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 \(L\left( t \right).\) This function, for example, may describe the current labour productivity of an employee. Let \({L_{\max }}\) be the maximum available value of \(L\left( t \right).\) 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:

\[\frac{{dL}}{{dt}} = k\left( {{L_{\max }} – L} \right),\]

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{{dL}}{{dt}} = k\left( {{L_{\max }} – L} \right),\;\;}\Rightarrow

{\frac{{dL}}{{{L_{\max }} – L}} = kdt,\;\;}\Rightarrow

{\int {\frac{{dL}}{{{L_{\max }} – L}}} = \int {kdt} ,\;\;}\Rightarrow

{{- \int {\frac{{d\left( {{L_{\max }} – L} \right)}}{{{L_{\max }} – L}}} }={ \int {kdt} ,\;\;}}\Rightarrow

{{- \ln \left( {{L_{\max }} – L} \right) }={ kt + \ln C,\;\;}}\Rightarrow

{{\ln \left( {{L_{\max }} – L} \right) }={ – kt + \ln C,\;\;}}\Rightarrow

{{\ln \left( {{L_{\max }} – L} \right) }={ \ln {e^{ – kt}} }+{ \ln C.}}

\]

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

\[{L_{\max }} – L = C{e^{ – kt}}.\]

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

\[{L\left( t \right) = {L_{\max }} }-{ \left( {{L_{\max }} – M} \right){e^{ – kt}}.}\]

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 \({L_{\max }}.\)

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A pharmacist in a drugstore must check \(1,000\) receipts per day. A new pharmacist was hired in the drugstore. In \(1\text{st}\) week, the pharmacist was able to check \(100\) receipts per day. Estimate the number of receipts the pharmacist can check for the second week.### Example 2

Suppose that a news is spread by mass media according to the law described by the learning curve. What should be the initial percentage of the population who know about this news so that it could reach \(50\%\) of the population in \(1\) week and \(90\%\) of the population in \(4\) weeks?### Example 1.

A pharmacist in a drugstore must check \(1,000\) receipts per day. A new pharmacist was hired in the drugstore. In \(1\text{st}\) week, the pharmacist was able to check \(100\) receipts per day. Estimate the number of receipts the pharmacist can check for the second week.Solution.

By setting the initial skill level to zero: \(M = 0,\) we can describe the learning process by the following law:

\[L\left( t \right) = {L_{\max }}\left( {1 – {e^{ – kt}}} \right).\]

Determine the parameter \(k\) knowing the number of receipts completed in \(1\) week:

\[

{100 = 1000\left( {1 – {e^{ – kt}}} \right),\;\;}\Rightarrow

{1 – {e^{ – kt}} = 0.1,\;\;}\Rightarrow

{{e^{ – kt}} = 0.9,\;\;}\Rightarrow

{- kt = \ln 0.9,\;\;}\Rightarrow

{k = – \frac{{\ln 0.9}}{t}.}

\]

By substituting \(t = 1\,\text{week},\) we find:

\[k = – \frac{{\ln 0.9}}{1} \approx – 0.105\]

Now we can calculate productivity of the new pharmacist for the second week:

\[

{L\left( {t = 2} \right) }={ 1000\left( {1 – {e^{ – 0.105 \cdot 2}}} \right) }

= {1000\left( {1 – {e^{ – 0.21}}} \right) }

\approx {1000\left( {1 – 0.811} \right) }

= {189\left[ {\frac{\text{receipts}}{\text{day}}} \right].}

\]