Differential equation problem: Modeling the spread of a rumor on campus

[Kolika28]

Homework Statement

A rumor is spread by a student 00.00 night to 12 of December at party. Two days later half of the students know the rumor. According to one model, the spread rate at time t (in hours after discovery) is proportional to the product of three factors

(1) the number of students knowing the rumor at time t, [i.e. how many that can disseminate information]

(2) the number of students who do not know the rumor at time t [i.e. How many who can receive the rumor] and

(3) 1 - sin (πt / 12) [students are least active at 06 in the morning].

We ignore that the number of students is an integer. We describe the process of a derivative function f (t) which describes the number of students who know the solution at time t (in hours). There are A> 1 students.

(a) Set up the initial value problem that describes the situation.

(b) Solve the initial value problem.

Relevant Equations

Differential equation

So this is what I have done:

$f'(t)=k*f(t)*(A-f(t))*(1-sin(\frac{pi*x}{12}))$

$\frac{1}{f(t)*(A-f(t))}=k*(1-sin(\frac{pi*x}{12}))$

I see that the left can be written as this (using partial fractions):

$1/A(\frac{1}{f(t)}-\frac{1}{A-f(t)})$ And then I take the integral on both sides and get

$1/A*(ln(f(t))-ln(A-f(t))=k(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)$

$ln(\frac{f(t)}{A-f(t)})=kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C)$

$\frac{f(t)}{A-f(t)}=e^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})+C}=Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}$

$f(t)=Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}*(A-f(t))$

$f(t)=A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}-f(t)*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}$
$f(t)+f(t)*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}=f(t)*(1+Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})})=A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}$

$f(t)=\frac{A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})}}{(1+Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*x}{12})})}$
I don't feel like I'm doing this right. I'm struggling with finding both C and k. What have I done wrong?

 

[hutchphd]

1. What at is x? (should be t?)
2. You have two constants and 2 boundary values (at t=0 and t= 2 days)

 

[tnich]

Aside from several notation errors (which I would knock off points for if I were grading your homework), this step is not correct. You can check your work by putting the two fractions over a common denominator.

$1/A(\frac{1}{f(t)}-\frac{1}{A-f(t)})$

You seem to have canceled out your error with another one in the next step, though.
So you have shown us the part you got right. You don't really need help with that. Can you show us the part you are stuck on?

 

[Kolika28]

Aside from several notation errors (which I would knock off points for if I were grading your homework), this step is not correct. You can check your work by putting the two fractions over a common denominator.
You seem to have canceled out your error with another one in the next step, though.
So you have shown us the part you got right. You don't really need help with that. Can you show us the part you are stuck on?

I know about the notation errors, I always mix up x and t, so sorry for that! Ok, so my last expression is right? I'm stuck on finding both k and C. Given that one person knows about the rumor 00.00, I guess that f(0)=1 and the text tells me that f(2*24)=A/2. And here I get stuck, because I'm not able to "isolate" c and k and find their value.

 

[tnich]

I know about the notation errors, I always mix up x and t, so sorry for that! Ok, so my last expression is right? I'm stuck on finding both k and C. Given that one person knows about the rumor 00.00, I guess that f(0)=1 and the text tells me that f(2*24)=A/2. And here I get stuck, because I'm not able to "isolate" c and k and find their value.

You may not be able to isolate them. You will get two equations to solve for two unknowns.

 

[Kolika28]

You may not be able to isolate them. You will get two equations to solve for two unknowns.

Ok, I will try. But when I solve the equation in GeoGebra, I don't get the same expression for f(t):

 

[hutchphd]

Remember your students are "continuous" (not integers) so the boundary condition at t=0 is probably 0 in that spirit...no that's not going to work...

 

[Kolika28]

Hmmm, I'm still not able to find k and c. I don't know what I'm doing wrong

 

[hutchphd]

What is your solution explicitly?

 

[tnich]

If we don't show us what you are trying, how can we figure out what you are doing wrong?

The GeoGebra solution is the same as yours, modulo a constant.

 

[epenguin]

A is the number of students. You can't derive that by math. If it is not given, you might as well talk about the fraction of the total population of students who know. (You might as well do this anyway.) Let this fraction be your variable, x, then isn't A = 1?

k is obtained from sentence 2 of your homework statement. Well it would be if you knew the initial fraction of students. If they don't tell you, make some reasonable assumption. E.g. Number of students on campus is 1000. Then at t=0, x = 0.001 . A smaller initial fraction should just make the graph steeper. The graph should be symmetrical re 180 deg. rotation about the midpoint. I hope you can see (in fact explain to us) why these should be so.When you have done it we would be interested to see a graph of your solutions, With two or three different initial fractions.

 

[Kolika28]

Thank you for everyones help, I think I found out of the k and C problem (or I hope so). Ok, so I think the problem was that I used the last expression when trying to find C and k. I have now found the answer using the fifth line I showed and making two equations.

$ln(\frac{f(t)}{A-f(t)})=kA(t+\frac{12}{\pi}*cos(\frac{\pi*t}{12})+C)$

1.$ln(\frac{1)}{A-1})=kA(0+\frac{12}{\pi}*cos(\frac{\pi*0}{12})+C), f(0)=1$
2.$ln(\frac{A/2}{A-A/2})=kA(48+\frac{12}{\pi}*cos(\frac{\pi*48}{12})+C), f(48)=\frac{A}{2}$

Looking at equation 2, I find out that $C=-48-\frac{12}{\pi}$. And plotting that value in equation 1, I get that $k=\frac{ln(A-1)}{48A}$. I am aware that this C is not equal to the C in this expression:

$f(t)=\frac{A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*t}{12})}}{(1+Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*t}{12})})}$

This due to my bad notation errors. So this is true for the C value I found now:
$f(t)=\frac{A*e^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*t}{12})+C}}{(1+e^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*t}{12})+C})}$

The only thing left is finding out how many know the rumor after 4 days, but here I'm stuck again. A friend told me that the answer is A-1. And when plugging in for differerent A values, I see that this is true. But I'm struggling with proving this.

$ln(\frac{f(96)}{A-f(96)})=(\frac{ln(A-1)}{48A})*A(96+\frac{12}{\pi}*cos(\frac{\pi*96}{12})-48-\frac{12}{\pi})$

$ln(\frac{f(96)}{A-f(96)})=ln(A-1)$

But after this I'm stuck. I don't see how the answer is A-1 given the expression above. Any suggestions?

 

[Kolika28]

Ohhh, I got it now.

$ln(\frac{f(96)}{A-f(96)})=ln(A-1)$

$\frac{f(96)}{A-f(96)}=A-1$. Must remember that (A-1) is a number.

$f(96)=(A-f(96)*(A-1)=A*(A-1)-f(96)*(A-1)$

$f(96)(1+(A-1))=A^2-A$

$f(96)=\frac{A^2-A}{A}=A-1$

Thank you so much for everyones help. I really appreciate it!

 

[epenguin]

Looking at equation 2, I find out that $C=-48-\frac{12}{\pi}$. And plotting that value in equation 1, I get that $k=\frac{ln(A-1)}{48A}$. I am aware that this C is not equal to the C in this expression:

$f(t)=\frac{A*Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*t}{12})}}{(1+Ce^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*t}{12})})}$

This due to my bad notation errors. So this is true for the C value I found now:
$f(t)=\frac{A*e^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*t}{12})+C}}{(1+e^{kA(t+\frac{12}{\pi}*cos(\frac{\pi*t}{12})+C})}$

The only thing left is finding out how many know the rumor after 4 days, but here I'm stuck again. A friend told me that the answer is A-1. And when plugging in for differerent A values, I see that this is true. But I'm struggling with proving this.

$ln(\frac{f(96)}{A-f(96)})=(\frac{ln(A-1)}{48A})*A(96+\frac{12}{\pi}*cos(\frac{\pi*96}{12})-48-\frac{12}{\pi})$

$ln(\frac{f(96)}{A-f(96)})=ln(A-1)$

But after this I'm stuck. I don't see how the answer is A-1 given the expression above. Any suggestions?

These two forms are equivalent, taking C as 'arbitrary constant', not the same in the end for the two forms, so maybe it would be clearer to use a different symbol e.g. capital K in one case. I think the first form is more commonly used than the second but I get the same result as you for your C.

It is not at all clear why you suddenly introduce the four days figure, what quite is its importance?

I think your friend is right. I could not follow yet your calculation but more simply I'd say, if the number of students in the know at t = 0 is 1 and the mid-point is at 2 days, then it follows from the symmetry I mentioned in #11 that the number of students not in the know, (A - 1), at 4 days is also 1.* However this adds nothing to ability to obtain any of the constants, A, k. Once you know, or assume, A, it would be possible to find the other relevant constant from f(0) and f(48).

I repeat that you cannot obtain A from the information that you have given, and the best you can do without more information is what I have suggested in #11. (Perhaps you have not transcribed all the information you were given? The first sentence of #1 reads a bit strange. And does it not sound incredible to you that you be asked to work out the total number of students on a campus from information like what you have given?).

This is something you ought to know, for it sounds that this exercise is taken not from a pure maths course but from a course on modelling. For this activity one of the things you need to know is what is the information you need in order to solve the problem. (And actually this question not rarely arises in practice. People have superstitious beliefs about what mathematicians can achieve in the way of making a silk purse out of a sow's ear (English language proverbial expression).So they bring problems to mathematicians asking them to magically solve with inadequate information. Even without the mathematical capacities they ought still to be able to work out what information is necessary, but it does happen that they don't.)

* Always with our idealisation of f being a precisely defined continuous functio.