Problem 53 - Calculus - AP Edition

Question

Use the differential \(d y\) to approximate \(\Delta y\) when \(x\) changes as indicated. $$y=\frac{x}{x^{2}+1} ; \text { from } x=2 \text { to } x=1.96$$

Step-by-Step Solution

Answer

\( \Delta y \approx 0.0048 \)

Step 1: Differentiate the function

The function given is \( y = \frac{x}{x^2 + 1} \). To find the differential \( dy \), we first need to find the derivative \( \frac{dy}{dx} \). We use the quotient rule for differentiation: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = x \) and \( v = x^2 + 1 \).

Step 2: Apply the quotient rule

Differentiate \( u \) and \( v \):- \( u' = 1 \)- \( v' = 2x \)Using the quotient rule:\[ \frac{dy}{dx} = \frac{1 \cdot (x^2 + 1) - x \cdot 2x}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} \]This simplifies to:\[ \frac{dy}{dx} = \frac{1 - x^2}{(x^2 + 1)^2} \]

Step 3: Calculate the differential \(dy\)

We use the derivative to find the differential \( dy \):\[ dy = \left( \frac{1 - x^2}{(x^2 + 1)^2} \right) dx \]The change in \( x \), noted as \( dx \), is equal to \(-0.04\) (from \(x = 2\) to \(x = 1.96\)).Substitute \( x = 2 \) into the derivative:\[ dy = \left( \frac{1 - 2^2}{(2^2 + 1)^2} \right) (-0.04) = \left( \frac{1 - 4}{25} \right)(-0.04) = \left( -\frac{3}{25} \right)(-0.04) \]

Step 4: Simplify and find \( dy \)

Calculate \( dy \):\[ dy = \frac{3}{25} \times 0.04 = \frac{3}{25} \times \frac{4}{100} = \frac{12}{2500} = \frac{6}{1250} \]Convert to decimal form:\[ dy = 0.0048 \]

Step 5: Approximate \( \Delta y \) using \( dy \)

With the calculation complete, we can approximate the change in \( y \), noted as \( \Delta y \), by using \( dy \). Therefore, \( \Delta y \approx dy = 0.0048 \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. When you differentiate a function, you essentially calculate its derivative. This derivative can then be used to find the differential, which is a small change in the function's value.Understanding differentiation involves:

• Recognizing a function and its variables
• Knowing the rules of differentiation, such as the power rule, product rule, and quotient rule

In our context, we are using the function \( y = \frac{x}{x^2 + 1} \). To differentiate this, we used the quotient rule. Differentiating allows us to find \( \frac{dy}{dx} \), which represents how \( y \) changes with respect to \( x \). Knowing \( \frac{dy}{dx} \) helps us determine \( dy \), the differential, which approximates small changes in \( y \) as \( x \) changes. When dealing with small changes, differentiation is crucial for both exact and approximate calculations.

Quotient Rule

The quotient rule is a technique in calculus used to differentiate functions that are expressed as a quotient of two functions. When faced with a fraction like \( y = \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), we apply the quotient rule to find the derivative.
The formula for the quotient rule is:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]Here, \( u' \) and \( v' \) represent the derivatives of \( u \) and \( v \), respectively. This rule helps manage the complexity of differentiating a fraction, breaking down the process into differentiating its individual parts.
In our example involving the function \( y = \frac{x}{x^2 + 1} \):

• Identify \( u = x \) and \( v = x^2 + 1 \)
• Differentiate both: \( u' = 1 \) and \( v' = 2x \)
• Plug these into the formula: \( \frac{dy}{dx} = \frac{1 \cdot (x^2 + 1) - x \cdot 2x}{(x^2 + 1)^2} \)

After simplifying, we find the derivative, which is a key step to calculating changes in the original function. The quotient rule is essential for handling rational functions and making accurate calculations of their rates of change.

Approximation Methods

Approximation methods in calculus help us estimate the changes in function values when exact calculations are either too complex or unnecessary. One common method is using differentials, which involves the derivative of a function to approximate small changes.
This process is useful in scenarios where the changes in the input are small, such as moving from \( x = 2 \) to \( x = 1.96 \), as seen in our example. By calculating the differential \( dy \), we approximate the actual change in the function, denoted as \( \Delta y \).
To approximate \( \Delta y \):

• Compute the derivative using differentiation methods (e.g., quotient rule)
• Obtain \( dy = \frac{dy}{dx} \cdot dx \), where \( dx \) is the small change in \( x \)
• Substitute values into \( dy \) to estimate \( \Delta y \)

In this specific problem, with \( dx = -0.04 \), we calculated:\[ dy = \left( -\frac{3}{25} \right) \times (-0.04) = 0.0048 \]Thus, \( \Delta y \approx dy \), giving us a practical means to estimate changes without complicated measurements. Such approximation methods are valuable in fields ranging from engineering to economics, wherever predictions based on small changes are required.