proportionality

Example Question #188 : Constant Of Proportionality

Example Question #189 : Constant Of Proportionality

Example Question #190 : Constant Of Proportionality

Example Question #191 : How To Find Constant Of Proportionality Of Rate

Example Question #1081 : Rate

All Calculus 1 Resources

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