Problem 19 - Calculus and its applications

Question

Differentiate two ways: first, by using the Quotient Rule; then, by dividing the expressions before differentiating. Compare your results as a check. Use a graphing calculator to check your results. \(y=\frac{t^{2}-16}{t+4}\)

Step-by-Step Solution

Answer

The derivative using the Quotient Rule is \( y' = \frac{t^2 + 8t + 16}{(t+4)^2} \). Simplified, it's \( y = t - 4 \) with derivative 1. Both check out, except at \( t = -4 \).

Step 1: Apply the Quotient Rule

The Quotient Rule states that if you have a function in the form of \( y = \frac{u}{v} \), then its derivative is \( y' = \frac{u'v - uv'}{v^2} \). Here, \( u = t^{2}-16 \) and \( v = t+4 \). First, compute \( u' = \frac{d}{dt}(t^2-16) = 2t \) and \( v' = \frac{d}{dt}(t+4) = 1 \). Thus, applying the Quotient Rule, we have:\[ y' = \frac{(2t)(t+4) - (t^2-16)(1)}{(t+4)^2} = \frac{2t^2 + 8t - t^2 + 16}{(t+4)^2} = \frac{t^2 + 8t + 16}{(t+4)^2} \].

Step 2: Simplify the Expression

Now, we simplify \( y = \frac{t^{2}-16}{t+4} \) by dividing \( t^{2}-16 \) by \( t+4 \). Notice that \( t^2 - 16 \) can be factored as \((t-4)(t+4)\). So we divide:\[ \frac{(t-4)(t+4)}{t+4} = t-4 \] for \( t eq -4 \). Thus, the simplified expression is \( y = t - 4 \).

Step 3: Differentiate the Simplified Expression

Differentiate the simplified expression \( y = t - 4 \). Its derivative is straightforward:\[ y' = \frac{d}{dt}(t - 4) = 1 \].

Step 4: Compare Results

Compare the result obtained using the Quotient Rule \( y' = \frac{t^2 + 8t + 16}{(t+4)^2} \) with the derivative of the simplified expression \( y' = 1 \). Notice that when \( t+4 eq 0 \), the derivative of \( t-4 \) is indeed 1. The derivation simplifies for \( t eq -4 \), and at \( t = -4 \), the expression is not valid due to division by zero.

Step 5: Verify with Graphing Calculator

Use a graphing calculator to plot both the original function and its derivatives from both methods. Confirm the derivatives match the slope produced by the graph of the function \( y = \frac{t^{2}-16}{t+4} \) for values of \( t eq -4 \). The graph should reflect a slope of 1 for \( y = t-4 \) which matches our derivative of 1, except at the point \( t = -4 \) where the function is not defined.

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental concept in calculus used to find the derivative of a quotient of two functions. It comes in handy when you're dealing with expressions of the form \( y = \frac{u}{v} \), where \( u \) and \( v \) are differentiable functions of a variable, usually \( t \) or \( x \). To derive this expression, the rule tells us to use:

• Find the derivative of the numerator \( u \), which we'll call \( u' \).
• Find the derivative of the denominator \( v \), labeled \( v' \).
• Plug these into the formula: \( y' = \frac{u'v - uv'}{v^2} \).

In the initial exercise given, the function \( y = \frac{t^2-16}{t+4} \) is differentiated using this rule. The derivatives of \( t^2 -16 \) (which simplifies to \( 2t \)) and \( t+4 \) (which becomes 1) are calculated and inserted into the quotient formula. This solution helps reveal the logic and simplicity behind differentiating using the Quotient Rule.

Function Simplification

Function simplification streamlines complex expressions to make calculating derivatives easier. In this context, simplifying involves expressing a complicated function in a more tractable form. The key is to look for factors that can be canceled out or restructured.
For the expression \( y = \frac{t^2-16}{t+4} \), note that the numerator \( t^2-16 \) can be factored into \((t-4)(t+4)\). This factorization makes it clear that \( t+4 \) is a common factor in both the numerator and the denominator, allowing the simplification:

• Cancel out \( t+4 \) from both parts.
• You're left with \( y = t-4 \) for \( t eq -4 \).

This simplifies the process of finding the derivative since differentiating \( y = t-4 \) is much easier than the original quotient form. It illustrates how simplification can significantly reduce the complexity of calculus problems.

Mathematical Derivatives

Mathematical derivatives represent the rate of change of functions. Calculating derivatives is a fundamental skill in calculus, providing insights into how functions behave. They're crucial for analyzing motion, growth, or changes over time in various contexts.
For the exercise, after simplifying \( y = \frac{t^2-16}{t+4} \) to \( y = t-4 \), the derivative becomes straightforward:

• Simply calculate: \( \frac{d}{dt}(t - 4) \).
• The result is \( 1 \), which indicates a constant rate of change.

Derivatives like this show how a function behaves over its domain, minus any undefined points—here, \( t = -4 \), where division by zero occurs. This distinction highlights the importance of examining the function's domain when performing differentiation.

Graphing Calculator Analysis

A graphing calculator is an invaluable tool for visual learners and anyone checking their calculus work. It allows you to see how a function behaves visually and to verify your analytical results.
For this particular problem, plotting both the original function \( y = \frac{t^2-16}{t+4} \) and its simplified version \( y = t-4 \) can provide a visual affirmation of their derivatives.

• By plotting, you should observe a straight-line graph with a constant slope of 1 for \( y = t-4 \).
• Make sure to note that the graph is undefined at \( t = -4 \) due to division by zero in the original formula.

This analysis confirms the analytical result that the derivative outside the undefined point matches the slope you'll observe on the graph, consolidating both learning and confidence in the calculus process.