2.3: More on Compounding Interest and Continuous Exponential Models
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Focus: More on Compound Interest
Many types of bank accounts utilize compound interest, such as a bank certificates of deposit (CD), savings accounts, and credit cards. The term \(compounding\) comes from the behavior that interest is earned not only on the original value, but on the accumulated value of the account. The time period, or \(compounding\) \(period\), over which interest is calculated may differ by the account. In one CD, interest could be calculated every month (such as 2% interest per month) whereas in another interest could be can calculated every year (such as 12% per year). In order to compare interest rates compounded over different time periods, interest rates are usually expressed as a nominal rate or an annual percentage rate (APR). Let's explore an example to understand how this effects how we calculate interest.
The annual percentage rate (APR) of an investment is the percentage of interest gained on the original principal, or amount invested, over a year.
You invest $1,000 into an investment account that pays a 6% APR (annual percentage rate) of interest.
a) If you compounded interest annually (or once per year), how much will be in the account in a year?
b) If you compounded interest quarterly, how much will be in the account in a year?
Solution
a) Since we are starting with $1,000, our initial value \(a = 1000\).
Our annual interest rate is 6%, so \(r = 0.06\), and our growth factor b = 100% + 6% =106% as a percent or b =1 +0.06 = 1.06 as a decimal. Each time we compound interest, we multiply the previous amount by our growth factor b = 1.06.
So in one year, we will have
\[A(1)=1000(1.06)^{1} =\$ 1060.\nonumber\]
b) Since we are compounding quarterly, we are compounding 4 times per year and need to break the annual interest rate into 4 parts. So the interest per quarter is \(r = \dfrac{6}{4}\) = 1.5% as a percent or as a decimal \(r = \dfrac{.06}{4}\) = 0.015.
So our quarterly growth factor b = 100% + 1.5% =101.5% as a percent or b =1 +0.015=1.015 as a decimal. Each time we compound interest, we multiply the previous amount by our growth factor b = 1.015.
So in one year, we will have
\[A(1)=1000(1.015)^{4} =\$ 1061.36.\nonumber\]
Can you explain why more interest is gained when we compounded interest quarterly? In the first quarter, we gained $15 in interest since
\[A=1000(1.015)^{1} =\$ 1015.\nonumber\]
In the second quarter, we gained interest on the original principal($1000) plus the first month's interest ($15). The interest on the original principal=\(1000(.015)=$15\) and the interest on the first months interest \((15)(.015)=$0.225\). More interest was gained in the second quarter, \(15+0.225=$15.225\) than the first quarter, since we are gaining interest on previous interest as well as the original principal.
When the interest is compounded more frequently than once a year, the account effectively earns more that the nominal rate or APR. We call this rate the account actually earned the \(effective\) \(annual\) \(rate\).
The effective annual rate is the percentage rate r that an account actually increased by over a year incorporating the effects of compounding interest.
In example 3.3.1 part b, the effective annual rate was \(r =\dfrac{61.36}{1000} = 0.06136\) or \(6.136\)%. Notice, that we could have obtained this by using the growth factor. After compounding four times, the account grows by a factor of \(b^{4}=(1.015)^{4}=1.06136.\) As a percent, the account is 106.136% or the original amount after a year. It has grown by 6.136%.
A 529 plan is a college savings plan in which a relative can invest money to pay for a child’s later college tuition. The account grows tax free. Lily wants to set up a 529 account for her new granddaughter so that the account grows to $40,000 over 18 years. She believes the account will earn a 4% APR compounded semi-annually (twice a year).
a) How much will Lily need to invest in the account?
b) What is the effective annual rate of her account?
Solution
a) Since the account is earning a 4% APR, the interest per compounding period (semi-annually)is r = \(\dfrac{4}{2}\)=2% or as a decimal r = \(\dfrac{0.04}{2}=0.02\) as a decimal.
Then the growth factor per period (semi-annually) is b = 100% + 2% =102% as a percent or b = 1 +0.02 = 1.02 as a decimal. Each time interest is compounded, we multiply the previous amount by our growth factor b = 1.02. The account will be compounding twice a year for 18 years, so the number of compoundings n = (18)(2) =36.
In this problem, we don’t know how much we are starting with, so we will be solving for a, the initial amount. We do know that we want the end amount to be $40,000. We want for the value of a so that \(A(18) = 40,000\).
\[\begin{array}{l} {40,000=A(18)=a(1.02)^{36} } \\ {40,000=a(2.039887)} \\ {a=\dfrac{40,000}{2.039887} \approx \$ 19,608.93} \end{array}\nonumber \]
Lily will need to invest $19,608.93 to have $40,000 in 18 years.
b) In one year, the account will compound twice. So we will multiply our initial amount by our growth factor twice \(b^{2}=1.02^{2}=1.0404\). Effectively our account will be 104.04% of the initial value. So the effective annual rate is 4.04%.
You owe $5000 on a credit card that charges a 12% APR of interest compounded monthly.
a) Find a formula for the amount owed, A, after t years if no payments are made.
b) Predict the amount owed in 5 years.
c) What is the effective annual rate of the credit card account?
- Answer
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a) Since the account is earning a 12% APR, the interest per month is r = \(\dfrac{12}{12}\)=1% as a percent or as a decimal r = \(\dfrac{0.12}{12}=0.01\) as a decimal. Then the growth factor per month is b = 100% + 1% =101% as a percent or b = 1 +0.01 = 1.01 as a decimal. Each time we compound interest, we multiply the previous amount by our growth factor b = 1.01.
The account will be compounding twelve times a year for t years. So the number of compoundings in t years is n = (12)(t) =12t.
The initial value, or principal, \(a=5000\). So a formula for the amount owed after t years is
\[A=A(t)=5000(1.01)^{12t} \nonumber\]
b) In 5 years, t = 5. So the amount in the owed will be
\[A=A(5)=5000(1.01)^{12(5)} = $9,083.48 \nonumber\]
c) In one year, the account will compound 12 times. So the initial amount will be multiplied by our growth factor b = 1.01 twelve times and increase by a factor of = \(b^{12}=(1.01)^{12}\approx 1.126825\). Effectively, the credit card debt will be 112.6825% of the initial value. So the effective annual rate is 12.6825%.
Generalizing this, we can form a general formula for compound interest. If the APR is written in decimal form as \(r\), and there are \(k\) compounding periods per year, then the interest per compounding period will be \(\dfrac{r}{k}\). So the growth factor per period will be b=\((1+\dfrac{r}{k})\). Likewise, if we are interested in the value after \(t\) years, then there will be \(n=kt\) compounding periods in that time.
Compound Interest can be calculated using the formula
\[A(t)=a\left(1+\dfrac{r}{k} \right)^{kt}\]
Where \(A(t)\) is the account value
- \(t\) is measured in years
- \(a\) is the starting amount of the account, often called the principal
- \(r\) is the annual percentage rate (APR), also called the nominal rate, written as a decimal
- \(k\) is the number of compounding periods in one year
Bank A offers an account paying 1.2% compounded quarterly. Bank B offers an account paying 1.1% compounded monthly. Which bank is offering a better rate?
- Answer
-
We can compare these rates using the effective annual percentage rate, the actual percent increase per year.
Bank A has a quarterly interest rate of r = \(\dfrac{0.012}{4}=0.003\) as a decimal. The growth factor per quarter is \(b=1+ 0.003=1.003\). In one year, the account will compound 4 times and grow by a factor of \(b^{4} =1.003^{4} \approx 1.012085411\). Therefore, the account will be 101.2085411% of the initial value. So the effective annual rate is 1.2085411%.
Bank B has a monthly interest rate of r = \(\dfrac{0.011}{12}=0.0009166667\) as a decimal. The growth factor per month is \(b=1+ 0.0009166667=1.0009166667\). In one year, the account will compound 12 times and grow by a factor of \(b^{12} =1.0009166667^{12} \approx 1.011661873\). Therefore, the account will be 101.1661873% of the initial value. So the effective annual rate is 1.1661873%.
Bank B’s monthly compounding is not enough to catch up with Bank A’s better APR. Bank A offers a better rate.
Focus: Periodic Exponential Growth versus Continuous Exponential Growth
Compounding interest is an example of periodic exponential growth, where the growth occurs at the end of a time period, such as 6% per year or 3% per quarter. Many other exponentially growing (or decaying) quantities grow (or decay) continuously through the time period. For example, the difference in temperature between a hot cup of coffee and its ambient environment decreases at every instant, not just at the end of the time period. Exponential growth (or decay) that occurs continuously through a time period is called continuous exponential growth (or decay). The population growth of a country is another example of continuous exponential growth since the growth in the number of people occurs continuously through the year not just at the end of the year. Be aware that we may communicate the growth of a continuously growing exponential quantity with a periodic growth rate, such as the population of a country growing by 1.12% per year. The exponential growth formula we have been working with so far, \(y=a(b)^{x}\), is related to the periodic growth rate. For example, if a quantity grows by 1.12% per year, the growth factor would be b = 100% + 1.12% = 101.12% as a percent or b = 1 + 0.0112 = 1.0112 as a decimal. For this reason, we call this the periodic exponential growth (or decay) formula.
Periodic Exponential Growth can be calculated using the formula
\[f(x)=ab^{x}\]
where
\(a\) is the starting amount,
\(r\) is the periodic growth rate written as a decimal,
and \(b\) is the growth factor where b = 1+ r.
How are periodic and continuous exponential growth related? Let's explore what happens to an account with compounded interest, which experiences periodic exponential growth, as we increase the frequency of the compoundings from yearly to quarterly to monthly to daily to hourly to every second as we approach continuous exponential growth. As we saw earlier, the amount we earn increases as we increase the compounding frequency. If we continue to increase the frequency of the compoundings, will the amount we earn keep increasing without bound or eventually level off? The table below shows that the increase from annual to quarterly compounding is larger than the increase from monthly to daily compounding. This might lead us to believe that, although increasing the frequency of compounding will increase our result, there is an upper limit to this process.
To see this, let us examine the value of $1000 invested at 6% interest for 10 years.
| Frequency | Interest per Period, r | Growth Factor per Period, b | Number of Compoundings in 10 years, n |
Amount in 10 years \[A=a(b)^{n} \nonumber \] |
|---|---|---|---|---|
| Annual | \(\dfrac{0.06}{1}=0.06\) | 1+0.06=1.06 (106%) | 10(1)=10 |
\(A=1000(1.06)^{10} \\ = 1790.85 \) |
| Quarterly | \(\dfrac{0.06}{4}=0.015\) | 1+0.015=1.015 | 10(4)=40 |
\(A=1000(1.015)^{40} \\ = 1814.02\) |
| Monthly | \(\dfrac{0.06}{12}=0.005\) | 1+ 0.005 = 1.005 | 10(12)=120 |
\(A=1000(1.005)^{120} \\ = 1819.40\) |
| Daily | \(\dfrac{0.06}{365}\approx0.000164384\) | 1.000164384 | 10(365)=3,650 | \(A=1000(1.000164384)^{3,650} \\=1822.03\) |
| Hourly | \(\dfrac{0.06}{8760}\approx0.0000068493\) | 1.0000068493 | 10(8760)=87,600 | \(A=1000(1.0000068493)^{87,600} \\=1822.11\) |
| Once per minute | 1822.12 | |||
| Once per second | 1822.12 |
These values do indeed appear to be approaching an upper limit of approximately $1822.12 as we approach continuous exponential growth with growth at each instant. In order to describe what this limiting value, we need to introduce a special number, which is so important that it gets represented by its own letter.
\(e\) is the letter used to represent the value that \(\left(1+\dfrac{1}{k} \right)^{k}\) approaches as \(k\) gets bigger and bigger.
\[e \approx 2.718282\nonumber\]
The number is also called the natural base.
\(e\) is a number that arises in continuous exponential growth similar to how number \(\pi\) arises when dealing with circles. Notice, that the limiting amount in our interest example can be expressed with \(e\).
\[A=1000(e)^{(0.06)(10)} \approx 1822.11\nonumber\]
Notice, the significance of each of the parameters in this equation: the initial value a = 1000, the time t = 10 years, and the percent increase r = 0.06. In this case, r = 0.06 = 6% is the continuous growth rate since growth is occurring at each instant.
In general, a continuously growing (or decaying) exponential quantity can be represented by the following formula.
Continuous Exponential Growth can be calculated using the formula
\[f(t)=ae^{rt}\]
where
\(a\) is the starting amount,
\(r\) is the continuous growth rate written as a decimal,
and \(t\) is the time.
This type of equation is commonly used when describing quantities that change more or less continuously, like chemical reactions, growth of large populations, and radioactive decay. In general, we can model an exponential quantity with either the periodic or continuous growth formulas, so y = \(ab^{t}=ae^{rt}\) where \(b=e^{r}\). However, we will see that there are some distinct advantages to using the continuous exponential models in some cases in section 3.4.
A virus is detected with 120 people infected initially and is spreading rapidly.
a) If the number of people infected is increasing by 25% per month, find a formula for the number of people infected, I, in t months. Then predict the number infected in 6 months.
b) If the number of people infected is increasing continuously by 25%, find a formula for the number of people infected, I, in t months. Then predict the number infected in 6 months.
Solution
a) In this scenario, we are given the periodic growth rate r = 25% = 0.25. Therefore, we should use the periodic exponential growth formula \(I = f(t) = ab^{t}\), where a = 120 and b = 100% + 25% = 125% as a percent or b = 1.25 as a decimal. So the number of people infected is given by
\[I=120(1.25)^{t}\nonumber\]
The number infected in 6 months is \(I=120(1.25)^{6}\approx 458\).
b) In this scenario, we are given the continuous growth rate r = 25% = 0.25. Therefore, we should use the continuous exponential growth formula \(I=ae^{rt}\), where a = 120. So the number of people infected is given by
\[I=120(e)^{0.25t}\nonumber\]
The number infected in 6 months is \(I=120(e)^{0.25(6)}\approx 538\).
Can you explain why the number infected was more with continuous growth than with periodic growth? It is similar to what happened with compounded interest where we gained interest on previous interest in the continuous model.
Interpret the following: \(S(t)=20e^{0.12t}\) if \(S(t)\) represents the growth of a substance in grams, and time is measured in days.
Solution
An initial substance weighing 20g is growing at a continuous rate of 12% per day.
Find the effective annual rate of an account which pays an 8% APR compounded continuously.
Solution
We can interpret the effective annual rate from the growth factor \(b = e^{r} = e^{0.08} \approx 1.08329\). After one year, the account will be 108.329% of the original amount. Therefore the effective annual rate is 8.329%.
Alternatively, after one year the account will have a value of \(A = ae^{(0.08)(1)} \approx a(1.08329) \) if an initial value of \(a\) is invested. Thus, the amount in one year will be 108.329% of the original amount. Therefore the effective annual rate is 8.329%.
A person ingests 20 mg of a medication.
a) If the amount of medication in a person's body decreases by 16% per hour, find a formula for the amount of medication, A, in the body after t hours. Then predict the amount of medication after 5 hours.
b) If the amount of medication in a person's body decreases continuously by 16%, find a formula for the amount of medication, A, in the body after t hours. Then predict the amount of medication after 5 hours.
- Answer
-
a) In this scenario, we are given the periodic decay rate r = -16% = -0.16. Therefore, we should use the periodic exponential formula \(A=ab^{t}\), where a = 20 and b = 100% - 16% = 84% as a percent or b = 0.84 as a decimal. So the amount of medication is given by
\[A=120(0.86)^{t}\nonumber\]
The amount of medication in 5 hours is \(A=20(0.86)^{5}\approx 9.41\) mg.
b) In this scenario, we are given the continuous decay rate r = -16% = 0.16. Therefore, we should use the continuous exponential growth formula \(A=ae^{rt}\), where a = 20. So the amount of medication is given by
\[A=20(e)^{-0.16t}\nonumber\]
The amount of medication in 5 hours is \(A=20(e)^{-0.16(5)}\approx 8.99\) mg.
Focus: Some issues with using exponential functions to make predictions
Case: Predict the input when the output is known.
In Example 3.3.1a, the amount in the account after t years was given by \(A=1000(1.06)^{t} \). How long until there is $5,000 in the account?
Solution
We are given A = 5,000. We want the time t. Substituting we have, \[\begin{array}{l} {5000=1000(1.06)^{t}} \\ {\dfrac{5000}{1000}=(1.06)^{t}} \\ {5=(1.06)^{t}} \end{array}\nonumber\]
How can we get the variable t out of the exponent? We need to develop more tools to solve this exponential equation and make predictions like this with exponential models.
Case: Modeling data with a continuous exponential function
Find a continuous exponential model, \(y=ae^{rt}\) that contains the points (0,4) and (2,12).
Solution
The initial value is a = 4. To find r, substitute in the other data point t = 2 and y = 12, so \[\begin{array}{l} {12=4(e)^{r(2)} } \\ {\dfrac{12}{3}=e^{2r} } \\ {4=e^{2r}} \end{array}\nonumber\]
We run into the same issue as example 5, we need to get the variable out of the exponent.
Important Topics of this Section
- Continuous Exponential Growth
- Compound Interest
- Effective Annual Rate