College Dating In The Colony

Posted By admin On 21.05.21

Summary: Previous to Ryan's current city of The Colony, TX, Ryan Grambling lived in El Paso TX. Joshua Waknine, Prentice Hill, Andrea Pernell, Wendy Harmon and Carmen Aponte, and many others are family members and associates of Ryan. Dating requires time management. College is nowhere near as structured and organized as high school is. The day doesn't start and end at the same time for everybody. Instead, people's days begin.

Learning Outcomes

Graph exponential growth and decay functions.
Solve problems involving radioactive decay, carbon dating, and half life.

Exponential Growth and Decay

In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:

[latex]y={A}_{0}{e}^{kt}[/latex]

where [latex]{A}_{0}[/latex] is equal to the value at time zero, e is Euler’s constant, and k is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.

On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the exponential decay model. Again, we have the form [latex]y={A}_{0}{e}^{-kt}[/latex] where [latex]{A}_{0}[/latex] is the starting value, and e is Euler’s constant. Now k is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating half-life, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.

In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in the graphs below. It is important to remember that, although parts of each of the two graphs seem to lie on the x-axis, they are really a tiny distance above the x-axis.

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A graph showing exponential growth. The equation is [latex]y=2{e}^{3x}[/latex]. Dating spot in new market va menu.

A graph showing exponential decay. The equation is [latex]y=3{e}^{-2x}[/latex].

Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The order of magnitude is the power of ten when the number is expressed in scientific notation with one digit to the left of the decimal. For example, the distance to the nearest star, Proxima Centauri, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is [latex]4.01134972times {10}^{13}[/latex]. We could describe this number as having order of magnitude [latex]{10}^{13}[/latex].

A General Note: Characteristics of the Exponential Function [latex]y=A_{0}e^{kt}[/latex]

An exponential function of the form [latex]y={A}_{0}{e}^{kt}[/latex] has the following characteristics:

one-to-one function
horizontal asymptote: y = 0
domain: [latex]left(-infty , infty right)[/latex]
range: [latex]left(0,infty right)[/latex]
x intercept: none
y-intercept: [latex]left(0,{A}_{0}right)[/latex]
increasing if k > 0
decreasing if k < 0

An exponential function models exponential growth when k > 0 and exponential decay when k < 0.

Example: Graphing Exponential Growth

A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

Show Solution

When an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[/latex] we use the fact that [latex]{A}_{0}[/latex] is the amount at time zero, so [latex]{A}_{0}=10[/latex]. To find k, use the fact that after one hour [latex]left(t=1right)[/latex] the population doubles from 10 to 20. The formula is derived as follows:

[latex]begin{array}{l}text{ }20=10{e}^{kcdot 1}hfill & hfill text{ }2={e}^{k}hfill & text{Divide both sides by 10}hfill mathrm{ln}2=khfill & text{Take the natural logarithm of both sides}hfill end{array}[/latex]

so [latex]k=mathrm{ln}left(2right)[/latex]. Thus the equation we want to graph is [latex]y=10{e}^{left(mathrm{ln}2right)t}=10{left({e}^{mathrm{ln}2}right)}^{t}=10cdot {2}^{t}[/latex]. The graph is shown below.

The graph of [latex]y=10{e}^{left(mathrm{ln}2right)t}[/latex].

Analysis of the Solution

The population of bacteria after ten hours is 10,240. We could describe this amount as being of the order of magnitude [latex]{10}^{4}[/latex]. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude [latex]{10}^{7}[/latex], so we could say that the population has increased by three orders of magnitude in ten hours.

Calculating Doubling Time

For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the doubling time.

Given the basic exponential growth equation [latex]A={A}_{0}{e}^{kt}[/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[/latex].

The formula is derived as follows:

[latex]begin{array}{l}2{A}_{0}={A}_{0}{e}^{kt}hfill & hfill 2={e}^{kt}hfill & text{Divide both sides by }{A}_{0}.hfill mathrm{ln}2=kthfill & text{Take the natural logarithm of both sides}.hfill t=frac{mathrm{ln}2}{k}hfill & text{Divide by the coefficient of }t.hfill end{array}[/latex]

Thus the doubling time is

[latex]t=frac{mathrm{ln}2}{k}[/latex]

Example: Finding a Function That Describes Exponential Growth

According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.

Show Solution

The formula is derived as follows: Sevierville best hookup site.

[latex]begin{array}{l}t=frac{mathrm{ln}2}{k}hfill & text{The doubling time formula}.hfill 2=frac{mathrm{ln}2}{k}hfill & text{Use a doubling time of two years}.hfill k=frac{mathrm{ln}2}{2}hfill & text{Multiply by }ktext{ and divide by 2}.hfill A={A}_{0}{e}^{frac{mathrm{ln}2}{2}t}hfill & text{Substitute }ktext{ into the continuous growth formula}.hfill end{array}[/latex]

The function is [latex]A={A}_{0}{e}^{frac{mathrm{ln}2}{2}t}[/latex].

Try It

Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.

Show Solution

[latex]fleft(tright)={A}_{0}{e}^{frac{mathrm{ln}2}{3}t}[/latex]

Half-Life

We now turn to exponential decay. One of the common terms associated with exponential decay, as stated above, is half-life, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.

To find the half-life of a function describing exponential decay, solve the following equation:

[latex]frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[/latex]

We find that the half-life depends only on the constant k and not on the starting quantity [latex]{A}_{0}[/latex].

The formula is derived as follows

[latex]begin{array}{l}frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}hfill & hfill frac{1}{2}={e}^{kt}hfill & text{Divide both sides by }{A}_{0}.hfill mathrm{ln}left(frac{1}{2}right)=kthfill & text{Take the natural log of both sides}.hfill -mathrm{ln}left(2right)=kthfill & text{Apply properties of logarithms}.hfill -frac{mathrm{ln}left(2right)}{k}=thfill & text{Divide by }k.hfill end{array}[/latex]

Since t, the time, is positive, k must, as expected, be negative. This gives us the half-life formula

[latex]t=-frac{mathrm{ln}left(2right)}{k}[/latex]

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. The table below lists the half-life for several of the more common radioactive substances.

Substance	Use	Half-life
gallium-67	nuclear medicine	80 hours
cobalt-60	manufacturing	5.3 years
technetium-99m	nuclear medicine	6 hours
americium-241	construction	432 years
carbon-14	archeological dating	5,715 years
uranium-235	atomic power	703,800,000 years

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

[latex]begin{array}{l}Aleft(tright)={A}_{0}{e}^{frac{mathrm{ln}left(0.5right)}{T}t}hfill Aleft(tright)={A}_{0}{e}^{mathrm{ln}left(0.5right)frac{t}{T}}hfill Aleft(tright)={A}_{0}{left({e}^{mathrm{ln}left(0.5right)}right)}^{frac{t}{T}}hfill Aleft(tright)={A}_{0}{left(frac{1}{2}right)}^{frac{t}{T}}hfill end{array}[/latex]

where

[latex]{A}_{0}[/latex] is the amount initially present
[latex]T[/latex] is the half-life of the substance
[latex]t[/latex] is the time period over which the substance is studied
[latex]A[/latex], or [latex]A(t)[/latex], is the amount of the substance present after time [latex]t[/latex]

How To: Given the half-life, find the decay rate

Write [latex]A={A}_{o}{e}^{kt}[/latex].
Replace A by [latex]frac{1}{2}{A}_{0}[/latex] and replace t by the given half-life.
Solve to find k. Express k as an exact value (do not round).

Note: It is also possible to find the decay rate using [latex]k=-frac{mathrm{ln}left(2right)}{t}[/latex].

Example: Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for 10% of a 1000-gram sample of uranium-235 to decay?

Show Solution

[latex]begin{array}{l}text{ }y=text{1000}efrac{mathrm{ln}left(0.5right)}{text{703,800,000}}thfill & hfill text{ }900=1000{e}^{frac{mathrm{ln}left(0.5right)}{text{703,800,000}}t}hfill & text{After 10% decays, 900 grams are left}.hfill text{ }0.9={e}^{frac{mathrm{ln}left(0.5right)}{text{703,800,000}}t}hfill & text{Divide by 1000}.hfill mathrm{ln}left(0.9right)=mathrm{ln}left({e}^{frac{mathrm{ln}left(0.5right)}{text{703,800,000}}t}right)hfill & text{Take ln of both sides}.hfill mathrm{ln}left(0.9right)=frac{mathrm{ln}left(0.5right)}{text{703,800,000}}thfill & text{ln}left({e}^{M}right)=Mhfill text{}text{}t=text{703,800,000}times frac{mathrm{ln}left(0.9right)}{mathrm{ln}left(0.5right)}text{years}hfill & text{Solve for }t.hfill text{}text{}tapprox text{106,979,777 years}hfill & hfill end{array}[/latex]

Analysis of the Solution

Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.

Try It

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

Show Solution

[latex]t=703,800,000times frac{mathrm{ln}left(0.8right)}{mathrm{ln}left(0.5right)}text{ years }approx text{ }226,572,993text{ years}[/latex].

Example: Finding the Function that Describes Radioactive Decay

The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t.

Show Solution

This formula is derived as follows.

[latex]begin{array}{l}text{}A={A}_{0}{e}^{kt}hfill & text{The continuous growth formula}.hfill 0.5{A}_{0}={A}_{0}{e}^{kcdot 5730}hfill & text{Substitute the half-life for }ttext{ and }0.5{A}_{0}text{ for }fleft(tright).hfill text{}0.5={e}^{5730k}hfill & text{Divide both sides by }{A}_{0}.hfill mathrm{ln}left(0.5right)=5730khfill & text{Take the natural log of both sides}.hfill text{}k=frac{mathrm{ln}left(0.5right)}{5730}hfill & text{Divide by the coefficient of }k.hfill text{}A={A}_{0}{e}^{left(frac{mathrm{ln}left(0.5right)}{5730}right)t}hfill & text{Substitute for }rtext{ in the continuous growth formula}.hfill end{array}[/latex]

The function that describes this continuous decay is [latex]fleft(tright)={A}_{0}{e}^{left(frac{mathrm{ln}left(0.5right)}{5730}right)t}[/latex]. We observe that the coefficient of t, [latex]frac{mathrm{ln}left(0.5right)}{5730}approx -1.2097[/latex] is negative, as expected in the case of exponential decay.

Try It

The half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of plutonium-244 remaining as a function of time measured in years.

Show Solution

[latex]fleft(tright)={A}_{0}{e}^{-0.0000000087t}[/latex]

Radiocarbon Dating

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The formula for radioactive decay is important in radiocarbon dating which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.

Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12 which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.

As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.

Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after t years is

[latex]Aapprox {A}_{0}{e}^{left(frac{mathrm{ln}left(0.5right)}{5730}right)t}[/latex]

where

[latex]A[/latex] is the amount of carbon-14 remaining
[latex]{A}_{0}[/latex] is the amount of carbon-14 when the plant or animal began decaying.

This formula is derived as follows:

[latex]begin{array}{l}text{ }A={A}_{0}{e}^{kt}hfill & text{The continuous growth formula}.hfill text{ }0.5{A}_{0}={A}_{0}{e}^{kcdot 5730}hfill & text{Substitute the half-life for }ttext{ and }0.5{A}_{0}text{ for }fleft(tright).hfill text{ }0.5={e}^{5730k}hfill & text{Divide both sides by }{A}_{0}.hfill mathrm{ln}left(0.5right)=5730khfill & text{Take the natural log of both sides}.hfill text{ }k=frac{mathrm{ln}left(0.5right)}{5730}hfill & text{Divide both sides by the coefficient of }k.hfill text{ }A={A}_{0}{e}^{left(frac{mathrm{ln}left(0.5right)}{5730}right)t}hfill & text{Substitute for }rtext{ in the continuous growth formula}.hfill end{array}[/latex]

To find the age of an object we solve this equation for t:

[latex]t=frac{mathrm{ln}left(frac{A}{{A}_{0}}right)}{-0.000121}[/latex]

Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let r be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated determined by a method called liquid scintillation. From the equation [latex]Aapprox {A}_{0}{e}^{-0.000121t}[/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=frac{A}{{A}_{0}}approx {e}^{-0.000121t}[/latex]. We solve this equation for t, to get

[latex]t=frac{mathrm{ln}left(rright)}{-0.000121}[/latex]

How To: Given the percentage of carbon-14 in an object, determine its age

Express the given percentage of carbon-14 as an equivalent decimal r.
Substitute for r in the equation [latex]t=frac{mathrm{ln}left(rright)}{-0.000121}[/latex] and solve for the age, t.

Example: Finding the Age of a Bone

A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?

Show Solution

We substitute 20% = 0.20 for r in the equation and solve for t:

[latex]begin{array}{l}t=frac{mathrm{ln}left(rright)}{-0.000121}hfill & text{Use the general form of the equation}.hfill =frac{mathrm{ln}left(0.20right)}{-0.000121}hfill & text{Substitute for }r.hfill approx 13301hfill & text{Round to the nearest year}.hfill end{array}[/latex]

The bone fragment is about 13,301 years old.

Analysis of the Solution

The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as [latex]text{13,301 years}pm text{1% or 13,301 years}pm text{133 years}[/latex].

Try It

Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?

Show Solution