Learning Curve

Home → Learning Curve

Mastering a new area or a skill always takes some time. In this section we try to model the learning process using a differential equation.

First of all, we introduce a measurable learning function $L (t) .$ This function, for example, may describe the current labour productivity of an employee. Let $L_{max}$ be the maximum available value of $L (t) .$ In many cases the following thumb rule is valid: the learning speed is proportional to the volume of remaining (unlearned) material. Mathematically, this can be represented by the equation:

\frac{d L}{d t} = k (L_{max} - L),

where $k$ is a coefficient of proportionality. The given differential equation is a separable equation, so it can be easily solved in general form:

\frac{d L}{d t} = k (L_{max} - L), \Rightarrow \frac{d L}{L_{max} - L} = k d t, \Rightarrow \int \frac{d L}{L_{max} - L} = \int k d t, \Rightarrow - \int \frac{d (L_{max} - L)}{L_{max} - L} = \int k d t, \Rightarrow - \ln (L_{max} - L) = k t + \ln C, \Rightarrow \ln (L_{max} - L) = - k t + \ln C, \Rightarrow \ln (L_{max} - L) = \ln e^{- k t} + \ln C .

After eliminating the logarithms, we obtain the general solution in the form:

L_{max} - L = C e^{- k t} .

The constant $C$ can be found from the initial condition: $L (t = 0) = M .$ Hence, $C = L_{max} - M .$ As a result, the learning curve is described by the formula

L (t) = L_{max} - (L_{max} - M) e^{- k t} .

The parameter $M$ is the last expression means the initial level of knowledge or skills. In the simplest case, we can suppose that $M = 0.$ The other parameter $k$ controls how fast the curve rises. View of the learning curves at different values of $M$ and $k$ is shown in Figure $1$ and $2,$ respectively.

As it can be seen, the learning level $L$ in all cases increases in the beginning of the process, and then the learning rate slows down as the level $L$ approaches the maximum value $L_{max} .$