# Thread: What are the practical meaning of the constants in the solution y=f(t)?

1. ## What are the practical meaning of the constants in the solution y=f(t)?

As you know, when a course ends, students start to forget the material they have learned. One model (called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant, . A. Let be the fraction of the original material remembered weeks after the course has ended. Set up a differential equation for , using as any constant of proportionality you may need (let ). Your equation will contain two constants; the constant (also positive) is less than for all .

What is the initial condition for your equation?

B. Solve the differential equation.

C. What are the practical meaning (in terms of the amount remembered) of the constants in the solution ? If after one week the student remembers 80 percent of the material learned in the semester, and after two weeks remembers 69 percent, how much will she or he remember after summer vacation (about 14 weeks)?

percent = _______________

^ I solved everything else just need to know C) Thank you!!

2. Originally Posted by ewkimchi
As you know, when a course ends, students start to forget the material they have learned. One model (called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant, . A. Let be the fraction of the original material remembered weeks after the course has ended. Set up a differential equation for , using as any constant of proportionality you may need (let ). Your equation will contain two constants; the constant (also positive) is less than for all .

What is the initial condition for your equation?

B. Solve the differential equation.

C. What are the practical meaning (in terms of the amount remembered) of the constants in the solution ? If after one week the student remembers 80 percent of the material learned in the semester, and after two weeks remembers 69 percent, how much will she or he remember after summer vacation (about 14 weeks)?

percent = _______________

^ I solved everything else just need to know C) Thank you!!
We can't say anything about the "practical meaning of the constants" in your solution without knowing what that solution is!

3. Originally Posted by ewkimchi
As you know, when a course ends, students start to forget the material they have learned. One model (called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant, . A. Let be the fraction of the original material remembered weeks after the course has ended. Set up a differential equation for , using as any constant of proportionality you may need (let ). Your equation will contain two constants; the constant (also positive) is less than for all .

What is the initial condition for your equation?

B. Solve the differential equation.

C. What are the practical meaning (in terms of the amount remembered) of the constants in the solution ? If after one week the student remembers 80 percent of the material learned in the semester, and after two weeks remembers 69 percent, how much will she or he remember after summer vacation (about 14 weeks)?

percent = _______________

^ I solved everything else just need to know C) Thank you!!
Let me assume that your solution looks something like this: $f(t)=a+(y_0-a)\cdot\mathrm{e}^{-kt}$. If so, one can interpret a as the amount of material that has been stably learned, i.e. that will never decay, since $\lim_{t\to\infty}f(t)=a$; and we can interpet the constant of decay k as indicating the speed with which that part of the material that will eventually be forgotten, actually decays in memory.

As to the practical application: for simplicity, let's assume that $y_0=1$, this simplifies f(t) to $f(t)=a+(1-a)\cdot\mathrm{e}^{-kt}$, and let's further assume that t equals the time in weeks.
Thus we are given that $f(1)=0.8$ and that $f(2)=0.69$ and are asked to deduce the value of $f(14)$.
The point is that a and k can be deduced from the givens; for example like this: from
$f(1)=a+(1-a)\cdot\mathrm{e}^{-k\cdot 1}=0.8$ and $f(2)=a+(1-a)\cdot\mathrm{e}^{-k\cdot 2}=0.69$
you can infer, that
$\mathrm{e}^{-k}=\frac{0.8-a}{1-a}$ and $\left(\mathrm{e}^{-k}\right)^2=\frac{0.69-a}{1-a}$
Combining these two equations gives you that
$\left(\frac{0.8-a}{1-a}\right)^2=\frac{0.69-a}{1-a}$
must hold. Now solve for a. After that, the value of k follows immediately
$k=-\ln\frac{0.8-a}{1-a}=\ln\frac{1-a}{0.8-a}$
We now have fully determined f(t) and can thus easily calculate the value of f(t) for any t, including, of course, for t=14.