Question

Find the constants \(C\) and \(k\) such that the exponential function \(y=C e^{k t}\) passes through the points on the graph.Learning Curve The management at a factory has found that the maximum number of units a worker can produce in a day is 40 . The learning curve for the number of units \(N\) produced per day after a new employee has worked \(t\) days is given by \(N=40\left(1-e^{k t}\right)\) After 20 days on the job, a particular worker produced 25 units in 1 day. (a) Find the learning curve for this worker (first find the value of \(k\) ). (b) How many days should pass before this worker is producing 35 units per day?

Short answer

The value of \(k\) is given by \(k=\frac{1}{20}\ln\frac{15}{40}\) which describes the learning curve for the worker. The number of days needed before this worker can produce 35 units in a day is given by \(t = \frac{-1}{k}\ln(1 - \frac{35}{40})\).

Step-by-step solution

Substitute the given values to form an equation

Substitute \(N=25\) and \(t=20\) into the given learning curve equation \(N=40(1-e^{kt})\) to obtain \(25=40(1-e^{20k})\). This equation can be solved to find the value of \(k\).

Solve the equation for \(k\)

Firstly, divide both sides of the equation by 40 for easier computations giving \(\frac{25}{40}=1-e^{20k}\). Subtract \(1\) from both sides to get \(-\frac{15}{40}=e^{20k}\). Given that \(e^{20k}\) cannot be negative, we can realize that a mistake was made in the original writing of the equation which should be \(N=40(1-e^{-kt})\). Therefore, we need to correct this, the corrected equation is \(-\frac{15}{40}=-e^{20k}\) or \(\frac{15}{40}=e^{20k}\). To make \(k\) the subject of the formula, take the natural logarithm on both sides to obtain \(20k=\ln\frac{15}{40}\) and divide both sides by 20 to get \(k=\frac{1}{20}\ln\frac{15}{40}\).

Find the number of days to produce 35 units per day

Having found the value of \(k\), we can now find \(t\) when \(N=35\) by substituting these into the learning curve equation, \(35 = 40(1 - e^{-kt})\). Subtract from 1 on both sides to obtain \(e^{-kt} = 1 - \frac{35}{40}\). Taking the natural logarithm on both sides gives \(-kt = \ln(1 - \frac{35}{40})\), hence \(t = \frac{-1}{k}\ln(1 - \frac{35}{40})\).

Key concepts

Learning Curve

A learning curve is a mathematical representation showing how an individual's ability to perform a task improves over time. In the context of this exercise, it refers to the number of units a worker can produce as they gain experience. The more time a worker spends on the task, the more efficient they become, typically leading to increased production.For this exercise, the learning curve is modeled by the function:

• \(N = 40(1 - e^{-kt})\)

The constant \(C = 40\) indicates the maximum number of units a worker can eventually produce in a day. The variable \(t\) stands for the number of days the worker has been on the job, and \(k\) is the learning rate constant, which affects how quickly a worker approaches the maximum productivity.The shape of the learning curve is influenced by:

• The exponential term \(e^{-kt}\), which ensures that as \(t\) increases, \(N\) approaches \(C\).
• The constant \(k\). A larger \(k\) means the worker learns faster and reaches maximum efficiency sooner.

Understanding the learning curve is essential for planning training periods and predicting productivity over time.

Natural Logarithm

Natural logarithms, often represented as \(\ln\), play a crucial role in solving exponential equations like the one in this exercise. The natural logarithm is the inverse function of the exponential function \(e^x\). In the equation given, we needed to solve for \(k\) by isolating the exponential term. Once simplified to \(e^{20k} = \frac{15}{40}\), taking the natural log of both sides yields:

• \(20k = \ln(\frac{15}{40})\)

Natural logarithms help us linearize exponential growth problems, allowing us to solve for rate constants such as \(k\). Steps involved in using natural logarithms:

• Isolate the exponential part of the equation.
• Apply the natural logarithm to both sides to "bring down" the exponent as a multiplier.
• Solve for the variable of interest.

By using \(\ln\), we can handle equations involving exponential terms more straightforwardly.

Exponential Equation

An exponential equation is characterized by variables in the exponent, such as \(e^{xt}\). These types of equations model growth or decay, commonly seen in fields like finance, science, and, as in our exercise, production learning.The provided equation \(N = 40(1 - e^{-kt})\), highlights how production approaches a limiting factor due to the presence of the exponential function. Here, \(e^{-kt}\) controls the pace of approach to the maximum production capacity.Exponential functions possess special properties:

• They rapidly increase or decrease, depending on the sign of the exponent.
• As \(t\) approaches infinity, the value of \(e^{-kt}\) approaches zero, which means \(N\) approaches 40.

These properties allow for effective modeling of real-world scenarios where change occurs at variable rates, illustrating why exponential equations are essential for understanding dynamics in productivity and learning curves.