1. ## Learning Curve

Hello Forumites,
The management at a certain factory has found that the maximum number of units a worker can produce in a day is 30. The rate of increase in the number of units N produced w.r.t. time t in days by a new employee is proportional to 30-N

A) I want to determine the differential equation describing the rate of change of performance w.r.t. time.
B)I want to solve the differential equation from part A)
C) I want to find the particular solution for a new employee who produced 10 units on the first day at the factory and 19 units on the twentieth day.

Solution.
I am browsing on internet for how to answer learning curve related questions. But meanwhille, if any forumite can answer these questions.

2. ## Re: Learning Curve

With N(t) as the number produced per day after t days, the "rate of change of performance w.r.t. time." is dN/dt. We are told that is "proportional to 30-N". dN/dt= k(30- N) where k is the "constant of proportion".

Write that equation as dN/(30- N)= kdt and integrate. There will be one constant of integration. Use the fact that there is a "new employee who produced 10 units on the first day at the factory and 19 units on the twentieth day" to determine both that constant and k.

3. ## Re: Learning Curve

Originally Posted by HallsofIvy
With N(t) as the number produced per day after t days, the "rate of change of performance w.r.t. time." is dN/dt. We are told that is "proportional to 30-N". dN/dt= k(30- N) where k is the "constant of proportion".

Write that equation as dN/(30- N)= kdt and integrate. There will be one constant of integration. Use the fact that there is a "new employee who produced 10 units on the first day at the factory and 19 units on the twentieth day" to determine both that constant and k.
Hello,
My answer to A) is $N'(t)=30k-Nk.$
Solving thid ODE, i got answer $n(t)=C_1e^{-kt}+30.$

For answering part C), i put in the above equation, Values for N and t as given in part C) now the equation became $-20=C_1e^{-k}...1$ and $-11=C_1e^{-k20}...2$
Now, Dividing the 2) equation by 1) equation we get $0.55=e^{-19k}\Rightarrow k=0.0314651053, C=-20.6392453791$

4. ## Re: Learning Curve

I agree the ODE may be written as:

$\displaystyle \frac{dN}{dt}=k(30-N)$

While this is separable, let's write it in standard linear form:

$\displaystyle \frac{dN}{dt}+kN=30k$

The integrating factor is thus:

$\displaystyle \mu(t)=\exp\left(k\int\,dt\right)=e^{kt}$

And so the ODE becomes:

$\displaystyle e^{kt}\frac{dN}{dt}+ke^{kt}N=30ke^{kt}$

$\displaystyle \frac{d}{dt}\left(e^{kt}N\right)=30ke^{kt}$

Integrate:

$\displaystyle e^{kt}N=30e^{kt}+c_1$

Hence:

$\displaystyle N(t)=30+c_1e^{-kt}$

Now, you have provided 2 data points:

$\displaystyle (1,10),\,(20,19)$

And this produces the system:

$\displaystyle 30+c_1e^{-k}=10$

$\displaystyle 30+c_1e^{-20k}=19$

Or:

$\displaystyle c_1e^{-k}=-20$

$\displaystyle c_1e^{-20k}=-11$

Solving this system, there results:

$\displaystyle \left(c_1,k\right)=\left(-20\left(\frac{20}{11}\right)^{\Large\frac{1}{19}}, \frac{1}{19}\ln\left(\frac{20}{11}\right)\right)$

And so the solution satisfying the IVP is:

$\displaystyle N(t)=30-20\left(\frac{11}{20}\right)^{ \Large\frac{t-1}{19}}$