Question

The percent \(P\) of defective parts produced by a new employee \(t\) days after the employee starts work can be modeled by \(P=\frac{t+1750}{50(t+2)}\) Find the rates of change of \(P\) when (a) \(t=1\) and (b) \(t=10\).

Short answer

The rate of change of \(P\) when \(t=1\) is approximately -38.84 and when \(t=10\) is approximately -3.35.

Step-by-step solution

Write down the function

The function given in the exercise is \(P=\frac{t+1750}{50(t+2)}\). We are interested in finding its rate of change, which means we need to find its derivative with respect to \(t\).

Find the derivative

To find the derivative we can use the quotient rule which states that the derivative of \(\frac{f(t)}{g(t)}\) is \(\frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2}\). Applying this to our function \(P\), we get \(P' = \frac{(1*50(t+2)) - ((t+1750)*50)}{(50(t+2))^2}\), which simplifies to \( P' = \frac{50t+100 - 50t - 87500}{2500(t^2+4t+4)}\),which further simplifies to \(P' = \frac{-87400}{2500(t^2+4t+4)}\).

Evaluate the derivative at \(t=1\)

Substituting \(t=1\) into the derivative gives \(P'(1) = \frac{-87400}{2500(1^2+4*1+4)}\), which simplifies to \(P'(1) = \frac{-87400}{2500*9}\), which reduces to \(P'(1) = -38.84\).

Evaluate the derivative at \(t=10\)

Substituting \(t=10\) into the derivative gives \(P'(10) = \frac{-87400}{2500(10^2+4*10+4)}\), which simplifies to \(P'(10) = \frac{-87400}{2500*104}\), which reduces to \(P'(10) = -3.35\) .

Key concepts

Derivative

The derivative is a fundamental concept in calculus, used to understand how a function changes. At its core, the derivative measures how a function's output value changes as its input value changes. For example, if you have a function that represents the position of an object over time, the derivative of this function can tell you the velocity, or how fast the position is changing every second.

Mathematically, the derivative of a function is denoted as \( f'(x) \) or \( \frac{df}{dx} \). It represents the slope of the function at any given point \( x \). A positive derivative indicates that the function is increasing, while a negative derivative shows that the function is decreasing. The larger the absolute value of the derivative, the steeper the slope and the faster the rate of change.

In our exercise, the goal is to find how the percentage of defective parts changes with time \( t \). This is done by calculating the derivative of the function \( P = \frac{t+1750}{50(t+2)} \) with respect to \( t \), thereby giving us a clearer picture of how the defects percentage changes as the number of days increases.

Rate of Change

The rate of change is a crucial concept that is closely tied to derivatives. It describes how one quantity changes in relation to another. In real-world scenarios, it helps answer questions like "How fast is the car driving?" or "At what rate is the population of a city increasing?"

In mathematical terms, it is often expressed as the derivative of a particular function. A constant derivative means the change is occurring steadily over time. A varying derivative indicates the change is more complex, such as speeding up or slowing down.

For the exercise in question, the rate of change refers to how quickly the percentage of defective parts changes as each day passes. By evaluating the derivative of the given function at specific days \( t = 1 \) and \( t = 10 \), we were able to find the rates of change \( -38.84 \) and \( -3.35 \) respectively. These values suggest that initially, the defects decrease rapidly, but the rate of improvement slows down over time.

Quotient Rule

The quotient rule is a technique used in calculus to find the derivative of a quotient of two functions. If you have a function expressed as a fraction, like \( \frac{f(t)}{g(t)} \), the quotient rule helps determine how this fraction changes as \( t \) changes.

The rule states:

• The derivative of \( \frac{f(t)}{g(t)} \) is \( \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2} \).

In essence, you differentiate the numerator \( f(t) \) and the denominator \( g(t) \) separately, apply the formula, and then simplify. This can be particularly useful when the function is complex, making other methods tedious.

In our case, the function \( P = \frac{t+1750}{50(t+2)} \) is a quotient, so we applied the quotient rule to find its derivative. The applying of the quotient rule gave us the derivative \( P' = \frac{-87400}{2500(t^2+4t+4)} \), which then allowed us to evaluate the rate of change at specific values of \( t \). Understanding this rule makes handling fractions in calculus much easier, adhering to a structured approach to differentiation.