As you know, when a course ends, students start to forget the material they have

Question

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. A. Let y=f(t) be the fraction of the original material remembered t weeks after the course has ended. Set up a differential equation for y, using k as any constant of proportionality you may need (let k>0). Your equation will contain two constants; the constant a (also positive) is less than y for all t. dy/dt= What is the initial condition for your equation? y(0)= B. Solve the differential equation. y= C. What are the practical meaning (in terms of the amount remembered) of the constants in the solution y=f(t)? If after one week the student remembers 85 percent of the material learned in the semester, and after two weeks remembers 75 percent, how much will she or he remember after summer vacation (about 14 weeks)? percent =

Explanation / Answer

I think that y(0) = 1 is correct and refers to the total amount that the student has learnt at the end of the course, which may not be 100% of the course material, but that does not matter. Your final equation came from ln(y - a) = -kt + ln(1 - a) and it is better to work with this. Put in the two given facts, t = 1, y = 0.75 and t = 2, y = 0.66 to get ln(0.75 - a) = -k + ln(1 - a) ln(0.66 - a) = -2k + ln(1 - a) Subtracting the second from the first gives ln(0.75 - a) - ln(0.66 - a) = k Substitute this inot the first equation above to get ln(0.75 - a) = - [ ln(0.75 - a) - ln(0.66 - a)] + ln(1 - a) This simplifies to 2*ln(0.75 - a) = ln(0.66 - a) + ln(1 - a) This is now an equation in one variable only which using the rules of logs should be fairly easy to solve. Can you take it from there? You should find a is a little over 0.6 and k is a little over 1. By the way, your final equation y = a + e^(-kt + ln(abs(1 - a))) can be simplified to y = a + (1 - a)*e^-kt

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