Exponential Growth
Growth and Decay Factors
In Calculus I, everything is about the rate of change. In a linear function, the rate of change is constant. That means that increasing $x$ by 1 causes a change in $y$ of the same amount, equal to the slope, $m$. In symbols, \[ f(x+1) = f(x) + m \]
Let $f(x) = 2x - 3$
\[ \begin{array}{r|l} x & f(x) \\ \hline -1 & -5 \\ 0 & -3 \\ 1 & -1 \\ 2 & 1 \\ 3 & 3 \\ \vdots & \vdots \\ \end{array} \]To get the next $y$, take the previous $y$ and add the slope, 2.
In an exponential function, increasing $x$ by 1 causes a change in $y$ by the same factor, called the growth rate.
Let $f(x) = 2^x$
\[ \begin{array}{r|l} x & f(x) \\ \hline -2 & 2^{-2} = \frac{1}{4} \\ -1 & 2^{-1} = \frac{1}{2} \\ 0 & 2^{0} = 1 \\ 1 & 2 \\ 2 & 4 \\ 3 & 8 \\ 4 & 16 \\ \vdots \end{array} \]To get the next $y$, take the previous $y$ and multiply by 2.
In a linear function, the rate of change is constant. In an exponential function the rate of change is proportional to the $y$-value.
\[ \begin{array}{r|l} \text{interval} & \frac{\Delta y}{\Delta x} \\ \hline [-2,-1] & \frac{1/2 - 1/4}{1} = \frac14 \\ [-1,0] & \frac{1 - 1/2}{1} = \frac12 \\ [0,1] & \frac{2 - 1}{1} = 1 \\ [1,2] & \frac{4 - 2}{1} = 2 \\ [2,3] & \frac{8 - 4}{1} = 4 \\ \vdots & \vdots \end{array} \]Exponential growth and decay occurs when the growth or decay rate proportional to the quantity. For example, if a collection of cells divide at regular intervals, then the rate at which new cells are being added in cells per unit time is proportional to how many cells there are at that time. If a radioactive isotope is decaying due to half=life, the rate at which the isotope decays is proportional to the quantity.
Because $f(0) = a \cdot b^0 = a$, $a$ represents the initial value of the function. The base is $b$, which represents the growth factor when $b > 1$ and the decay factor when $0 < b < 1$. The value of the function at $x + 1$ is $b$ times the value of the function at $x$.
A population of 100,000 cells are growing by 20% each day. Write an equation for the population $y = P(t)$ as a function of time $t$ in days. Then write an equation the population $y = P(w)$ as a function of time $w$ in weeks, and $y = P(h)$ as a function of time $h$ in hours. Write all three equations in the form $y = a \cdot b^x$.
An increase of 20% is a growth factor of $1.20$. Think of it as 100% plus the increase of 20%, which is 120%, or $1.2$ as a decimal. Therefore, when $t$ is time in days, $b = 1.2$ and the equation is \[ y = P(t) = 100,000 \cdot (1.2)^t \]
Because there are 7 days in a week, $t = 7w$, so by substitution, $y = P(m) = 100,000 \cdot (1.2)^{7w}$. This is not quite in the form $y = a \cdot b^{x}$ because the exponent contains more than just the input variable. However, because \[ \left(b^u\right)^v = b^{uv} = b^{vu} = \left(b^v\right)^u \] $(1.2)^{7w}$ can be written as $\left(1.2^{7}\right)^w$. Therefore, \[ y = P(w) = 100,000 \cdot \left(1.2^{7}\right)^w \]
Because there are 24 hours in a day, $t = d/24$, and $y = P(h) = 100,000 \cdot (1.2)^{h/24}$ (for example, when $h = 24$, there is one day and $24/24 = 1$). Therefore, \[ y = P(h) = 100,000 \cdot \left(1.2^{1/24}\right)^h \]
A population of 100,000 bacteria that has been treated with a biocide is declining at a rate of 10% each hour. Write an equation for the population $y = P(t)$ as a function of time $t$ in hours, $y = P(m)$ as a function of time $m$ in minutes, and $y = P(d)$ as a function of time $d$ in days. Write all three equations in the form $y = a \cdot b^x$.
A decrease of 10% is a decay factor of $0.90$. Think of it as 100% minus the decrease of 10%. Therefore, when $t$ is time in hours, $b = 0.90$ and the equation is \[ y = P(t) = 100,000 \cdot (0.90)^t \]
If $t$ is the number of hours and $m$ is the number of minutes, $t = m/60$, and by substitution \[ y = P(m) = 100,000 \cdot \left(0.90^{1/60}\right)^{m} \] Similarly, if $d$ is the number of days, then $t = 24d$ and \[ y = P(d) = 100,000 \cdot \left(0.90^{24}\right)^d \]
Growth and Decay Rates
The growth or decay factor, $b$, can be written as $b = 1 + r$ where $r$ is the growth or decay rate. When $r > 0$, there is exponential growth, and when $-1 < r < 0$, there is exponential decay. $r$ cannot be $-1$ or less. If $r = -1$, then the decay rate would be 100% and there wouldn't be anything left.
A population of 100,000 cells are growing by 20% each day. What are the weekly and hourly decay rates? Give your answers as percentages precise to two decimal places.
Because the population $P$ as a function of time in weeks is $y = \left(1.2^{7}\right)^w$, the weekly growth factor is the base $b = 1.2^{7} \approx 3.5831808$. Because $b = 1 + r$, $r = b - 1$, so the weekly growth rate is $2.5831808$, or $258.32\%$. The population is increasing at a rate of about 258% each week.
The hourly growth factor is $1.2^{1/24} \approx 1.00762566$. The hour growth rate is $0.00762566$, or $0.76\%$. The population is growing at a bit more than three quarters of a percent each hour, with a net result of 20% per day. To check, observe that $1.0076^{24} \approx 1.1992668$.
A population of 100,000 baceteria treated with a biocide are declining by 10% each hour. What are the decay rates per day and per minute? Give your answers as percentages precise to two decimal places.
The daily decay factor is $0.90^{24} \approx 0.079766$. The rate is 1 minus the factor, so the daily decay rate is $1 - 0.079766 = 0.920234$, or $92.02\%$. About 92% of the baceteria are gone after one day.
The decay factor each minute is $0.90^{1/60} \approx 0.9982455$, so the decay rate each minute is $1 - 0.9982455 = 0.0017545$, or 0.18\%$. About two-tenths of a percent are gone each minute (that is, about 99.82% that were present at the beginning of the minute remain at the end of the minute).
Money
You can solve compound interest problems this way. The only difference is that if the annual percentage rate interest is $r$, and interest is compounded monthly (for example), then the monthly interest is $r/12$ rather than $1 - (1+r)^{1/12}$.
An investor invests some amount of money at 6\% compounded monthly. What are the monthly, annual, and daily growth rates?
The monthly growth rate is $0.06/12 = 0.005$, so the monthly growth factor is $1.005$. The annual growth factor is $1.005^{12} \approx 1.060677812$. The annual growth rate is $6.07\%$. It is slightly more than 6\% because of compounding.
The daily growth factor is $1.005^{1/30} \approx 1.000166265$, so the daily growth rate is about $0.17\%$.