Exponential growth (and decay) [1.7]
Half-life and doubling time
The doubling time of an exponentially increasing quantity is the time required for the quantity to double.
The half-life of an exponentially decaying quantity is the time required for the quantity to be reduced by a factor of one half.
Our prototypical exponential function: $$P(t)=P_0a^t.$$
Time that it takes for an investment of \$1000 at 5% annual interest to double?
We know that the value of investment as a function of time is: $$P(t)=1000 (1.05)^t.$$ You can see (by setting $t=0$) that the initial value of the investment is \$1000. So, to find the doubling time, $t_d$, we could set the value of the investment to \$2000 and then solve for the time $$\begineq P(t=t_d)=2000 &= 1000 (1.05)^{t_d}\\ \frac{2000}{1000}&=1.05^{t_d}\\ 2&=1.05^{t_d}\\ \ln 2 &= \ln[1.05^{t_d}]=t_d\ln 1.05\\ \frac{\ln 2}{\ln 1.05}&=t_d=14.2\text{ years} \endeq $$
Finding $a$ from the doubling time
An investment doubles in value every 8 years. What is the interest rate?
Hint: Start from our prototypical exponential like this: $$\begineq P(t_\text{double})&=P_0a^{t_\text{double}}\\ 2P_0 &= P_0a^8\endeq$$ and solve to find $a$ and then use $a=(1+r)$. Answer: $r$ is close to 9%...
So the interest rate must be $r=0.0905=9.05\%$.
Or maybe this is simpler?
An investment doubles in value every 8 years. What is the interest rate?
Start from our exponential expression $P=P_0(1+r)^t$ like this: $$\begineq 2P_0&=P_0(1+r)^{t_\text{double}}=P_0(1+r)^8\\ 2&=(1+r)^8\\ [2]^{\frac 18}&=\left[(1+r)^8\right]^\frac 18=(1+r)\\ 1.0905&=1+r\\ 1.0905-1=0.0905&=r \endeq$$ So, again, the rate is 9.05%.
Re-writing exponential growth / decay
Another way of writing exponential growth or decay: Since $P(t)=2P_0$ when $t=t_d$, apparently $$\begineq a^{t_d}&=2 \\ [a^{t_d}]^{\frac{1}{t_d}} &=2^{\frac{1}{t_d}} \\ a &= 2^{\frac{1}{t_d}} \\ \endeq$$
This means that we could rewrite exponential growth in terms of the doubling time: $$P(t)=P_0a^t =P_0\left(2^{\frac{1}{t_d}}\right)^t = P_0 2^{\frac{t}{t_d}}$$
And exponential decay in terms of the half-life $t_h$, becomes $$P(t)=P_0\left(\frac 12\right)^{\frac{t}{t_h}}.$$
Exponential relation connecting two points
A common problem with straight lines is to find the equation of a line that connects two given points.
Solve a similar problem: Find the unique exponential function that connects two points:
[Variation on problem 20]: The number of people living with HIV infections increased worldwide approximately exponentially from 2.5 million in 1985 to 37.8 million in 2003. Find an exponential formula for the number of HIV cases, $H(t)$, as a function of the number of years, $t$, since 1983. (In the process, you'll be "backcasting" to find how many cases there were in 1983.)
Hint: To solve this, start by writing out expressions for $P(2)$ and $P(20)$.
- In 1985 (2 years after 1983) we had $P(2)=2.5$ (million) cases=$P_0a^2$.
- In 2003 (20 years after 1983) we had $P(20)=37.8$ (million) cases=$P_0a^{20}$.
Keep going...Now we have two equations, and we don't know $P_0$ and we don't know $a$. But we can eliminate $P_0$ by dividing one equation by the other...
$$\begineq \frac{P(20)}{P(2)}&=\frac{P_0 a^{20}}{P_0 a^{2}}\\ \frac{37.8}{2.5}=15.12&=\frac{a^{20}}{a^{2}}=a^{20-2}=a^{18}\\ \endeq$$ Now, we raise both sides to the $1/18$ power: $$\begineq \left[15.12\right]^{\frac 1{18}}&=\left[a^{18}\right]^\frac 1{18}\\ 1.163&=a\\ \endeq$$ Now that we know what $a$ is, we can substitute it back in to one or the other of our two initial equations to find $P_0$. For example, the 1985 data point (for $t=2$), was: $$\begineq P(2)=2.5 &=P_0a^2=P_0(1.163)^2\\ 2.5&=P_0 1.353.\endeq$$ So, $P_0=2.5/1.353=1.85$.
...
Putting this together:
$$P(t)=1.85(1.163)^t.$$
This means that in 1983 (t=0) there were 1.85 million cases. And the average rate of increase of infections was 16.3% more each year.