Problem 74 - Brief Applied Calculus

Question

BUSINESS: Consumer Credit From 2006 to 2010 , outstanding consumer credit was approximately \(C(x)=11.475\left(x^{2}-11.54 x+38.8\right)\left(x^{2}+0.79 x+5.56\right)\) billion dollars where \(x\) is the number of years after 2005 . Differentiating using the product rule, find \(\mathrm{C}^{\prime}(2)\) and \(\mathrm{C}^{\prime}(4)\) and interpret your answers.

Step-by-Step Solution

Answer

In 2007, consumer credit was decreasing at about $200.55 billion/year, and in 2009, at about $960.36 billion/year.

Step 1: Understand the Function

The function given is \(C(x) = 11.475(x^2 - 11.54x + 38.8)(x^2 + 0.79x + 5.56)\). This function represents the consumer credit in billions of dollars over time, where \(x\) is the number of years after 2005.

Step 2: Apply the Product Rule for Differentiation

To find \(C'(x)\), use the product rule of differentiation. If you have a product of two functions \(u(x)v(x)\), the derivative is given by \((uv)' = u'v + uv'\). Here, let \(u(x) = x^2 - 11.54x + 38.8\) and \(v(x) = x^2 + 0.79x + 5.56\).

Step 3: Differentiate Each Component

Differentiate \(u(x) = x^2 - 11.54x + 38.8\) to get \(u'(x) = 2x - 11.54\). Differentiate \(v(x) = x^2 + 0.79x + 5.56\) to get \(v'(x) = 2x + 0.79\).

Step 4: Combine Results using the Product Rule

Using the results from Step 3, the derivative of \(C(x)\) is \[C'(x) = 11.475 [(2x - 11.54)(x^2 + 0.79x + 5.56) + (x^2 - 11.54x + 38.8)(2x + 0.79)]\].

Step 5: Evaluate at \(x = 2\)

Substitute \(x = 2\) into the derivative function to find \(C'(2)\): 1. Calculate values: - \(u(2) = 2^2 - 11.54 \, (2) + 38.8 = 23.72\) - \(v(2) = 2^2 + 0.79 \, (2) + 5.56 = 12.94\) - \(u'(2) = 2(2) - 11.54 = -7.54\) - \(v'(2) = 2(2) + 0.79 = 4.79\)2. Compute: \[C'(2) = 11.475 [(-7.54)(12.94) + (23.72)(4.79)] \approx -200.55\].

Step 6: Evaluate at \(x = 4\)

Substitute \(x = 4\) into the derivative function to find \(C'(4)\): 1. Calculate values: - \(u(4) = 4^2 - 11.54 \, (4) + 38.8 = 0.04\) - \(v(4) = 4^2 + 0.79 \, (4) + 5.56 = 25.72\) - \(u'(4) = 2(4) - 11.54 = -3.54\) - \(v'(4) = 2(4) + 0.79 = 8.79\)2. Compute: \[C'(4) = 11.475 [(-3.54)(25.72) + (0.04)(8.79)] \approx -960.36\].

Step 7: Interpret the Results

\(C'(2)\approx -200.55\) indicates that in 2007, the consumer credit was decreasing at approximately 200.55 billion dollars per annum. Similarly, \(C'(4)\approx -960.36\) implies that in 2009, the consumer credit was decreasing at a faster rate of about 960.36 billion dollars per annum.

Key Concepts

Understanding Consumer Credit

Consumer credit refers to the amount of credit extended to individuals by financial institutions. This concept is crucial in understanding consumer spending and financial health. It includes credit cards, personal loans, and other forms of borrowings that consumers use to purchase goods and services.
For businesses and economists, tracking consumer credit trends helps in assessing economic conditions. If consumer credit is increasing, it might indicate higher consumer confidence and spending. Conversely, a decrease suggests reduced consumer spending or a more cautious approach to borrowing and lending.

• This exercise calculated consumer credit over several years to observe how much it changed annually.
• Such calculations are essential for predicting economic trends and making financial decisions, both for businesses and policymakers.

Applying the Product Rule in Calculus Differentiation

In calculus, the product rule is a formula used to find the derivative of a product of two functions. It's a foundational tool, particularly in applied calculus, which involves practical applications of mathematical concepts to solve real-world problems.
This rule states that if you have two functions, say \( u(x) \) and \( v(x) \), then their derivative \((uv)'\) is given by \( u'v + uv' \).

• In this exercise, the problem involved differentiating a consumer credit function represented as a product of two quadratic expressions.
• Applying the product rule allowed us to calculate how the consumer credit was changing over time.

By understanding how to apply the product rule, students can tackle complex functions that model real-world phenomena, grasping both the techniques and the implications of differentiation.

Exploring Applied Calculus

Applied calculus is a branch of mathematics that explores how calculus concepts can be used to solve practical problems in fields such as economics, engineering, and the natural sciences. The primary aim of applied calculus is not just to understand the "how" of calculus, but to see the "why" behind its application.
In this exercise, applied calculus was used to differentiate the consumer credit equation, illustrating:

• How a mathematical model describes financial trends over time.
• How differentiating the model offers insights into rate changes, which is critical for financial analysis.

By employing differentiation, specifically through the product rule, we gain significant insights into how consumer credit trends evolve.
This empowers economists and business analysts to make informed judgments based on these mathematical insights. When students see how calculus connects with daily applications, it enhances their understanding and appreciation for mathematics.