Tutorial: The number e and exponential growth and decay
Adaptive game version
(This topic is also in Section 2.3 in Finite Mathematics and Applied Calculus) #[I don't like this new tutorial. Take me back to the older tutorial!][No me gusta este nueve tutorial. ¡Regresame al tutorial más viejo (solo en inglés)!]#Resources
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The number e
Suppose we invest &D&1 in the bank for 1 year at 100% interest. If the interest is compounded (that is, added to the account) once a year, then we will have &D&2 at the end of the year.If it is compounded twice a year, then after 6 months we will have half a year's interest added: $1+\frac{1}{2}$, and then half of this quantity added at the end of the year:
$\displaystyle \left(1 + \frac{1}{2}\right) + \frac{1}{2}\left(1 + \frac{1}{2}\right)$ \t $\displaystyle = \left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{2}\right)$ \gap[40] \t #[Factor out][Factorizar]# $\left(1 + \frac{1}{2}\right)$.
\\ \t $\displaystyle =\left(1 + \frac{1}{2}\right)^2$.
If the interest is compounded three times a year, a similar argument will show that we will have
- $\displaystyle {}=\left(1 + \frac{1}{3}\right)^3$ dollars at the end of the year
- $\displaystyle {}=\left(1 + \frac{1}{m}\right)^m \qquad$ This is the compound interest formula for a 100% rate of interest.
-
$e = 2.71828182845904523536028747135266...$
%%A: #[If you look at the %%compoundInteresttut you will find that the expression $\left(1 + \frac{1}{m}\right)^m$, for &D&1 invested for one year at 100% interest compounded $m$ times a year comes from a more general formula: ][Si a el %%compoundInteresttut, encontrarás que la expresión $\left(1 + \frac{1}{m}\right)^m$, para &D&1 invertido durante un año al 100% de interés compuesto $m$ veces por año proviene de una fórmula mas general:]# #[Compound interest formula][Formula para interés compuesto]#
#[If an investment earns interest an annual interest rate of $r$ compounded $m$ times per year for $t$ years, then the value of the investment after $t$ years will be ][Si una inversión gana intereses a una tasa anual de $r$ pero se compone $m$ veces por año, entonces el valor de la inversión después de $t$ años será]#
$P\left(1+\frac{r}{m}\right)^{mt}\qquad$ \t #[Invest $P$ for $t$ years at $r$% compounded $m$ times per year.][Invertir $P$ para $t$ al $r$% compuesto $m$ veces por año.]#
#[We can use a little algebra rewrite this expression to show the the expression leading to $e$ more clearly:][Podemos usar un poco de álgebra para reescribir esta expresión para mostrar la expresión que conduce a $e$ mas claramente:]#
$P\left(1+\frac{r}{m}\right)^{mt}$ \t ${}= P\left(1+\frac{1}{m/r}\right)^{\frac{m}{r}.rt}$
\\ \t ${}=P\left(1+\frac{1}{M}\right)^{M.rt}$ \t #[Write $\frac{m}{r}$ as $M$.][Escribe $\frac{m}{r}$ como $M$.]#
\\ \t ${}=P\left[\left(1+\frac{1}{M}\right)^{M}\right]^{rt}$ \t #[Law of exponents:][Ley de exponentes:]# $b^{Mq} = (b^M)^q$
#[So, as $m$ gets larger an larger, so does $M$, and the quantity in square brackets appraches $e$. We therefore have the following:][Así, a medida que $m$ crece cada vez más, también lo hace $M$, y la cantidad en corchetes se aproxima a $e$. Por lo tanto, tenemos lo siguiente:]#
The number e and continuous compounding
The number e and continuous compounding
#[The number $e$ is the limiting value of the quantities $\left(1 + \dfrac{1}{m}\right)^m$ as $m$ gets larger and larger and has the value 2.71828182845904523536….][El número $e$ es el valor límite de las cantidades $\left(1 + \dfrac{1}{m}\right)^m$ a medida que $m$ se hace cada vez más grande y tiene el valor 2,71828182845904523536….]#
#[If $\$P$ is invested at an annual interest rate $r$ compounded continuously, the accumulated amount after $t$ years is][Si $\$P$ se invierte a una tasa de interés anual $r$ compuesta continuamente, la cantidad acumulada después de $t$ años es]#
$A(t) = Pe^{rt}.\qquad$ P*e^(r*t) #[or][o]# P*EXP(r*t)
Natural logarithm
The logarithm with base $e$ is called the natural logarithm and is often written as $\ln$. Thus, for instance, $\ln 4$ means $\log_e 4$.
%%Examples
1. If &D&500 is invested at an annual rate of 4% compounded continuously, then at the end of 15 years the investment will be worth
If you invest &D&1,000 ar 5% interest compounded continuously, how many years will it take for the investment to reach one million dollars? Putting all this information in the formula gives
$A(15)^{ }$\t $= 500e^{(0.04)(15)}\qquad$\t 500*e^(0.04*15) %%or 500*EXP(0.04*15)
\\ \t $\approx \$911.06$
2. If &D&1 is invested at an annual rate of 100% compounded continuously, then at the end of $x$ years the investment will be worth
$A(x)= e^{x}$ #[dollars][dólares]#.
#[One for you][Uno para t]#
3. How long to invest
If you invest &D&1,000 ar 5% interest compounded continuously, how many years will it take for the investment to reach one million dollars? Putting all this information in the formula gives
$1{,}000{,}000 = 1{,}000e^{0.05t}\qquad$\t #[The unknown is $t$.][Lo desconocido es $t$.]#
\\ $1{,}000 =e^{0.05t}\qquad$\t #[Divide both sides by 1,000.][Dividir ambos lados por 1,000.]#
\\ $0.05t = \ln 1{,}000 \approx 6.90776$ \t #[Rewrite in logarithmic form.][Reescribir en forma logarítmica.]#
\\ $t \approx \dfrac{6.90776}{0.05} \approx 138$ #[years][años]#
#[One for you][Uno para t]#
The number e and exponential functions
If we write the continuous compounding formula $A(t) = Pe^{rt}$ as $A(t) = P(e^r)^t$, we see that $A(t)$ is an exponential function of $t$, where the base is $b = e^r$, so we have really not introduced a new kind of function. In fact, every exponential function $f(t) = Ab^t$ can be written in this way:
$Ab^t$ \t ${}= A(e^{\ln b})^t \qquad$ \t #[When you raise $e$ to $\log_e b$ you obtain $b$.][Cuando elevas $e$ al $\log_e b$ obtienes $b$.]# \t #[(See the logarithm identities in the %%logstut.)][(Ve los identidades logarítmicas en el %%logstut.)]#
\\ \t ${}=Ae^{(\ln b)t}$, \t #[Laws of exponents][Leyes de los exponentes]#
#[which has the form][que tiene la forma]# $Ae^{rt}$, #[where][donde]# $r = \ln b$
Exponential functions expressed in terms of e
#[We can write any exponential function in the following form:][Podemos escribir cualquier función exponencial de la siguiente forma:]#
$f(x) = Ae^{rx}$,
#[where $A$ and $r$ are constants. If $r$ is positive, $f$ models exponential growth; if $r$ is negative, $f$ models exponential decay.][donde $A$ y $r$ son constantes. Si $r$ es positivo, $f$ modela un crecimiento exponencial; si $r$ es negativo, $f$ modela el decaimiento exponencial.]#
#[If $r$ is positive, it is called the growth constant. If $r$ is negative, its absolute value is called the decay constant.][Si $r$ es positivo, se denomina constante de crecimiento. Si $r$ es negativo, su valor absoluto se denomina constante de desintegración.]#
%%Examples
1. $f(x)$ \t ${}= 100e^{0.15x}\qquad $ \t #[Exponential growth with growth constant 0.15.][Crecimeinto exponencial con constante de crecimiento 0.15]#
\\ 2. $f(t) $ \t ${}=Ae^{-0.00012101t}\qquad$ \t #[Exponential decay of carbon 14 with decay constant $0.00012101$][Desintegración exponencial del carbono 14 con constante de desintegración $-0.00012101$]#
\\ 3. $f(t) $ \t ${}=100e^{0.15t}= 100(e^{0.15})^t\qquad$
\\ \t ${}= 100(1.1618)^t$ \t #[Converting $Ae^{rt}$ to the form $Ab^t$][Convertiendo $Ae^{rt}$ a la forma $Ab^t$]#
Half-life and doubling time
#[We have seen that in exponential functions $f(t)$, every increase of $t$ by the same fixed amount results in an increase of $f(t)$ by the same fixed multiple, so if $Q(t) = Q_0e^{kt}$ is an exponential growth functions and doubles after a certain time $t= t_d$, then it will continue to double for each subsequent inrcease of time by that same amount $t_d$. So, we call $t_d$ the doubling time..][Hemos visto que en funciones exponenciales $f(t) = Ab^t $, cada aumento de $t$ por la misma cantidad fija da como resultado un aumento de $f(t)$ por el mismo múltiplo fijo, entonces si $f(t)$ es una función de crecimiento exponencial y se duplica después de un cierto tiempo $t = t_d $, entonces continuará duplicándose para cada aumento subsiguiente de tiempo en la misma cantidad $t_d$. Entonces, llamamos $t_d$ el tiempo de duplicación .]#
#[To calculate the doubling time, we use the fact that $Q(t + t_d)$ must be twice $Q(t)$:][Para calcular el tiempo de duplicación, usamos el hecho que $Q(t + t_d)$ debe ser el doble de $Q(t)$:]#
\t $Q(t+t_d) = 2Q(t)$
\\ \t $Q_0e^{k(t+t_d)} = 2Q_0e^{kt}$
\\ \t $Q_0e^{kt}e^{kt_d} = 2Q_0e^{kt} \qquad $ \t #[Laws of exponents][Leyes de los exponentes]#
\\ \t $e^{kt_d} = 2$ \t #[Cancel the][Cancela la]# $Q_0e^{kt}$.
\\ \t $kt_d = \ln 2$ \t #[Logarithmic form][Forma logarítmico]#
\\ \t $t_d = \dfrac{\ln 2}{k}$ \t #[Formula for doubling time][Formula para tiempo de duplicación]#
#[Similarly, if $Q(t)$ is exponential decay $Q(t) = Q_0e^{-kt}$, then there is a fixed half-life $t_h$, the time it takes for the quantity to halve in size. A similar calculationto that above shows that][De manera similar, si $Q(t)$ es un decaimiento exponencial $Q(t) = Q_0e^{-kt}$, entonces hay una vida media $t_h$ fija, el tiempo que tarda la cantidad a reducir a la mitad. Un cálculo similar al anterior muestra que]#
\\ \t $t_h = \dfrac{\ln 2}{k}$ \t #[Formula for half-life][Formula para tiempo medio]#
#[Multilying b $k$ gives the following.][Multiplicar por $k$ da el siguiente.]#
Half-life and doubling time
For a quantity undergoing exponential growth, $Q(t)=Q_0e^{kt}$, the value of $Q$ doubles every $t_d$ units of time, where #
$t_dk = \ln2$. \t \gap[40] Doubling time × Growth constant = ln 2
For a quantity undergoing exponential decay, $Q(t)=Q_0e^{-kt}$, the value of $Q$ halves every $t_h$ units of time, where the half-life an decay constant are related by #
$t_hk = \ln2$. \t \gap[40] Half-life × Decay constant = ln 2
Now try some of the exercises in Section 2.3 in Finite Mathematics and Applied Calculus.
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