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# Exponential Decay

## Rational functions with x as an exponent in the denominator

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Exponential Decay

You observe that the glass of water you took out of the microwave dropped from 190 deg to 170 deg in 1 min. In another minute, the temperature has dropped to 152 deg., and after 3 mins, it is down to 136 deg.

What type of function would allow you to calculate the projected temperature of the water over time?

### Exponential Decay

A decreasing quantity can be modeled with an exponential function in much the same way as an increasing (growing) quantity can. This kind of situation is referred to as exponential decay.

Perhaps the most common example of exponential decay is that of radioactive decay, which refers to the transformation of an atom of one type into an atom of a different type, when the nucleus of the atom loses energy. The rate of radioactive decay is usually measured in terms of “half-life,” or the time it takes for half of the atoms in a sample to decay. For example Carbon-14 is a radioactive isotope that is used in “carbon dating,” a method of determining the age of organic materials. The half-life of Carbon-14 is 5730 years. This means that if we have a sample of Carbon-14, it will take 5730 years for half of the sample to decay. Then it will take another 5730 years for half of the remaining sample to decay, and so on.

We can model decay using the same form of equation we use to model growth, except that the exponent in the equation is negative: A(t) = A0 e-kt.

Newton’s Law of Cooling is an exponential decay model. The Law of Cooling allows us to determine the temperature of a cooling (or warming) object, based on the temperature of the surroundings and the time since the object entered the surroundings. The general form of the cooling function is T(x) = T5 + (T0 - T5) e-kx, where T5, is the surrounding temperature, T0 is the initial temperature, and x represents the time since the object began cooling or warming.

The first graph shows a situation in which an object is cooling. The graph has a horizontal asymptote at y = 70. This tells us that the object is cooling to \begin{align*}70^{\circ}\end{align*}F. The second graph has a horizontal asymptote at y = 70 as well, but in this situation, the object is warming up to \begin{align*}70^{\circ}\end{align*}F. We can use the general form of the function to answer questions about cooling (or warming) situations.

### Examples

#### Example 1

Earlier, you were asked to find the type of function that would allow you to calculate the projected temperature of water over time.

The temperature drop of the water over time is an example of exponential decay.

To predict the temperature of the water over time, you would use a function like \begin{align*}T_f = T_i (.8947)^t\end{align*} where \begin{align*}T_f\end{align*} and \begin{align*}T_i\end{align*} are starting and ending temperature and \begin{align*}t\end{align*} is time in minutes.

#### Example 2

You have a sample of Carbon-14. How much time will pass before 75% of the original sample remains?

We can use the half-life of 5730 years to determine the value of k:

\begin{align*}A(t) = A_0e^{-kt}\end{align*}
\begin{align*}\frac{1} {2} = 1e^{-k\cdot 5730}\end{align*} We do not know the value of A0, so we use “1” as 100%. (1/2) of the sample remains when t = 5730 years
\begin{align*}ln \frac{1} {2} = ln e^{-k\cdot 5730}\end{align*} Take the ln of both sides
\begin{align*}ln \frac{1} {2} = -5730 klne\end{align*} Use the power property of logs
\begin{align*}ln \frac{1} {2} = -5730k\end{align*} \begin{align*}ln(e) = 1\end{align*}
\begin{align*}-ln 2 = -5730k\end{align*} \begin{align*}ln(1/2) = ln(2^{-1})=-ln2\end{align*}
\begin{align*}ln 2 = 5730k\end{align*}
\begin{align*}k = \frac{ln2} {5730}\end{align*} Isolate k

Now we can determine when the amount of Carbon-14 remaining is 75% of the original:

\begin{align*}0.75 = 1 e^{\frac{-ln 2} {5730}t}\end{align*}

\begin{align*}0.75 = 1 e^{\frac{-ln 2} {5730}t}\end{align*}

\begin{align*}ln (0.75) = ln e^{\frac{-ln 2} {5730}t}\end{align*}

\begin{align*}ln (0.75) = \frac{-ln 2} {5730}t\end{align*}

\begin{align*}t = \frac{5730 ln (0.75)} {-ln 2} \approx 2378\end{align*}

Therefore it would take about 2,378 years for 75% of the original sample to be remaining.

#### Example 3

You are baking a casserole in a dish, and the oven is set to \begin{align*}325^{\circ}\end{align*} F. You take the pan out of the oven and put it on a cooling rack in your kitchen which is \begin{align*}70^{\circ}\end{align*} F, and after 10 minutes the pan has cooled to \begin{align*}300^{\circ}\end{align*}F. How long will it take for the pan to cool to 200F?

We can use the general form of the equation and the information given in the problem to find the value of k:

\begin{align*}T(x) = T_s + (T_0 -T_s) e^{-kx}\end{align*}

\begin{align*}T(x) = 70 + (325 - 70)e^{-kx}\end{align*}

\begin{align*}T(x) = 70 + (255)e^{-kx}\end{align*}

\begin{align*}T(10) = 70 + 255e^{-10k} = 300\end{align*}

\begin{align*}255e^{-10k} = 230\end{align*}

\begin{align*}e^{-10k} = \frac{230} {255}\end{align*}

\begin{align*}ln e^{-10k} = ln \left (\frac{230} {255}\right )\end{align*}

\begin{align*}-10k = ln \left (\frac{230} {255}\right )\end{align*}

\begin{align*}k = \frac{ln \left (\frac{230} {255}\right )} {-10} \approx 0.0103\end{align*}

Now we can determine the amount of time it takes for the pan to cool to 200 degrees:

\begin{align*}T(x) = 70 + (255)e^{-.0103}x\end{align*}

\begin{align*}T(x) = 70 + (255)e^{-.0103}x\end{align*}

\begin{align*}200 = 70 + (255)e^{-.0103}x\end{align*}

\begin{align*}130 = (255)e^{-.0103}x\end{align*}

\begin{align*}\frac{130} {255} = e^{-.0103}x\end{align*}

\begin{align*}ln \left (\frac{130} {255}\right ) = ln e^{-.0103}x\end{align*}

\begin{align*}ln \left (\frac{130} {255}\right ) = -.0103 x\end{align*}

\begin{align*}x = \frac{ln \left (\frac{130} {255}\right )} {-.0103} \approx 65\end{align*}

Therefore, in the given surroundings, it would take about an hour for the pan to cool to 200 degrees.

#### Example 4

When doctors prescribe medicine, they consider how quickly the drug’s effectiveness decreases as time passes in order to calculate the time to administer the next dose. If a drug is only 85% as effective each hour as it was the previous hour, at some point the patient must be given another dose. If the initial dose was 350 mg, how long will it take for the initial dose to reach the minimum level of 83 mg, to the nearest hour?

Use the general decay formula: A(t) = A0 rt, to solve for t, time in hours.

Substituting gives: 83 = 350 (.85)t

Dividing both sides by 350 gives: .237 = .85t

Using logs: log .237 = log .85t

Properties of logs: log .237 = t(log .85)

Dividing to isolate t: t = log .237 / log .85

Using a calculator: t = (-.625) / (-.07) = 9 hours (apx.)

For the following examples, use the model below:

Exponential Decay Model \begin{align*}A_f = A_i (1 - r)^t\end{align*}
\begin{align*}A_f\end{align*} = final amount
\begin{align*}A_i\end{align*} = initial amount
\begin{align*}r\end{align*} = decay rate (percent as a decimal)
\begin{align*}t\end{align*} = time
\begin{align*}(1 - r)\end{align*} = decay factor

#### Example 5

If a particular Egyptian artifact originally contained 21 grams of carbon-14, and if carbon-14 decays at a rate modeled by \begin{align*}A_f = A_i (e)^{-0.000121t}\end{align*} how much carbon-14 should be present after 25000 years?

\begin{align*}A_f = A_i (e)^{-0.000121t}\end{align*}

\begin{align*}A_f = 21e^{-0.000121(25000)}\end{align*} substitute the given values

\begin{align*}A_f = 21e^{-3.025}\end{align*} simplify

\begin{align*}A_f = 21 \cdot .048557821\end{align*} with a calculator

\begin{align*}A_f \approx 1.019714\end{align*}

\begin{align*}\therefore\end{align*} there will be approximately 1.019714 grams of carbon-14 left after 25,000 years.

#### Example 6

thorium-234 decays with a half-life of 24 days to protactinium-234. If the decay rate can be calculated by \begin{align*}t_{\frac{1}{2}} = \frac{ln 2}{r}\end{align*} where \begin{align*}t_{\frac{1}{2}}\end{align*} is the half-life of the material and \begin{align*}r\end{align*} is the rate of decay, what is the rate of decay of thorium-234?

To calculate decay rate, begin with the formula given in the problem:

\begin{align*}t_{\frac{1}{2}} = \frac{ln 2}{r}\end{align*}

\begin{align*}24 = \frac{ln 2}{r}\end{align*} : substitute the given half-life value

\begin{align*}24 = \frac{.69}{r}\end{align*} : substitute the value of \begin{align*}ln 2\end{align*} with a calculator

\begin{align*}r = \frac{.69}{24}\end{align*} : re-arrange to solve for \begin{align*}r\end{align*}

\begin{align*}r \approx .029\end{align*}

\begin{align*}\therefore\end{align*} the rate of decay is \begin{align*}\approx 2.9\%\end{align*} per day

#### Example 7

If thorium-234 has a decay rate of \begin{align*}\approx .0289\end{align*} as calculated above, and if you begin with a sample of thorium-234 weighing 37 grams, how long will it take before you have only 23 grams?

To calculate time, begin with the exponential decay formula:

\begin{align*}A_f = A_i (1 - r)^t\end{align*}

\begin{align*}23 = 34 (.971)^t\end{align*} : substitute the given values

\begin{align*}.6765 = .971^t\end{align*} : divide both sides by 34

\begin{align*}log .6765 = log .971^t\end{align*} : take the log of both sides

\begin{align*}log .6765 = (t) log .971\end{align*} : using \begin{align*}log x^y = y log x\end{align*}

\begin{align*}\frac{log .6765}{log .971} = t\end{align*} : divide both sides by \begin{align*}log .971\end{align*}

\begin{align*}13.28 = t\end{align*} : with a calculator

\begin{align*}\therefore\end{align*} it will be approximately 13.28 days before the 37 gram sample of thorium-234 decays into a 23 gram sample of thorium-234.

### Review

Are the following equations examples of exponential decay?

1. \begin{align*}y = 0.7(1.1)^t\end{align*}
2. \begin{align*}y = 0.95(0.3t)^2\end{align*}
3. Write an exponential decay model for a new car valued at $28,000 which depreciates at a rate of 8% per year. 4. Write an exponential decay model for a stock in a new dyeing company is valued at$500.00. Its value decreases by 7% each year.
5. What is the value of a car after 5 years, if you buy it for 12,000 and it depreciates at a rate of 10% annually? Identify the initial amount and the decay rate in the exponential function. 1. \begin{align*}y = 10(1 - 0.2)^t\end{align*} 2. \begin{align*} y = 18(0.11)^t\end{align*} 3. \begin{align*}y = 2(\frac{1}{4})^t\end{align*} Write an exponential function to model the situation: 1. A32,000 boat depreciates at a rate of 5.2% every other year.
2. The population of a small country is 185,000 people and it increases by 1.75% semi-annually.
3. The initial investment of $732.00 in a mutual fund is losing value at a rate of 5% every 4mos. 4. You have owned a car for 6 years. You purchased the car for$16,000. If it has depreciated at a rate of 6.5% annually, how much is it worth today?

An artifact originally had 10 grams of carbon-14 present. The decay model \begin{align*}A = 10e^{-0.000121t}\end{align*} describes the amount of carbon after \begin{align*}t\end{align*} years.

1. How many grams of carbon-14 will be present in this artifact after 17,000 years?
2. What is the half-life of carbon-14?

A certain substance decays at a daily relative rate of .495%. The initial amount of the quantity is 300 mg.

1. Express the quantity as a function of time (in days) and find how much of the quantity will be left after one week.

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### Vocabulary Language: English

Exponential decay

Exponential decay occurs when a quantity decreases by the same proportion in each given time period.

Exponential Function

An exponential function is a function whose variable is in the exponent. The general form is $y=a \cdot b^{x-h}+k$.

Half-life

Half-life refers to the time required for a radioactive material to decay to one-half of its initial concentration.

Model

A model is a mathematical expression or function used to describe a physical item or situation.

Newton's Law of Cooling

Newton's law of cooling is used to calculate changes in temperature of an object immersed in a fluid of a different temperature.