Question

explain 3 modeling interest compounded continuously
read explain 3 part a and complete your turn #1 (adapted from lesson 13.4).
recall the model for interest compounded more than once a year is $a(t)=p(1 +\frac{r}{n})^{nt}$. let $m =\frac{r}{n}$, so $\frac{r}{n}=\frac{1}{m}$ and $nt = mrt$. so after substituting values into $a(t)$, $a(t)=p(1+\frac{1}{m})^{mrt}$. in 13.3 explore section, it was shown that $(1 +\frac{1}{x})^{x}\to e$ as $x\to\infty$. so, $(1+\frac{1}{m})^{m}\to e$ as $x\to\infty$.
function for annual interest compounded continuously
thus, $a(t)=pe^{rt}$. this model gives the value of an investment with principal $p$ and annual interest rate $r$ when interest is compounded continuously. this means the principal amount invested keeps earning interest on the interest earned without stopping.
continuously compounded interest: $a(t)=pe^{rt}$
$p$: initial value
$r$: constant percent
t: unit of time
example 3 find the function that corresponds with the given situation. then use the graph of the function to make a prediction.
a. a person invests $5,000 in an account that earns 3.5% annual interest compound continuously. when will the value of the investment reach $12,000? graph the function on a graphing calculator or a desmos.com and use the graph to make the prediction.
step 1 - find the function $a(t)$ that represents the value of the investment
$p = 5000$
the person invested $5,000.
r = 0.035$
the increase in value is 3.5%. must convert to a decimal before substituting.
substitute the values for $a(t)$.
$a(t)=pe^{rt}$
$a(t)=5000e^{0.035t}$
step 2 - use the graph of the function to make a prediction
since the question is when will the investment be worth $12,000, $a(t)=12000$. to find the correct value of $t$, graph the following functions on a graphing calculator or desmos.com/calculator.
$a(t)=5000e^{0.035t}$
$a(t)=12000$
the two functions are shown to cross at $(25.013,12000)$ meaning $t = 25.013$.
thus, the investment is worth $12,000 after about 25 years.
your turn

1. a person invests $1550 in an account that earns 4% annual interest compounded continuously. find when the value of the investment reaches $2000.

Explanation:

Step1: Identify values of $P$ and $r$

$P = 1550$, $r=0.04$

Step2: Write the compound - continuous interest formula

$A(t)=Pe^{rt}$, so $A(t)=1550e^{0.04t}$

Step3: Set $A(t)$ equal to the target amount

Since $A(t) = 2000$, we have the equation $2000=1550e^{0.04t}$

Step4: Solve for $t$

First, divide both sides by 1550: $\frac{2000}{1550}=e^{0.04t}$, or $\frac{40}{31}=e^{0.04t}$.
Then take the natural logarithm of both sides: $\ln(\frac{40}{31})=\ln(e^{0.04t})$.
Since $\ln(e^{x}) = x$, we get $\ln(\frac{40}{31}) = 0.04t$.
$t=\frac{\ln(\frac{40}{31})}{0.04}\approx\frac{0.2506}{0.04}\approx 6.27$ years.

Answer:

$t\approx 6.27$ years