# Thread: Advanced Periodic and Exponential Functions

1. ## Advanced Periodic and Exponential Functions

At 9am, 5% of the students in a school have heard a rumour concerning a student-tracher relationship. By 11am, 20% of the students have
heard the rumour. Find the time when 90% have heard the rumour.

2. ## Re: Advanced Periodic and Exponential Functions

What model are you assuming? Since you are given "two data points" (t= 9, s= .05 and t= 11, s= .20), the simplest model would be linear: s= at+ b. t= 9, s= .05 gives .05= 9a+ b and t=11, s= .20 gives .20= 11a+ b. That gives you two linear equations to solve for a and b.

But you do not necessarily apply the same mathematics to any situation. There might be good reason to think the model was "exponential": $s= ba^t$. Then we would have $.05= b(a^9)$ and $.20= b(a^{11})$ which we can, again, solve for a and b.
Since you titled this "exponential functions", I imagine this is the one you want.

Since there is, presumably, an upper bound on the number of students in the school and so an upper bound on the value of s, the most reasonable model might be the "logistic function", $s= \frac{b}{1+ a^{-t}}$. Now we would have $.05= \frac{b}{1+ a^{-9}}$ and $.20= \frac{b}{1+ a^{-11}}$, again two equations to solve for a and b.