Question

Learning Curve The management at a certain factory has found that the maximum number of units a worker can produce in a day is \(75 .\) The rate of increase in the number of units \(N\) produced with respect to time \(t\) in days by a new employee is proportional to \(75-N .\) (a) Determine the differential equation describing the rate of change of performance with respect to time. (b) Solve the differential equation from part (a). (c) Find the particular solution for a new employee who produced 20 units on the first day at the factory and 35 units on the twentieth day.

Short answer

The differential equation describing the rate of change of performance with respect to time is given by \(\frac{dN}{dt} = k(75 - N)\). The general solution of this differential equation is \(N(t) = 75 - Ce^{-kt}\). The specific solution for the worker who produced 20 units on the first day and 35 units on the twentieth day can be found by solving the simultaneous equations: \(20 = 75 - Ce^{-k}\) and \(35 = 75 - Ce^{-20k}\) for \(C\) and \(k\).

Step-by-step solution

Formulate the Differential Equation

The problem states that the rate of increase of produced units \(N\) with respect to time \(t\) is proportional to \((75 - N)\). This can be represented by the differential equation: \(\frac{dN}{dt} = k(75 - N)\), where \(k\) is the constant of proportionality.

Solve the Differential Equation

This is a first order linear differential equation and can be solved using variables separable method. Rearrange the terms to get: \(\frac{dN}{75 - N} = kdt\). Integrate both sides, \(-ln|75 - N| = kt + C\), where \(C\) is the constant of integration. To remove the modulus sign, divide both sides by -1 and exponentiate to get the solution: \(75 - N = Ce^{-kt}\), rearrange the equation to obtain \(N(t) = 75 - Ce^{-kt}\).

Calculate the Specific Solution

Now, find the value of \(C\) and \(k\) for the specific instance where the worker produced 20 units on the first day and 35 units on the twentieth day. Put these values into the equation \(N(t) = 75 - Ce^{-kt}\) to get two equations: \(1st equation: N(1) = 20 = 75 - Ce^{-k}\) and \(2nd equation: N(20) = 35 = 75 - Ce^{-20k}\). Solve these two equations to find \(C\) and \(k\).