4.7: Exponential and Logarithmic Models
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Learning Objectives
- Compound and Continuous Interest applications
- Exponential growth and decay models.
- Newton’s Law of Heating and Cooling.
- Logistic-growth models.
In previous sections we were able to solve some applications that were modeled with exponential equations. Now that we have so many more techniques to solve these equations, we are able to solve more applications.
Compound and Continuous Interest Formulas
We will again use the Compound Interest Formulas and so we list them here for reference.
Definition 4.7.1: Compound Interest
For a principal, P, invested at an interest rate, r, for t years, the new balance, A is:
A=P(1+rn)nt when compounded n times a year. A=Pert when compounded continuously.
Recall that compound interest occurs when interest accumulated for one period is added to the principal investment before calculating interest for the next period. The amount A accrued in this manner over time t is modeled by the compound interest formula A(t)=P(1+rn)nt where the initial principal P is accumulating compound interest at an annual rate r where the value n represents the number of times the interest is compounded in a year.
Example 4.7.1r: Compounded interest - Find time and total amount
Susan invested $500 in an account earning 412% annual interest that is compounded monthly.
a. How much will be in the account after 3 years? b. How long will it take for the amount to grow to $750?
Solution
State variable values and formula. In this example, the principal P= $500, the interest rate r=412% =0.045, and because the interest is compounded monthly, n=12. The investment can be modeled by the following formula:
A(t)=P(1+rn)ntFormulaA(t)=500(1+0.04512)12tA(t)=500(1.00375)12tModel equation
a. Use this model equation (4.7.1) to calculate the amount in the account after t=3 years.
A(3)=500(1.00375)12(3)=500(1.00375)36≈572.12
Rounded off to the nearest cent, after 3 years, the amount accumulated will be $572.12.
b. To calculate the time it takes to accumulate $750, set A(t)=750 and solve for t using Equation (4.7.1).
A(t)=500(1.00375)12t750=500(1.00375)12t
This results in an exponential equation that can be solved by first isolating the exponential expression.
750=500(1.00375)12t750500=(1.00375)12t1.5=(1.00375)12t
At this point take the common logarithm of both sides, apply the power rule for logarithms, and then solve for t.
log(1.5)=log(1.00375)12tlog(1.5)=12tlog(1.00375)log(1.5)12log(1.00375)=12tlog(1.00375)12log(1.00375)log(1.5)12log(1.00375)=t
Using a calculator we can approximate the time it takes.
t=log(1.5)/(12∗log(1.00375))≈9 years
Answers:
a. $572.12 b. Approximately 9 years
Example 4.7.2r: Compounded interest - Find time
Mario invested $1000 in an account earning 6.3% annual interest, that is compounded semi-annually. How long will it take the investment to double?
Solution
State variable values and formula. Here the principal P= $1,000, the interest rate r=6.3% =0.063, and because the interest is compounded semi-annually n=2. This investment can be modeled as follows:
A(t)=P(1+rn)ntFormulaA(t)=1,000(1+0.0632)2tA(t)=1,000(1.0315)2tModel equation
Since we are looking for the time it takes to double $1,000, substitute $2,000 for the resulting amount A(t) and then solve for t.
2,000=1,000(1.0315)2t2,0001,000=(1.0315)2t2=(1.0315)2t
At this point we take the common logarithm of both sides.
2=(1.0315)2tlog2=log(1.0315)2tlog2=2tlog(1.0315)log22log(1.0315)=t
Using a calculator we can approximate the time it takes: t=log(2)/(2∗log(1.0315))≈11.17.
So the answer is it takes approximately 11.17 years to double at 6.3%.
If the investment in the previous example was one million dollars, how long would it take to double? To answer this we would use P= $1,000,000 and A(t)= $2,000,000:
A(t)=1,000(1.0315)2t2,000,000=1,000,000(1.0315)2t
Dividing both sides by 1,000,000 we obtain the same exponential function as before.
2=(1.0315)2t
Hence, the result will be the same, about 11.17 years. In fact, doubling time is independent of the initial investment P.
Interest is typically compounded semi-annually (n=2), quarterly (n=4), monthly (n=12), or daily (n=365). However if interest is compounded every instant we obtain a formula for continuously compounding interest: A(t)=Pert where P represents the initial principal amount invested, r represents the annual interest rate, and t represents the time in years the investment is allowed to accrue continuously compounded interest.
Example 4.7.3r: Continuous compounding - Find time
Mary invested $200 in an account earning 534% annual interest that is compounded continuously. How long will it take the investment to grow to $350?
Solution
State variable values and formula. Here the principal P= $200 and the interest rate r=534% =5.75% =0.0575. Since the interest is compounded continuously, use the formula A(t)=Pert. Hence, the investment can be modeled by the following,
A(t)=PertFormulaA(t)=200e0.0575tModel equation
To calculate the time it takes to accumulate to $350, set A(t)=350 and solve for t.
A(t)=200e0.0575t350=200e0.0575t
Begin by isolating the exponential expression.
350200=e0.0575t74=e0.0575t1.75=e0.0575t
Because this exponential has base e, we choose to take the natural logarithm of both sides and then solve for t.
ln(1.75)=lne0.0575tApplythepowerruleforlogarithms.ln(1.75)=0.0575tlneRecallthatlne=1.ln(1.75)=0.0575t⋅1ln(1.75)0.0575=t
Using a calculator we can approximate the time it takes:
t=ln(1.75)/0.0575≈9.73 years
Answer:
It will approximately 9.73 years.
When solving applications involving compound interest, look for the keyword “continuous,” or the keywords that indicate the number of annual compoundings. It is these keywords that determine which formula to choose.
Example 4.7.4m: Continuously compounded - find interest rate
Jermael’s parents put $10,000 in investments for his college expenses on his first birthday. They hope the investments will be worth $50,000 when he turns 18. If the interest compounds continuously, approximately what rate of growth will they need to achieve their goal?
Solution:
Identify the variables and the formula. A=$50,000, P=$10,000, r=?, t=17 years.
A(t)=PertFormula50,000=10,000er⋅17Model equation with substitutions made
Solve for r.
A(t)=PertFormula50,000=10,000er⋅17Model equation with substitutions madeNow solve for r5=e17rDivide each side by 10,000 ln5=lne17rTake the natural log of each side. ln5=17rlneUse the Power Property. ln5=17rSimplify. ln517=rDivide each side by 17. r≈0.095Approximate the answer.
They need the rate of growth (or interest rate) to be approximately 9.5%.
Try It 4.7.4rm
- Mario invested $1,000 in an account earning 6.3% annual interest that is compounded continuously. How long will it take the investment to double?
- Hector invests $10,000 at age 21. He hopes the investments will be worth $150,000 when he turns 50. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal?
- Rachel invests $15,000 at age 25. She hopes the investments will be worth $90,000 when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?
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Exponential Growth and Decay
In real-world applications, we need to model the behavior of a function. Two common examples are instances of rapid growth, where the exponential growth function could be used, or situations where a quantity is falling rapidly toward zero, where the exponential decay model may be a good choice.
EXPONENTIAL Growth and Decay Model, y=a0ekt
The function used to model exponential growth or decay is A(t)=A0ekt where
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![]() exponential growth (k>0) and exponential decay (k<0) |
Exponential Growth
Example 4.7.5: Find growth rate and model equation
A population of bacteria doubles every hour. If the culture started with 10 bacteria, construct an equation that models the population as a function of time.
Solution
When an amount grows at a fixed percent per unit time, the growth is exponential. To find A0 we use the fact that A0 is the amount at time zero, so A0=10. To find k, use the fact that after one hour (t=1) the population doubles from 10 to 20.The formula is derived as follows
y=A0ektFormula20=10ek⋅1Substitute given values2=ekDivide by 10ln2=kTake the natural logarithm
The growth rate is k=ln(2). ln(2)≈0.6931 so the equation could be written y=10e0.6931t. However, a more exact version of the equation can be obtained by writing the equation in terms of ek, the growth factor. Using some log properties we get y=10e(ln2)t=10(eln2)t=10·2t
Model Equations: using growth rate k approximation: y=10e0.6931t or more exactly: y=10·2t
Analysis. The population of bacteria after ten hours using y=10·2t is 10,240. We could describe this amount is being of the order of magnitude 104. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude 107, so we could say that the population has increased by three orders of magnitude in ten hours.
A cautionary word about rounding. If the rounded approximation k=0.6931 had been used, the result would be 10,235. If the full precision of your calculator is used to represent k by using the [2nd][ANS] feature or [STO] and [MEMVAR] functions, for example, the result of 10,240 would still be obtained. This illustrates that using rounded intermediate results produces rounding errors, so avoid getting answers that sometimes are significantly "off." Always use exact values or the full precision of your calculator in calculations to reduce or eliminate rounding errors.
Example 4.7.6r: Find model equation, total amount and time, given a growth rate
It is estimated that the population of a certain small town is 93,000 people with an annual growth rate of 2.6%. If the population continues to increase exponentially at this rate:
- Estimate the population in 7 years' time.
- Estimate the time it will take for the population to reach 120,000 people.
Solution
We begin by constructing a mathematical model based on the given information. Here the initial population P0=93,000 people and the growth rate r=2.6% =0.026. The following model gives population in terms of time measured in years:
A(t)=A0ektFormulaP(t)=93,000e0.026tModel equation
a. Use this model equation (4.7.4) to estimate the population in t=7 years.
P(t)=93,000e0006(7)=93,000e0.182≈111,564people
b. Use this model equation (4.7.4) to determine the time it takes to reach P(t)=120,000 people.
P(t)=93,000e0.026t120,000=93,000e0.026t120,00093,000=e0.026t4031=e0.026t
Take the natural logarithm of both sides and then solve for t.
ln(4031)=lne0.026t
ln(4031)=0.026tlne
ln(4031)=0.026t⋅1
ln(4031)0.026=t
Using a calculator,
t=ln(40/31)/0.026≈9.8years
Answer:
- 111,564 people
- 9.8 years
Often the growth rate k is not given. In this case, we look for some other information so that we can determine it and then construct a mathematical model. The general steps are outlined in the following example.
Example 4.7.7: Find growth rate, model equation, and total amount
Under optimal conditions Escherichia coli (E. coli) bacteria will grow exponentially with a doubling time of 20 minutes. If 1,000 E. coli cells are placed in a Petri dish and maintained under optimal conditions, how many E. coli cells will be present in 2 hours?
Solution
The goal is to use the given information to construct a mathematical model based on the formula A(t)=A0ekt.
Step 1: Find the growth rate k. Use the fact that the initial amount, A0=1,000 cells, doubles in 20 minutes. That is, A(t)=2,000 cells when t=20 minutes.
A(t)=A0ekt2,000=1,000ek20
Solve for the only variable k.
2,000=1,000ek202,0001,000=ek202=ek20ln(2)=lnek20ln(2)=k20lneln(2)=k20⋅1ln(2)20=k
Step 2: Write a mathematical model based on the given information for the number of E. coli cells in terms of time in minutes. Here k≈0.0347, which is about 3.5% growth rate per minute. So we could use the equation A(t)=1,000e0.0347t for our model. However, we will use the exact value for k in our model. This will allow us to avoid round-off error in the final result. Use A0=1,000 and k=ln(2)/20:
A(t)=1,000e(ln(2)/20)t=1,000eln(2)20t=1,000eln(2)t20=1,000(eln(2))t/20A(t)=1,000(2)t/20Alternate form of Model equation
This equation models the number of E. coli cells in terms of time in minutes.
Step 3: Use the function to answer the questions. In this case, we are asked to find the number of cells present in 2 hours. Because time is measured in minutes, use t=120 minutes to calculate the number of E. coli cells.
A(120)=1,000⋅(2)120/20=1,000⋅26=64,000 cells
Answer:
In two hours 64,000 cells will be present. (The answer obtained when k=0.0347 is 64,328 cells.)
We can now solve applications that give us enough information to determine the rate of growth. We can then use that rate of growth to predict other situations.
Example 4.7.8m: Find growth rate, model equation, and total amount
Researchers recorded that a certain bacteria population grew from 100 to 300 in 3 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?
Solution:
This problem requires two main steps. First we must find the unknown rate, k. Then we use that value of k to help us find the unknown number of bacteria. Use the formula A(t)=A0ekt.
Step 1: Find the growth rate k. Identify values for the variables in the formula. A=300, A0=100, k=?, t=3 hours.
A(t)=A0ektformula300=100ek⋅3substitute. only one variable left - solve for k3=e3kln3=lne3kTake the natural log of each side.ln3=3klneUse the Power Property. ln3=3klne ln33=kln e = 1 k≈0.366
Step 2: Write the model equation using A0=100 and k=ln33. We use this revised equation to predict the number of bacteria A(t) there will be in t hours.
A(t)=100eln33t or A(t)=100e0.366t Model equations
Step 3: Use the function to answer the questions. How many bacteria will there be t hours from the start of the experiment? Substitute in the value t=24 and evaluate.
A(24)=100eln33⋅24
A(24)=656,100
At this rate of growth, they can expect 656,100 bacteria. (The answer obtained when k=0.366 is 652,894 bacteria.)
Example 4.7.9: Find model if not given initial amount
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.
Solution
Start with the formula Use the formula A(t)=A0ekt.
Step 1. Find the growth rate k. Identify values for the variables in the formula. A=2A0, k=?, t=2 years.
A(t)=A0ektformula2A0=A0ek⋅22=e2kDIvide both sides by A0ln2=lne2kTake the natural log of each side.ln2=2klneUse the Power Property. ln22=kln e = 1
The function is A(t)=A0eln22t or A(t)=A0(2)t/2.
Try It 4.7.9ma
- Researchers recorded that a certain bacteria population grew from 100 to 500 in 6 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?
- 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.
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Exponential Decay
When the growth rate is negative the function models exponential decay. With exponential decay, quantities are declining at a rapid rate. The rate of decay k, is directly proportional to the size of the quantity and is negative. The main example of exponential decay is radioactive decay. Radioactive elements and isotopes spontaneously emit subatomic particles, and this process gradually changes the substance into a different isotope. For example, the radioactive isotope Uranium-238 eventually decays into the stable isotope Lead-206. This is a random process for individual atoms, but overall the mass of the substance decreases according to the exponential decay formula. One of the common terms associated with exponential decay 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.
Example 4.7.10n: Exponential decay - find the equation
Suppose that a certain population of bacteria has an annual decay rate of 10%. Starting with 20,000 bacteria, how many bacteria will be left in 3 years?
Solution
Start with the formula A(t)=A0ekt.
Identify values for the variables in the formula. A=?, A0=20,000, k=10% decay ⟶k=−0.10, t=3 years.
A(t)=A0ektformulaA(3)=20000e−0.10⋅3A(3)≈148.1636grams
Example 4.7.11c: Exponential decay - find the equation and original amount
The loudness of a sound dies down at an exponential rate of 6.91. If it takes .2 seconds for a loud clap of thunder to diminish to the sound level of a whisper (of 30 decibels), how loud was the clap of thunder originally?
Solution
Start with the formula A(t)=A0ekt.
Identify values for the variables in the formula. A=30, A0=?, |k|=6.91 decay ⟶k=−6.91, t=.2 seconds.
A(t)=A0ektformula30=A0e−6.91⋅.230e6.91⋅.2=A0=A0≈119.5decibels
Example 4.7.12c: Exponential decay - find the equation and elapsed time
A lawmaker once introduced legislation that would force Americans to cut back on their consumption of salt. It never passed the House, but it proposed that each year restaurants would be mandated to decrease sodium levels by two and a half percent annually. If all restaurants started out using a collective total of 5,000 kilograms of salt initially annually, and they were asked to reduce their consumption by two and a half percent each year, how long would it take to reduce their collective consumption to 3000 kg per year?
Start with the formula A(t)=A0ekt.
Identify values for the variables in the formula. A=3000, A0=5000, k=2.5% decrease ⟶k=−.025, t=? seconds.
A(t)=A0ektFormula3000=5000e−.025tSubstitute values. Now solve for t30005000=e−.025tln(.6)=ln(e−.025t)Take ln of both sidesln(.6)=−.025tln(e)Power ruleln(.6)=−.025tln(e)=1t=ln(.6)−.025t≈20.4years
Half-Life
Radioactive substances decay or decompose according to the exponential decay formula. The amount of time it takes for the substance to decay to half of its original amount is called the half-life of the substance. As with exponential growth processes that have a fixed doubling time. Similarly, exponential decay processes have a fixed half-life, the time in which one-half the original amount decays.
Example 4.7.13: Radioactive Decay (half life) - find decay rate and equation
The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t.
Solution
Start with the formula A(t)=A0ekt.
Identify values for the variables in the formula. A=12A0, A0=?, k=?, t=5730 years.
A=A0ektThe continuous growth formula0.5A0=A0ek⋅5730Substitute the half-life for t and 0.5A0 for f(t)0.5=e5730kDivide by A0ln(0.5)=5730kTake the natural log of both sidesk=ln(0.5)5730Divide by the coefficient of kk≈−1.2097×10−4
The function that describes this continuous decay is A(t)=A0e(ln(0.5)5730)t. We observe that k is negative, as expected in the case of exponential decay. Other forms of this function include A(t)=A0e.00012t and A(t)=A0(12)t/5730
Example 4.7.14m: Exponential Decay - find decay rate, equation, amount remaining
The half-life of radium-226 is 1,590 years. How much of a 100 mg sample will be left in 500 years?
Solution:
Start with the formula A(t)=A0ekt. Identify values for the variables in the formula.
Step 1. Identify values of the variables needed to find the decay constant k: At time t=1590 years, A(1590)=12A0, A0=?, k=?, t=1590 years. Since the next part of the question asks how much of a 100 mg sample will be left in 500 years, we may also choose to use A(1590)=50 mg, and A0=100 mg instead.
A(t)=A0ektThe continuous growth formula50=100ek⋅1590Substitute values. Now solve for k0.5=e1590k ln0.5=lne1590kTake the natural log of each side. ln0.5=1590klneUse the Power Property. ln0.5=1590kln(e)=1k=ln0.51590exact answerk≈−.00044approximate answer
Step 2: Write the model equation that predicts the amount of radium-226 A(t) there will be left in t years. The decay factor is \boldsymbol{e^k = e^{\ce(\ln 0.5)/1590}} = (e^{\ln 0.5})^{\ce{1/1590}} =(.5) ^ {\frac{1}{1590}}}. Also ekt=(ek)t
A(t)=100(.5)t/1590 or A(t)≈100e−.00044t Model Equations
Step 3: Use the model equation to answer the question. How much of a 100 mg sample will be left in 500 years? Here the values of the variables are A(500)=? mg and t=500 years.
A(500)=100(.5)500/1590≈80.41
In 500 years there would be approximately 80.4 mg remaining. (The answer obtained when k=−.00044 is 80.3 mg.)
Example 4.7.15r: Exponential Decay - find decay rate, model equation, and elapsed time
Due to radioactive decay, caesium-137 has a half-life of 30 years. How long will it take a 50-milligram sample to decay to 10 milligrams?
Solution:
Start with the formula A(t)=A0ekt. Identify values for the variables in the formula.
Step 1: Find the growth rate k. Identify values for the variables in the formula: A(30)=25 mg, A0=50 mg, k=?, t=30 years.
A(t)=A0ekt25=50ek⋅302550=e30kln(12)=lne30kln(.5)=30klneRecall that lne=1.k=ln(.5)30k≈−0.0231049
Step 2: Write the model equation to predict the amount of caesium-137 that will be left in a 50-milligram sample after t years. Note that k≈−0.0231 is negative. However, we will use the exact value to construct a model that gives the amount of cesium-137 with respect to time in years.
A(t)=50e(ln(.5)/30)t=50eln(.5)t/30=50(e(ln(.5))t/30Use the Inverse Property elnM=M.A(t)=50(.5)t/30or alternatively A(t)=50e−0.0231t Model equations
Step 3: Use the model to find t when A(t)=10 milligrams.
10=50(.5)t/301050=(.5)t/30ln(.2)=ln(.5)t/30ln(.2)=t/30ln(.5)ln(.2)=ln(.5)30tt=ln(.2)⋅30ln(.5)t=30ln(.2)ln.5t≈69.66
Answer: It will take t≈69.66 years to decay to 10 milligrams. (The answer obtained when k=−0.0231 is t≈69.67.)
Radiocarbon Dating
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. 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. While the carbon-14 technique only works on plants and animals, there are other similar dating techniques, using other radioactive isotopes, that are used to date rocks and other inorganic matter.
Example 4.7.16r: Carbon dating - find decay rate, model equation, and elapsed time
An ancient bone tool is found to contain 25% of the carbon-14 normally found in bone. Given that carbon-14 has a half-life of 5,730 years, estimate the age of the tool.
Solution
Start with the formula A(t)=A0ekt. Identify values for the variables in the formula.
Step 1: Find the growth rate k. Identify values for the variables in the formula: A(5,730)=12A0 mg, A0=? mg, k=?, t=5,730 years. Notice below that when both sides are divided by A0, the remaining exponential equation is only in terms of k. This shows that half-life is independent of the initial amount.
A(t)=A0ekt12A0=A0ek5,730.5=ek5,730ln(.5)=lnek5,730ln(.5)=5,730klnek=ln(.5)5,730k≈−.00012
Step 2: Write the model equation that predicts the amount of carbon-14 remaining in the bone t years after death of the object.
A(t)=A0e−.00012torA(t)=A0eln(.5)5,730t=A0(eln(.5))t5,730=A0(.5)t/5730 Model equations
Step 3: Use the model to find the time t it takes the carbon-14 to decay to 25% of the initial amount, or A(t)=0.25A0
A(t)=A0(.5)t/57300.25A0=A0(.5)t/57300.25=(.5)t/5730ln(0.25)=ln(.5)t/5730ln(0.25)=t/5730(ln(.5))t=5,730ln(0.25)ln(.5)t≈11,460
Answer: The tool is approximately 11,460 years old. (The answer obtained when k=−.00012 is 11,552 years old).
Try It 4.7.16aA
a. Researchers recorded that a certain bacteria population declined from 700,000 to 400,000 in 5 hours after the administration of medication. At this rate of decay, how many bacteria will there be 24 hours from the start of the experiment?
b. The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of carbon-14 remaining as a function of time, measured in years.
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Try It 4.7.16bm
a. The half-life of strontium-90 is about 28 years. How long will it take a 36 milligram sample of strontium-90 to decay to 30 milligrams?
b. Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, how long will it take until only 1 milligram remains?
c. The half-life of magnesium-27 is 9.45 minutes. How much of a 10-mg sample will be left in 6 minutes?
d. The half-life of radioactive iodine is 60 days. How much of a 50-mg sample will be left in 40 days?
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Newton’s Law of Cooling
Exponential decay can also be applied to temperature. When a hot object is put into a place that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature.
NEWTON’S LAW OF COOLING
The temperature of an object, T, in surrounding air with temperature S will behave according to the formula
T(t)=Aekt+S{t is timeA is the difference between the initial temperature of the object and the surroundingsk is a constant, the continuous rate of cooling of the object
Example 4.7.17: Using Newton’s Law of Cooling
A cheesecake is taken out of the oven with an ideal internal temperature of 165°F, and is placed into a 35°F refrigerator. After 10 minutes, the cheesecake has cooled to 150°F. If we must wait until the cheesecake has cooled to 70°F before we eat it, how long will we have to wait?
Solution
Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation
T(t)=Ae^{kt}+35
Step 1. Find A. We know the initial temperature was 165, so T(0)=165.
\begin{align*} 165&= Ae^{k0}+35 &&\text{Substitute } (0,165)\\ 165&= A+35 \\ A&= 130 &&\text{Solve for A} \end{align*}
At the conclusion of this step, we now have a modifed formula, T(t)=130e^{kt}+35
Step 2. Find k. To do this, use our modified formula, T(t)=130e^{kt}+35, and the fact that T(10)=150.
\begin{align*} T(t)&=130e^{kt}+35 \\ 150&= 130e^{k10}+35 \qquad &&\text{Substitute } (10, 150)\\ 115&= 130e^{k10} \qquad &&\text{Subtract 35}\\ \dfrac{115}{130}&= e^{10k} \qquad &&\text{Divide by 130}\\ \ln\left (\dfrac{115}{130} \right )&= 10k \qquad &&\text{Take the natural log of both sides}\\ k&= \dfrac{\ln \left (\dfrac{115}{130} \right )}{10}\\ &= -0.0123 \qquad &&\text{Divide by the coefficient of k} \end{align*}
At the end of this step we now have the equation for the cooling of the cheesecake: T(t)=130e^{–0.0123t}+35.
Another form this equation can take is T(t)=130e^{\frac{\ln \left( \tfrac{115}{130} \right) }{10}}+35 which simplifies to T(t)=130 \left( \frac{115}{130} \right )^\tfrac{t}{10}+35.
Step 3. Now we can solve for the time it will take for the temperature to cool to 70 degrees.
\begin{align*} 70&= 130e^{-0.0123t}+35 \qquad \text{Substitute in 70 for } T(t)\\ 35&= 130e^{-0.0123t} \qquad \text{Subtract 35}\\ \dfrac{35}{130}&= e^{-0.0123t} \qquad \text{Divide by 130}\\ \ln \left (\dfrac{35}{130} \right )&= -0.0123t \qquad \text{Take the natural log of both sides}\\ t&= \dfrac{\ln \left (\dfrac{35}{130} \right )}{-0.0123}\\ &\approx 106.68 \qquad \text{Divide by the coefficient of t} \end{align*}
It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to 70°F. (The answer, rounded to the nearest minute, obtained when \(k=\frac{\ln \left (\tfrac{115}{130} \right ) \) is the same).
Try It \PageIndex{17}
A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?
- Answer
-
6.026 hours
Logistic Growth Models
Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.
The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity. For constants a, b, and c, the logistic growth of a population over time x is represented by the model
f(x)=\dfrac{c}{1+ae^{−bx}}
The graph in Figure \PageIndex{6} shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.
LOGISTIC GROWTH
The logistic growth model is f(x)=\dfrac{c}{1+ae^{−bx}} where
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Example \PageIndex{18}: Using the Logistic-Growth Model
An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.
For example, at time t=0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is b=0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.
Solution
We substitute the given data into the logistic growth model f(x)=\dfrac{c}{1+ae^{−bx}}
Step 1. Find one of the constants by evaluating f when x \rightarrow \infty . This results in e^{-bx} approaching zero, so f(x \rightarrow \infty) = c. Because at most the entire population of the community, 1,000 people, can eventually get the flu, we know the limiting value is c=1000. Step 2. Find one of the constants by evaluating f when x=0. We are given f(0) = 1. Therefore \dfrac{c}{1+a}=1, from which it follows that a=999. Step 3. Given that the logistic growth constant is b=0.6030, we obtain the model equation f(x)=\dfrac{1000}{1+999e^{−0.6030x}}. Step 4. Use the model. This model predicts that, after ten days, the number of people who have had the flu is f(10)=\dfrac{1000}{1+999e^{−0.6030\cdot 10}}≈293.8. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, c=1000. |
Figure \PageIndex{7}: The graph of f(x)=\dfrac{1000}{1+999e^{−0.6030x}} After a long time has passed, the entire community of 1000 people will have had this flu. |
Analysis. Remember that, because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values. The graph in Figure \PageIndex{7} gives a good picture of how this model fits the data.
Try It \PageIndex{18}
Using the model in the Example above, estimate the number of cases of flu on day 15.
- Answer
-
895 \; cases on day 15
Key Concepts
- f(x)=ab^x. If b>1, we have exponential growth; if 0<b<1, we have exponential decay. The basic exponential function is
- We can also write this formula in terms of continuous growth as A=A_0e^{kx}, where A_0 is the starting value. If A_0 is positive, then we have exponential growth when k>0 and exponential decay when k<0.
- In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay.
- We can find the age of an organic artifact by measuring the amount of carbon-14 (which is subject to exponential decay) remaining in the artifact.
- Given a substance’s doubling time or half-time, we can find a function that represents its exponential growth or decay.
- We can use Newton’s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time.
- We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.
- Any exponential function with the form y=A_0b^x can be rewritten as an equivalent exponential function with the form y=A_0e^{kx} where k=\ln b.
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.