We just started learning about differential equations (just first order, ordinary differential equations so far). I have to solve an exercise whose translation is:
There is a rumor among the students of this class that this problem
will be on the final exam in June. If there are 70 students enrolled
and the rumor is spreading in proportion to the number of students who
have not yet heard about it, how long will it take for 60 students to know if by the second day 40 already know about it? Note: it is assumed that in t = 0 no student knows the rumor.
So I've tried to come up with a solution, but I realized that the number of unaware students doesn't come up in it. For starters, I'm working with a function S(t) (number of students who have heard about the rumor depending on how much time, in days, has passed). I followed this logic:
The number of students who know about it must be equal to the initial number of aware students (zero) plus $\frac{dS(t)}{dt} t$ (the number of students who find out each day times the days passed). After solving, I get $A(t)=kt$, being $k$ a constant which should equal 20.
Did I go wrong somewhere? Thanks in advance!
Best Answer
Let $S(t)$ be the number of students who have not heard the rumor at time $t$, measured in days. We could, as you have done, set up the problem so that the function measures the number of students who have heard the rumor, but it strikes me that it might save a little computational headache if we approach things this way.
We are told that the rate at which the rumor spreads is proportional to the number of students who have not yet heard the rumor. In notation, this means that there is some constant $k$ such that $$ \frac{\mathrm{d}}{\mathrm{d}t} S(t) = kS(t). $$ We are also given the information that $S(0) = 70$, and that $S(2) = 30$ (that is, 70 students are unaware of the rumor on day zero, and $70-40=30$ students are unaware of the rumor by day two).
This gives us the initial value problem $$\begin{cases} \dfrac{\mathrm{d}}{\mathrm{d}t} S(t) = kS(t) \\ S(0) = 70,\quad S(2) = 30. \end{cases} $$ Via some abstract nonsense (this feels like an intro level differential equations question, so we don't really need to get into the details), we know that such an IVP has a unique solution of the form $$ S(t) = S(0) \mathrm{e}^{kt} = 70 \mathrm{e}^{kt}. $$ Note that we have used one of the two initial values to get this formula. All that is left is to determine a value for $k$. This can be done using the second initial value: $$ 30 = S(2) = 70 \mathrm{e}^{2k} \implies \mathrm{e}^{2k} = \frac{3}{7} \implies 2k = \log\left( \frac{3}{7} \right) \implies k = \frac{1}{2}\log\left( \frac{3}{7} \right). $$ Observe that $\frac{3}{7} < 1$, and so its logarithm is negative. This makes sense, since the number of students who have not heard the rumor is decaying to zero, thus the constant of decay should be negative. Huzzah! We haven't completely borked things up yet.
Finally, we want to know how long it will take until 60 students have heard the rumor. That is, we want to find $t$ such that $S(t) = 10$ (i.e. 10 students have not heard the rumor). So we just plug-and-chug:
\begin{align} 10 = S(t) = 70 \mathrm{e}^{\frac{1}{2}\log\left( \frac{3}{7} \right) t} &\implies \mathrm{e}^{\frac{1}{2}\log\left( \frac{3}{7} \right) t} \frac{1}{7} \\ &\implies \frac{1}{2}\log\left( \frac{3}{7} \right) t = \log\left(\frac{1}{7}\right) \\ &\implies t = \frac{\log\left(\frac{1}{7}\right)}{\frac{1}{2}\log\left( \frac{3}{7} \right)} \approx 4.59. \end{align} Thus it will take about four and a half days for 60 students to hear the rumor (or, if you prefer, 60 students will have heard the rumor by the end of the day 5).