Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent line to \(y=1 /\left(x^{2}+4\right)\) at the point \((1,1 / 5)\).

Step-by-Step Solution

Answer

The tangent line is \(y = -\frac{2}{25}x + \frac{7}{25}\).

Step 1: Differentiate the function

We need the derivative of the given function to find the slope of the tangent line at the point \((1, 1/5)\). The function is \(y = \frac{1}{x^2 + 4}\). We can use the quotient rule for differentiation which states that if \(y = \frac{u}{v}\), then \(y' = \frac{v\cdot u' - u\cdot v'}{v^2}\). In this case, \(u = 1\) and \(v = x^2 + 4\). Thus, \(u' = 0\) and \(v' = 2x\). Now, we differentiate: \[ y' = \frac{(x^2 + 4)(0) - 1(2x)}{(x^2 + 4)^2} = \frac{-2x}{(x^2 + 4)^2} \]

Step 2: Evaluate the derivative at the point

Now that we have the derivative \(y' = \frac{-2x}{(x^2 + 4)^2}\), we need to find the slope of the tangent line at \(x = 1\). Substitute \(x = 1\) into \(y'\):\[ y'(1) = \frac{-2(1)}{(1^2 + 4)^2} = \frac{-2}{25} \]So, the slope \(m\) of the tangent line at the point \((1, 1/5)\) is \(-\frac{2}{25}\).

Step 3: Use the point-slope form of a line

The point-slope form of the equation of a line is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope. For our problem, \((x_1, y_1) = (1, 1/5)\) and \(m = -\frac{2}{25}\). Plug these into the point-slope formula:\[ y - \frac{1}{5} = -\frac{2}{25}(x - 1) \]Simplify the equation to find the tangent line.

Step 4: Simplify the tangent line equation

Distribute \(-\frac{2}{25}\):\[ y - \frac{1}{5} = -\frac{2}{25}x + \frac{2}{25} \]Add \(\frac{1}{5}\) to each side to solve for \(y\):\[ y = -\frac{2}{25}x + \frac{2}{25} + \frac{1}{5} \]Convert \(\frac{1}{5}\) to \(\frac{5}{25}\) so the fractions have a common denominator:\[ y = -\frac{2}{25}x + \frac{2}{25} + \frac{5}{25} = -\frac{2}{25}x + \frac{7}{25} \]Thus, the equation of the tangent line is \(y = -\frac{2}{25}x + \frac{7}{25}\).

Key Concepts

Quotient Rule

The quotient rule is a fundamental tool in calculus used to differentiate functions that are ratios of two other functions. When faced with a function like \( y = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions of \( x \), the quotient rule provides an efficient way to find the derivative. The formula is given by:

• \( y' = \frac{v \cdot u' - u \cdot v'}{v^2} \)

In our example, we have \( y = \frac{1}{x^2+4} \), where \( u = 1 \) and \( v = x^2 + 4 \). The derivative of \( u \), \( u' \), is 0 because \( u \) is a constant. For \( v \), the derivative \( v' = 2x \). Applying the quotient rule:

• \( y' = \frac{(x^2 + 4) \cdot 0 - 1 \cdot 2x}{(x^2 + 4)^2} = \frac{-2x}{(x^2 + 4)^2} \)

This tells us how the function changes at any point along the curve, providing crucial information for finding the slope of the tangent line.

Derivative Evaluation

After finding the derivative using the quotient rule, the next step is to evaluate it at a specific point. This process is called derivative evaluation and gives us the slope of the tangent line at that particular point.
In this exercise, we evaluate the derivative \( y' = \frac{-2x}{(x^2 + 4)^2} \) at \( x = 1 \). By substituting \( x = 1 \) into the derivative, we get:

• \( y'(1) = \frac{-2 \cdot 1}{(1^2 + 4)^2} = \frac{-2}{25} \)

This value, \( -\frac{2}{25} \), represents the slope of the tangent line at the point \( (1, \frac{1}{5}) \). Derivative evaluation is essential as it transforms the derivative function into a numerical value, making it possible to write the equation of the tangent line.

Point-Slope Form

The point-slope form is a linear equation format used to determine the equation of a line given a point and the slope. It is usually expressed as:

• \( y - y_1 = m(x - x_1) \)

Here, \( (x_1, y_1) \) is the known point on the line, and \( m \) is the slope obtained from the derivative evaluation.
For our problem, we have the slope \( m = -\frac{2}{25} \) and the point \( (1, \frac{1}{5}) \). Inserting these into the point-slope form gives:

• \( y - \frac{1}{5} = -\frac{2}{25}(x - 1) \)

This equation enables us to express the linear function representing the tangent line at a specific point. It is a straightforward method of translating calculus’ derivative information into familiar algebraic terms.

Function Differentiation

Function differentiation is the process of finding the derivative of a function, which is a fundamental aspect of calculus. The derivative of a function gives us the rate at which the function's value is changing at any given point.
In this example, the function \( y = \frac{1}{x^2 + 4} \) required the application of the quotient rule to differentiate it, showcasing how rules of differentiation can simplify finding the derivative in more complex functions involving division.
Differentiation transforms the function from a simple algebraic expression to a tool for analyzing behavior at points, understanding function behavior, and, as seen here, to find the equation of tangent lines. By differentiating, we grasp more about the steepness and orientation of the function graph, setting the foundation for further exploration in calculus of motion, optimization, and modeling.