Rate at which flu spreads

Question

Here's the question:

Say the rate at which the flu spreads is proportional to the number of people currently have the flu.
Assuming an unlimited population, if one person is sick at 9:00 AM, and two people are sick at 3:00 PM, how many people will be sick at 9:00 AM the next morning (rounding to the nearest whole number)?

Here's my answer:

9 am - 1
3 pm - 2
9 pm - 4
3 am - 16
9 am - 256

My understanding is that proportional means exponentially.

Please let me know if my logic is correct and if my answer is right.

Answer 1

"Proportionally" does mean "exponentially" - but you have misunderstood what "exponentially" means.

"Proportional to the number of people currently sick" means that each sick person has the same chance to infect someone else. The data given tell you that it takes six hours for the number of infected people to double. That means your list should proceed $$ 1, 2, 4, 8, 16, \ldots . $$

You have squared the infecteds. That can't make any sense if you think about the numbers as counting people, since you can't have a square person.

(PS I find reading and responding to such a ridiculous "word problem" very painful. That's your teacher's fault, not yours.)

Answer 2

You have a solid intuition about this problem: let's make it rigorous.

If we define $F(t)$ to be the number of people infected at time $t$, then we know that $$\frac{dF}{dt} = kF$$ where $k$ is a constant.

Hence, $$\frac{dF}F=k\cdot dt$$$$\int \frac{dF}F = \int k\cdot dt$$$$\ln |F| = kt + C$$ $$F(t) = e^{kt+C} = Ae^{kt}$$.

This demonstrates that $F(t)$ should be an exponential function.

However, your data is wrong, since the relationship between your numbers seems to be proportional to $F(t)=t^2$, not an exponential function.

So since you know $F(1)=1$ and $F(2)=2$, we can plug those into our function.

$$Ae^{k\cdot1}=1$$$$Ae^{k\cdot2}=2$$

Substituting the first equation in the second gives

$$\frac{e^{k\cdot2}}{e^{k\cdot1}}=2$$ $$e^k = 2$$ $$k = \ln 2$$ Now, going back to the first equation

$$Ae^{\ln 2} = 1$$ $$2A = 1$$ $$A = \frac 1 2$$

Hence, your final function is $$F(t) = \frac {e^{t\cdot\ln 2}}2$$