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Biology application:  Suppose the number of bacteria grow at an average rate or r = 10\% per hour.  If the initial population is 100 bacteria, find the number of bacteria after 1 hour, after 2 hours, after $T$ hours.

\vskip 5pt \hrule\vskip -2pt

Identical application, but in Finance:

Let $P(t) =$ amount in an account at time $t$ (in years).

Ex 1:  Suppose \$100 is deposited in the account earning an interest rate of
 $r = 10\%$ per year.  Find the amount in the account after 1 year, after 2 years, after 3 years, and after $T$ years. 



$t = 0$:  $P(0) =$ \$100

$t = 1$:  $P(1) =$$100(1 + 0.1) = 100(1.1)$ = \$110

$t = 2$:  $P(2) =$$100(1 + 0.1)(1 + 0.1) = 100(1 + 0.1)^2 = 100(1.1)^2$ = \$121

$t = 3$:  $P(3) =$$100(1 + 0.1)^3 = \$100(1.1)^3 =133.10$

\d

$t = T$:  $P(T) =$$100(1 + 0.1)^T = \$100(1.1)^T$

The average interest rate earned is $10\%$ per year.
\v

\line{The average rate of change in the account btwn year $0$ and year $1$:}
\v
\centerline{${P(1) - P(0) \over 1 }$ =
${100(1.1)- 100}$  = $100(0.1)$ = \$10/year.}

The average rate of change between year $t$ and year $t+1$:

\v
\centerline{${P(t + 1) - P(t) \over 1 }$ =
${100(1.1)^{t+1} - 100(1.1)^t}$  = \$$100(1.1)^{t} (0.1)$/year.}

Instantaneous rate of change at time $t$:

\v
\centerline{$P'(t) = [100(1.1)^t]' = 100ln(1.1)(1.1)^t = (9.53102...) \cdot (1.1)^t$}

At $t = 1$:  $P'(1) = 100ln(1.1)(1.1) = 10.48...$
\r


Ex 2:  Suppose \$100 is deposited in the account earning an interest rate of $r = 10\%$ per year, compounded monthly.  Find the amount in the account after 1 month, 1 year,  2 years, and $T$ years.
\s\s
$t = 0$:  $P(0) =$ \$100
\s
$t = 1$ month:    $P({1 \over 12}) =100(1 + {0.1 \over 12}) $ = \$100.83
\s
$t = 1$ year:  $P(1) =100(1 + {0.1 \over 12})^{12}$ = \$110.47
\s
$t = 2$ years:  $P(2) =100(1 + {0.1 \over 12})^{12 \cdot 2}$ = \$122.04

\s
\d
\s

$t = T$ years:   $P(T) =100(1 + {0.1 \over 12})^{12 \cdot T} = \$100(1.1047...)^T$

The average interest rate earned is ${10 \over 12}\%$ per month.
\s
The average interest rate earned is $10.47...\%$ per year.
\s
%%\line{The average rate of change in the account btwn year $0$ and year $1$:}\v
%%\centerline{${P(1) - P(0) \over 1 }$ =${100(1.1)- 100}$  = $100(0.1)$ = \$10/year.}

The average rate of change between year $t$ and year $t+1$:
\v\s
\centerline{${P(t + 1) - P(t) \over 1 }$ =
$100(1 + {0.1 \over 12})^{12(t+1)} - 100(1 + {0.1 \over 12})^{12t}$ }
\v
\centerline{~~~~~~~~~~~~~~~~~= 
\$$100(1 + {0.1 \over 12})^{12t}[(1 + {0.1 \over 12})^{12} - 1]$/year.}

\s
The approximate average rate of change between year $t$ and year $t+1$:
\v\s
\centerline{${P(t + 1) - P(t) \over 1 }$ =
${100(1.1047)^{t+1} - 100(1.1047)^t}$}
 \v
\centerline{ ~~~~~~~~~~~~= 
\$$100(1.1047)^{t} (0.1047)$/year.}
\s

The instantaneous rate of change at time $t$:
\s
\v
\centerline{$P'(t) = [100(1 + {0.1 \over 12})^{12 \cdot t} ]' = 100ln[(1 + {0.1 \over 12})^{12}]\cdot [(1 + {0.1 \over 12})^{12}]^t$}

\v
\rightline{$= 1200ln(1 + {0.1 \over 12}) \cdot (1 + {0.1 \over 12})^{12t}$~~~~~~~~~~}

\v
\rightline{$= (9.95856...) \cdot (1 + {0.1 \over 12})^{12t}$~~~~~~~~~~}

At $t = 1:$ $P'(1) = 1200ln(1 + {0.1 \over 12}) \cdot (1 + {0.1 \over 12})^{12} = 11.001...$

At $t = {1 \over 12}$,  $P'( {1 \over 12}) = 1200ln(1 + {0.1 \over 12}) \cdot (1 + {0.1 \over 12}) = 10.0416$

\r

\eject

Ex 3:  Suppose \$100 is deposited in the account earning an interest rate of $r = 10\%$ per year, compounded daily.  Find the amount in the account after 1 day, 1 year, 2 years, and $T$ years.


$t = 0$:  $P(0) =$ \$100

$t = 1$ day:    $P({1 \over 356}) =100(1 + {0.1 \over 365}) $ = \$100.03

$t = 1$ year:  $P(1) =100(1 + {0.1 \over 365})^{365}$ = \$110.52

$t = 2$ years:  $P(2) =100(1 + {0.1 \over 365})^{365 \cdot 2}$ = \$122.14

\d


$t = T$ years:   $P(T) =100(1 + {0.1 \over 365})^{365 \cdot T} = \$100(1.10515578...)^{T} $


The average interest rate earned is ${10 \over 365}\%$ per day.

The average interest rate earned is $10.515578...\%$ per year.

%%\line{The average rate of change in the account btwn year $0$ and year $1$:}\v
%%\centerline{${P(1) - P(0) \over 1 }$ =${100(1.1)- 100}$  = $100(0.1)$ = \$10/year.}

The average rate of change between year $t$ and year $t+1$:
\v
\centerline{${P(t + 1) - P(t) \over 1 }$ =
$100(1 + {0.1 \over 365})^{365(t+1)} - 100(1 + {0.1 \over 365})^{365t}$ }
\v
\centerline{~~~~~~~~~~~~~~~~~= 
\$$100(1 + {0.1 \over 365})^{365t}[(1 + {0.1 \over 365})^{365} - 1]$/year.}

The instantaneous rate of change at time $t$:

\v
\centerline{$P'(t) = [100(1 + {0.1 \over 365})^{365 \cdot t} ]' = 100ln[(1 + {0.1 \over 365})^{365}]\cdot [(1 + {0.1 \over 365})^{365}]^t$}

\v
\rightline{$= 36500ln(1 + {0.1 \over 365}) \cdot (1 + {0.1 \over 365})^{365t}$~~~~~~~~~~}

\v
\rightline{$= (9.99863... )\cdot (1 + {0.1 \over 365})^{365t}$~~~~~~~~~~}


At $t = 1:$ $P'(1) = 36500ln(1 + {0.1 \over 365}) \cdot (1 + {0.1 \over 365})^{365}= 11.05...$

At $t = {1 \over 365}$,  $P'( {1 \over 365}) = 36500ln(1 + {0.1 \over 365}) \cdot (1 + {0.1 \over 365}) = 10.00...$

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%%Suppose \$100 is deposited in the account earning an interest rate of $r = 10\%$ per year, %%compounded $n$ times per year.  Find the amount in the account after $T$ years.

%%$t = T$ years:   $P(T) = 100(1 + {0.1 \over n})^{n \cdot T}$

\r

Ex 4:  Suppose \$100 is deposited in the account earning an interest rate of $r = 10\%$ per year, compounded $n$ times per year.  Find the amount in the account after $T$ years.

$t = T$ years:   $P(T) = 100(1 + {0.1 \over n})^{n \cdot T}$

\l

Ex 5:  Suppose \$100 is deposited in the account earning an interest rate of $r = 10\%$ per year, compounded continuously.  Find the amount in the account after $T$ years.

$t = T$ years:   $P(T) = lim_{n \rightarrow \infty} 100(1 + {0.1 \over n})^{n \cdot T} = 100 e^{0.1T}$


%%Interest rate of $r = 10\%$ per year, compounded continuously = instantaneous growth rate %%of 10\% per year.

%%Suppose $P(t)$ = amount of money in account at time $t$

%%Hence slope of tangent line is ${{\hbox{change in P}} \over{\hbox{change in time}}}$ = %%${\$0.1P \over years}$

\l

Definition $e = lim_{n \rightarrow \infty} (1 + {1 \over n})^{n} = 2.7...$
\l
FYI :  By Taylor series approximation from Calculus II

$e^{0.1T} = lim_{n \rightarrow \infty} (1 + {1 \over n})^{n (0.1)T} 
=lim_{n \rightarrow \infty} [(1 + {1 \over n})^{0.1}]^{nT} $

~~~~~~~~$= lim_{n \rightarrow \infty}[ 1 + {0.1 \over n} - {0.045 \over n^2} + {0.0285 \over n^3} - ...]^{nT} $
$= lim_{n \rightarrow \infty}[ 1 + {0.1 \over n} ]^{nT} $

\l

Ex 6:  Suppose \$$P_0$ is deposited in the account earning an interest rate of $r = s \%$ per year 
($r = {s \over 100}$), compounded continuously.  
\u
$t$ years:   $P(t) = lim_{n \rightarrow \infty} P_0(1 + {r \over n})^{n \cdot t} = P_0 e^{rt}$

\l

Ex 7:  Suppose \$1 is deposited in the account earning an interest rate of $r = 10\%$ per year 
($r = {100 \over 100} = 1$), compounded continuously.  
\u
$t$ years:   $P(t) = lim_{n \rightarrow \infty} (1 + {.1 \over n})^{n \cdot t} = e^{0.1 t}$
\v\u
\centerline{Note the instantaneous rate of change is 10\% = $ 0.1e^{0.1t}$}
\v\u
\centerline{That is $P'(t) = [e^{0.1t}]' = 0.1e^{0.1t}$}

\l

Ex 8:  Suppose \$1 is deposited in the account earning an interest rate of $r = 100\%$ per year 
($r = {100 \over 100} = 1$), compounded continuously.  
\u
$t$ years:   $P(t) = lim_{n \rightarrow \infty} (1 + {1 \over n})^{n \cdot t} = e^{t}$
\v\u
\centerline{Note the instantaneous rate of change is 100\% = $ e^{t}$}
\v\u
\centerline{That is $P'(t) = [e^t]' = e^{t}$}

















\end
\l
\vfill
Hence   $lim_{n \rightarrow \infty} 100(1 + {0.1 \over n})^{n \cdot T}$

$ =   \$100 lim_{n \rightarrow \infty} [(1 + {0.1 \over n})^{n}]^T$

$lim_{n \rightarrow \infty} [(1 + {0.1 \over n})^{nt}] =  lim_{n \rightarrow \infty} (1 + {1 \over n})^{0.1nt}$


$lim_{n \rightarrow \infty} [(1 + {0.1 \over n})] =  lim_{n \rightarrow \infty} (1 + {1 \over n})^{0.1}=  e^{lim_{n \rightarrow \infty}ln (1 + {1 \over n})^{0.1}}$

$=  e^{0.1lim_{n \rightarrow \infty}ln (1 + {1 \over n})}$

$= [ e^{lim_{n \rightarrow \infty}ln (1 + {1 \over n})}]^{0.1}$

$= [ {lim_{n \rightarrow \infty}(1 + {1 \over n})}]^{0.1}$


$100 e^T$
\end

\r

Identical application, but in Finance:
Suppose \$100 is deposited in an account earning an interest rate of
 $r = 10\%$ per year.  Find the amount in the account after 1 year, after 2 years, after $T$ years.



Biology application:  Suppose the number of bacteria grow at an average rate or r = 10\% per hour.  If the initial population is 100 bacteria, find the number of bacteria after 1 hour, after 2 hours, after $T$ hours.
\r
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