3.12: Exponential Decay
You observe that the glass of water you took out of the microwave dropped from 190deg to 170deg in 1 min. In another minute, the temperature has dropped to 152deg., and after 3 mins, it is down to 136deg.
What type of function would allow you to calculate the projected temperature of the water over time?
Watch This
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 James Sousa: Exponential Decay Models
Guidance
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 “halflife,” or the time it takes for half of the atoms in a sample to decay. For example Carbon14 is a radioactive isotope that is used in “carbon dating,” a method of determining the age of organic materials. The halflife of Carbon14 is 5730 years. This means that if we have a sample of Carbon14, 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) = A_{0} 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) = T_{5} + (T_{0}  T_{5}) e^{}^{kx}, where T_{5}, is the surrounding temperature, T_{0} 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*}
Example A
You have a sample of Carbon14. How much time will pass before 75% of the original sample remains?
Solution:
We can use the halflife of 5730 years to determine the value of k:

\begin{align*}A(t) = A_0e^{kt}\end{align*} A(t)=A0e−kt \begin{align*}\frac{1} {2} = 1e^{k\cdot 5730}\end{align*} 12=1e−k⋅5730 We do not know the value of A_{0}, 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*} ln12=lne−k⋅5730 Take the ln of both sides \begin{align*}ln \frac{1} {2} = 5730 klne\end{align*} ln12=−5730klne Use the power property of logs \begin{align*}ln \frac{1} {2} = 5730k\end{align*} ln12=−5730k \begin{align*}ln(e) = 1\end{align*} ln(e)=1 \begin{align*}ln 2 = 5730k\end{align*} −ln2=−5730k \begin{align*}ln(1/2) = ln(2^{1})=ln2\end{align*} ln(1/2)=ln(2−1)=−ln2 \begin{align*}ln 2 = 5730k\end{align*} ln2=5730k \begin{align*}k = \frac{ln2} {5730}\end{align*} k=ln25730 Isolate k
Now we can determine when the amount of Carbon14 remaining is 75% of the original:


\begin{align*}0.75 = 1 e^{\frac{ln 2} {5730}t}\end{align*}
0.75=1e−ln25730t 
\begin{align*}0.75 = 1 e^{\frac{ln 2} {5730}t}\end{align*}
0.75=1e−ln25730t 
\begin{align*}ln (0.75) = ln e^{\frac{ln 2} {5730}t}\end{align*}
ln(0.75)=lne−ln25730t 
\begin{align*}ln (0.75) = \frac{ln 2} {5730}t\end{align*}
ln(0.75)=−ln25730t 
\begin{align*}t = \frac{5730 ln (0.75)} {ln 2} \approx 2378\end{align*}
t=5730ln(0.75)−ln2≈2378

\begin{align*}0.75 = 1 e^{\frac{ln 2} {5730}t}\end{align*}
'Therefore it would take about 2,378 years for 75% of the original sample to be remaining.'
Example B
You are baking a casserole in a dish, and the oven is set to \begin{align*}325^{\circ}\end{align*}
Solution:
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*}
T(x)=Ts+(T0−Ts)e−kx 
\begin{align*}T(x) = 70 + (325  70)e^{kx}\end{align*}
T(x)=70+(325−70)e−kx 
\begin{align*}T(x) = 70 + (255)e^{kx}\end{align*}
T(x)=70+(255)e−kx 
\begin{align*}T(10) = 70 + 255e^{10k} = 300\end{align*}
T(10)=70+255e−10k=300 
\begin{align*}255e^{10k} = 230\end{align*}
255e−10k=230 
\begin{align*}e^{10k} = \frac{230} {255}\end{align*}
e−10k=230255 
\begin{align*}ln e^{10k} = ln \left (\frac{230} {255}\right )\end{align*}
lne−10k=ln(230255) 
\begin{align*}10k = ln \left (\frac{230} {255}\right )\end{align*}
−10k=ln(230255) 
\begin{align*}k = \frac{ln \left (\frac{230} {255}\right )} {10} \approx 0.0103\end{align*}
k=ln(230255)−10≈0.0103

\begin{align*}T(x) = T_s + (T_0 T_s) e^{kx}\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 C
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?
Solution
Use the general decay formula: A(t) = A_{0} r^{t}, to solve for t, time in hours.
Substituting gives: 83 = 350 (.85)^{t}
Dividing both sides by 350 gives: .237 = .85^{t}
Using logs: log .237 = log .85^{t}
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.)
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. 

Vocabulary
Exponential decay occurs when a quantity decreases by the same proportion in each given time period.
Halflife refers to the time required for a radioactive material to decay to onehalf of its initial concentration.
Newton's Law of Cooling is used to calculate changes in temperature of an object immersed in a fluid of a different temperature.
Guided Practice
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 
1) If a particular Egyptian artifact originally contained 21 grams of carbon14, and if carbon14 decays at a rate modeled by \begin{align*}A_f = A_i (e)^{0.000121t}\end{align*} how much carbon14 should be present after 25000 years?
2) thorium234 decays with a halflife of 24 days to protactinium234 (actual figure). 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 halflife of the material and \begin{align*}r\end{align*} is the rate of decay, what is the rate of decay of thorium234?
3) If thorium234 has a decay rate of \begin{align*}\approx .0289\end{align*} as calculated above, and if you begin with a sample of thorium234 weighing 37 grams, how long will it take before you have only 23 grams?
Answers
1) To calculate the remaining carbon14, start with the model:
 \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 carbon14 left after 25,000 years.
2) 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 halflife 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*} : rearrange 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
3) 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 thorium234 decays into a 23 gram sample of thorium234.
Practice
Are the following equations examples of exponential decay?
 \begin{align*}y = 0.7(1.1)^t\end{align*}
 \begin{align*}y = 0.95(0.3t)^2\end{align*}
 Write an exponential decay model for a new car valued at $28,000 which depreciates at a rate of 8% per year.
 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.
 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.
 \begin{align*}y = 10(1  0.2)^t\end{align*}
 \begin{align*} y = 18(0.11)^t\end{align*}
 \begin{align*}y = 2(\frac{1}{4})^t\end{align*}
Write an exponential function to model the situation:
 A $32,000 boat depreciates at a rate of 5.2% every other year.
 The population of a small country is 185,000 people and it increases by 1.75% semiannually.
 The initial investment of $732.00 in a mutual fund is losing value at a rate of 5% every 4mos.
 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 carbon14 present. The decay model \begin{align*}A = 10e^{0.000121t}\end{align*} describes the amount of carbon after \begin{align*}t\end{align*} years.
 How many grams of carbon14 will be present in this artifact after 17,000 years?
 What is the halflife of carbon14?
A certain substance decays at a daily relative rate of .495%. The initial amount of the quantity is 300 mg.
 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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Exponential decay
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