Continuous Compounding Formula (appreciation and depreciation): $$A = P{e^{rt}}$$
CONTINUOUS COMPOUND INTEREST: Round all answers to two decimal places.
Hometown Bank offers a CD that earns 1.58% compounded continuously. If \$10,000 is invested in this CD, how much will it be worth in 3 years?
$A = 10,000{e^{.0158(3)}}$
$A = \$ 10,485.41$
The account will be worth approximately \$$10,485.41$ in three years.
Hometown Bank offers a CD that earns 1.58% compounded continuously. If \$10,000 is invested in this CD, how long will it take the account to be worth $11,000?
$11,000 = 10,000{e^{.0158t}}$
$\frac{{11,000}}{{10,000}} = \frac{{10,000{e^{.0158t}}}}{{10,000}}$
$\frac{{11}}{{10}} = {e^{.0158t}}$
$\ln \frac{{11}}{{10}} = \ln {e^{.0158t}}$
$\ln \frac{{11}}{{10}} = .0158t$
$\frac{{\ln \frac{{11}}{{10}}}}{{.0158}} = \frac{{.0158t}}{{.0158}}$
$6.03 = t$
It will take approximately 6.03 years for the account to be worth $11,000.
Doubling Time: How long will it take money to double if it is invested at 5% compounded continuously?
$A = P{e^{rt}}$
$2 = 1{e^{.05t}}$
$2 = {e^{0.05t}}$
$\ln 2 = \ln {e^{0.05t}}$
$\ln 2 = 0.05t$
$\frac{{\ln 2}}{{0.05}} = t$
$t = 13.86$
It will take approximately 13.86 years for the initial investment to double.
Doubling Rate: At what nominal rate compounded continuously must money be invested to double in 8 years?
$A = P{e^{rt}}$
$2 = 1{e^{r(8)}}$
$2 = {e^{8r}}$
$\ln 2 = \ln {e^{8r}}$
$\ln 2 = 8r$
$\frac{{\ln 2}}{8} = r$
$0.0866 = r$
In order for the initial investment to double in 8 years, the money must be invested in an account with a nominal rate of approximately 8.7% compounded continuously.
How long will it take money to triple if it is invested at 10.5% compounded continuously?
$A = P{e^{rt}}$
$3 = 1{e^{.105t}}$
$\ln 3 = \ln {e^{.105t}}$
$\ln 3 = .105t$
$\frac{{\ln 3}}{{.105}} = t$
$10.46=t$
It will take approximately 10.46 years for the initial investment to triple.
Radioactive Decay: A mathematical model for the decay of radioactive substances is given by $$Q = {Q_0}\;{e^{rt}}.$$ The continuous compound rate of decay of carbon-14 per year is $r = -0.0001238.$ How long will it take a certain amount of carbon-14 to decay to half the original amount?
$\frac{1}{2} = 1{e^{ - 0.0001238t}}$
$\ln .5 = \ln {e^{ - 0.0001238t}}$
$\ln .5 = - 0.0001238t$
$\frac{{\ln .5}}{{ - 0.0001238}} = t$
$t = 5598.93$
It will take approximately 5598.93 years for the carbon-14 to decay to half the original amount.
The estimated resale value R (in dollars) of a company car after t years is given by: $$R(t) = 20000{(0.86)^t}.$$ What will be the resale value of the car after 2 years? How long will it take the car to depreciate to half the original value?
$R(2) = 20,000{(0.86)^2} = \$ 14,792$
The resale value of the car after two years will be $14.792.
$\frac{{10,000}}{{20,000}} = \frac{{20,000{{(0.86)}^t}}}{{20,000}}$
$.5 = {0.86^t}$
$\ln .5 = \ln {0.86^t}$
$\ln .5 = t\ln 0.86$
$\frac{{\ln .5}}{{\ln 0.86}} = \frac{{t\ln 0.86}}{{\ln 0.86}}$
$\frac{{\ln .5}}{{\ln 0.86}} = t$
$t = 4.5957$
It will take approximately 4.6 years for the car to depreciate to half its original value.
If you invested $1,000 in an account paying an annual percentage rate (quoted rate) of 2%, compounded continuously, how much would you have in your account at the end of
$A = 1000{e^{.02(1)}} = 1020.20$
At the end of one year, there will be $1020.20 in the account.
$A = 1000{e^{.02(10)}} = 1221.40$
At the end of ten years, there will be $1221.40 in the account.
$A = 1000{e^{.02(20)}} = 1491.82$
At the end of twenty years, there will be $1491.82 in the account.
$A = 1000{e^{.02(50)}} = 2718.28$
At the end of fifty years, there will be $2718.28 in the account.
A $1,000 investment is made in a trust fund at an annual percentage rate of 12%, compounded continuously. How long will it take the investment to
$2000 = 1000{e^{0.12t}}$
$2 = {e^{0.12t}}$
$\ln 2 = \ln {e^{0.12t}}$
$\ln 2 = 0.12t$
$\frac{{\ln 2}}{{0.12}} = t$
$t = 5.78$
The investment will double in approximately 5.78 years.
$3000 = 1000{e^{0.12t}}\quad$
$3 = {e^{0.12t}}$
$\ln 3 = \ln {e^{0.12t}}$
$\ln 3 = 0.12t$
$\frac{{\ln 3}}{{0.12}} = t$
$t = 9.16$
The investment will triple in approximately 9.16 years.
If $500 is invested in an account which offers 0.75%, compounded continuously find:
$A = 500{e^{.0075t}}$
$A(5) = 500{e^{.0075(5)}}=519.11$
After 5 years, $519.11 will be in the account.
$A(10) = 500{e^{.0075(10)}}=538.94$
After 10 years, $538.94 will be in the account.
$A(30) = 500{e^{.0075(30)}}=626.16$
After 30 years, $626.16 will be in the account.
$A(35) = 500{e^{.0075(35)}}=650.09$
After 35 years, $650.09 will be in the account.
$1000 = 500{e^{.0075t}}$
$2 = {e^{.0075t}}$
$\ln 2 = \ln {e^{.0075t}}$
$\ln 2 = .0075t$
$\frac{{\ln 2}}{{.0075}} = t$
It will take approximately 92.42 years for the initial investment to double.
4th year $A(4) = 500{e^{.0075(4)}} = \$515.23$
5th year $A(5) = 500{e^{.0075(5)}} = \$519.11$
$\frac{A(5)-A(4)}{5-4}=\frac{519.11-515.23}{1}=3.88$
The balance in the account is increasing by an average of $3.88 per year.
If $5000 is invested in an account which offers 2.125%, compounded continuously, find:
$A(t) = 5000{e^{.02125t}}$
$A(5) = 5000{e^{.02125(5)}} = 5560.50$
After 5 years, $5560.50 will be in the account.
$A(10) = 5000{e^{.02125(10)}} = 6183.83$
After 10 years, $6183.83 will be in the account.
$A(30) = 5000{e^{.02125(30)}} = 9458.73$
After 30 years, $9458.73 will be in the account.
$A(35) = 5000{e^{.02125(35)}} =10519.05$
After 35 years, $10519.05 will be in the account.
$10,000 = 5000{e^{.02125t}}$
$2 = {e^{.02125t}}$
$\ln 2 = \ln {e^{.02125t}}$
$\ln 2 = .02125t$
$\frac{{\ln 2}}{{.02125}} = t$
It will take approximately 32.62 years for the initial investment to double.
$A(4) = 5000{e^{.02125(4)}}$
$A(4) = \$ 5443.59$
$A(5) = 5000{e^{.02125(5)}}$
$A(5) = \$ 5560.50$
$\frac{A(5)-A(4)}{5-4}=\frac{\displaystyle5560.50-5443.59}{5-4}=\frac{\displaystyle116.91}1=116.91$
The balance in the account is increasing by an average of $116.91 per year.
How much money needs to be invested now to obtain \$5000 in 10 years if the interest rate in a CD is 2.25%, compounded continuously?
$A = P{e^{rt}}$
$5000 = P{e^{0.0225(10)}}$
$5000 = P{e^{0.225}}$
$\frac{{5000}}{{{e^{0.225}}}} = \frac{{P{e^{0.225}}}}{{{e^{0.225}}}}$
$\$ 3992.58 = P$
\$3992.58 needs to be invested now, in order to have $5000 in 10 years.
A mathematical model for depreciation of a car is given by $A = P{(1-r)^t}$, where A is defined as the value of the car after t years, P is defined as the original value of the car, and r is the rate of depreciation per year. The cost of a new car is $32,000. It depreciates at a rate of 15% per year. This means that it loses 15% of its value each year.
$A = 32,000{\left( {1-0.15} \right)^t}$
$A = 32,000{\left( {0.85} \right)^t}$
$A = 32,000{\left( {.85} \right)^4}$
$A = 16,704.20$
The car is worth approximately $16,704.20 when it is 4 years old.
A mathematical model for depreciation of an ATV (all-terrain vehicle) is given by $A = P{(1-r)^t}$, where A is defined as the value of the vehicle after t years, P is defined as the original value of the vehicle, and r is the rate of depreciation per year. The cost of a new ATV (all-terrain vehicle) is \$7200. It depreciates at 18% per year.
$A = 7200{\left( {1-0.18} \right)^t}$
$A = 7200{\left( {.82} \right)^t}$
$A = 7200{\left( {.82} \right)^{10}}$
$A = 989.63$
The value of the ATV when it is 10 years old will be $989.63.
Michigan’s population is declining at a rate of 0.5% per year. In 2004, the state had a population of 10,112,620.
$y = 10,112,620{\left( {.995} \right)^t}$
$y = 10,112,620{\left( {.995} \right)^8}$
In 2012 the population of Michigan will be approximately 9,715,124 people.
$9,900,000 = 10,112,620{\left( {.995} \right)^t}$
$\frac{{9,900,000}}{{10,112,620}} = {.995^t}$
$\ln (\frac{{9,900,000}}{{10,112,620}}) = \ln {.995^t}$
$\ln (\frac{{9,900,000}}{{10,112,620}}) = t\;\ln .995$
$\frac{{\ln (\frac{{9,900,000}}{{10,112,60}})}}{{\ln .995}} = t$
$t = 4.24$ years
The population of Michigan will be 9,900,000 people in March of 2008.
$y = 10,112,620{\left( {.995} \right)^{ - 4}}$
$y = 10,317,426.06$
According to this model, the population of Michigan in 2000 was 10,317,426 people.
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