Problem 59 - Precalculus with Limits

Question

The predicted cost \(C\) (in thousands of dollars) for a company to remove \(p \%\) of a chemical from its waste water is given by the model $$C=\frac{120 p}{10,000-p^{2}}, \quad 0 \leq p<100$$ Write the partial fraction decomposition for the rational function. Verify your result by using a graphing utility to create a table comparing the original function with the partial fractions.

Step-by-Step Solution

Answer

The partial fraction decomposition of the function \(C=\frac{120 p}{10,000-p^{2}}\) is \(C = \frac{0.6}{100+p} + \frac{0.6}{100-p}\).

Step 1: Recognizing the Type of Rational Function

Rational functions are classified into two: proper and improper. Proper rational functions are those where the degree of the numerator is less than or equal to the degree of the denominator. In this case, \(\frac{120 p}{10,000 - p^2}\) is a proper rational function since the degree of the numerator \(120p\) which is 1 is less than the degree of the denominator \(10,000 - p^2\) which is 2.

Step 2: Factoring the Denominator

Factor the denominator \(10,000 - p^2\). This simplifies the differential equation to \(C = \frac{120 p}{(100+p) (100-p)}\) since this difference of squares factors into the expression indicated.

Step 3: Decomposing the Fraction

A proper rational function can be represented as a sum of simpler fractions using a method called partial fractions. This process is called decomposing the fraction. For the given function, it can be decomposed as: \(\frac{120 p}{(100+p) (100-p)} = \frac{A}{(100+p)} + \frac{B}{(100-p)}\).

Step 4: Finding A and B

To find the values of A and B, different values of \(p\) that make A vanish and B vanish should be substituted in separately to solve for each variable. When \(p = 100\), the equation simplifies to \(A = 0.6\). Similarly, when \(p = -100\), the equation simplifies to \(B = 0.6\). Thus the original rational function can be represented in a decomposed form as \(C = \frac{0.6}{100+p} + \frac{0.6}{100-p}\).

Key Concepts

Proper Rational Function

When we work with rational functions, it's crucial to understand whether we're dealing with a proper or improper rational function. A proper rational function is one where the degree of the numerator is less than the degree of the denominator. Consider the example of the cost to remove a chemical from wastewater, which is represented by the function
\[ C = \frac{120 p}{10,000 - p^2}, \quad 0 \leq p < 100 \]
In this model, the numerator is 120p, a linear term (degree 1), and the denominator is a quadratic term (degree 2). Since the numerator's degree is less than the denominator's degree, this is a proper rational function. Understanding this concept helps us predict that we can perform partial fraction decomposition on this function. Simplifying complex functions into easier parts is akin to breaking a meal into bite-sized pieces — it makes digestion (or in this case, calculation) much easier!

Factorizing Polynomials

Factorizing polynomials is like uncovering the building blocks of an algebraic expression. For our wastewater treatment cost model,
\[ C = \frac{120 p}{10,000 - p^2} \]
, the denominator \(10,000 - p^2\) appears complex at first glance. However, by recognizing it as a difference of squares, we can factor it into \((100 + p)(100 - p)\). This step transforms our unwieldy denominator into two linear factors, making it far easier to work with.

In fact, factorizing is so powerful that it's the linchpin in the process of partial fraction decomposition. It's also a practical skill beyond academia—it's akin to breaking down a problem into smaller, more manageable parts, which is a useful strategy in almost any context!

Graphing Utility

A graphing utility can serve as a powerful ally in verifying algebraic manipulations. After finding the partial fractions form of our function, which is
\[ C = \frac{0.6}{100+p} + \frac{0.6}{100-p} \],
we can use a graphing utility to plot both the original function and the decomposed one. These utilities often allow us to create tables of values for each function. By comparing these tables, we can observe that the y-values for the graphs are the same for corresponding x-values, which verifies our algebraic decomposition.

Utilizing such technological aids not only confirms our understanding but also provides a visual representation, which can enhance comprehension. Remember, in mathematics and in life, always seek to verify your work — after all, even the best of us can make a calculation slip now and then!