Problem 23 - Calculus Early Transcendentals

Question

(a) If $$F(x)=5 x /\left(1+x^{2}\right),$$ find \(F^{\prime}(2)$$ and use it to find an equation of the tangent line to the curve $$y=5 x /\left(1+x^{2}\right)$$ at the point \)(2,2)$ . (b) Illustrate part (a) by graphing the curve and the tangentlines on the same screen.

Step-by-Step Solution

Answer

Tangent line: \( y = -\frac{3}{5}x + \frac{16}{5} \).

Step 1: Find the Derivative

We need to find the derivative of the function \( F(x) = \frac{5x}{1+x^2} \). Here, we will use the quotient rule for differentiation, which is given by \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \). Let \( u = 5x \) and \( v = 1 + x^2 \). The derivative of \( u \) is \( u' = 5 \) and the derivative of \( v \) is \( v' = 2x \). Using the quotient rule, we have:\[ F'(x) = \frac{(5)(1+x^2) - (5x)(2x)}{(1+x^2)^2} = \frac{5 + 5x^2 - 10x^2}{(1+x^2)^2} = \frac{5 - 5x^2}{(1+x^2)^2}. \]

Step 2: Calculate the Derivative at x = 2

Now, we substitute \( x = 2 \) into the derivative we found:\[ F'(2) = \frac{5 - 5(2)^2}{(1+(2)^2)^2} = \frac{5 - 20}{(1+4)^2} = \frac{-15}{25} = -\frac{3}{5}. \]

Step 3: Find the Tangent Line Equation

To find the equation of the tangent line, we use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope (derivative at the point) and \( (x_1, y_1) \) is the point given, \( (2, 2) \).Using \( F'(2) = -\frac{3}{5} \) as the slope, the equation becomes:\[ y - 2 = -\frac{3}{5}(x - 2). \]

Step 4: Simplify the Tangent Line Equation

Expanding and simplifying the equation:\[ y - 2 = -\frac{3}{5}x + \frac{6}{5}. \]Adding 2 to both sides gives:\[ y = -\frac{3}{5}x + \frac{6}{5} + 2 = -\frac{3}{5}x + \frac{16}{5}. \]This is the equation of the tangent line.

Step 5: Visualize the Graph

Graph the function \( y = \frac{5x}{1+x^2} \) and the tangent line \( y = -\frac{3}{5}x + \frac{16}{5} \) on the same coordinate plane. The curve should be plotted along with a straight line intersecting it at the point \( (2, 2) \). The tangent line will touch the curve only at this single point, showing that it is the line that approximates the slope of the curve at \( x=2 \).

Key Concepts

Quotient Rule

The Quotient Rule is a formula in calculus for finding the derivative of a function that is the quotient of two other functions. In simpler terms, if you have a function expressed as one function divided by another, the Quotient Rule helps you find how this function changes. This is crucial for problems where you need to differentiate fractions or rational functions.

To apply the Quotient Rule, suppose you have a function written as \( \frac{u(x)}{v(x)} \). The rule states that the derivative, \( \left(\frac{u}{v}\right)' \), is calculated as:

• Differentiate \( u(x) \) to get \( u'(x) \).
• Differentiate \( v(x) \) to get \( v'(x) \).
• Use the formula \( \frac{u'v - uv'}{v^2} \).

For the exercise given, \( u = 5x \) and \( v = 1 + x^2 \). Thus, \( u' = 5 \) and \( v' = 2x \). Plug these into the Quotient Rule to find the derivative of the function, which is an essential step for solving the problem.

Derivative Calculation

To find the precise change or slope of a curve at a particular point, we calculate its derivative. The derivative tells us how a function changes as its input changes. In the solution, once the Quotient Rule was applied, we found the general form of the derivative to be \( F'(x) = \frac{5 - 5x^2}{(1+x^2)^2} \).

Next, to find the derivative at a specific point, \( x = 2 \), we substitute \( 2 \) into this formula to get:

• Calculate \( 5 - 5(2)^2 = 5 - 20 = -15 \).
• Compute the denominator \( (1+4)^2 = 5^2 = 25 \).
• Divide to get \( F'(2) = \frac{-15}{25} = -\frac{3}{5} \).

This result, \( -\frac{3}{5} \), is the slope of the tangent to the curve at \( x = 2 \). Calculating derivatives like this is vital to understand the behavior of functions at particular points.

Point-Slope Form

The Point-Slope Form is a simple way to write the equation of a line. It's especially useful when you know a point on the line and the slope at that point. The formula is given as \( y - y_1 = m(x - x_1) \), where:

• \( (x_1, y_1) \) is a known point on the line.
• \( m \) is the slope of the line.

For this exercise, the point given is \( (2, 2) \), and the slope \( m \) we found is \( -\frac{3}{5} \).

Plugging these into the Point-Slope Form gives:

• First, \( y - 2 = -\frac{3}{5}(x - 2) \).
• Expanding and simplifying, we get \( y = -\frac{3}{5}x + \frac{16}{5} \).

This equation represents the tangent line to the function at the given point.

Graphing Functions

Graphing is a visual method to understand behaviors of functions and lines on a coordinate plane. When given an intricate function like \( y = \frac{5x}{1+x^2} \), graphing helps visualize how the function behaves over different values of \( x \).

To illustrate part (a) of the problem, graph both the function and its tangent line. The goal is to see:

• The curve \( y = \frac{5x}{1+x^2} \) and how it behaves across the plane.
• The tangent line \( y = -\frac{3}{5}x + \frac{16}{5} \) touching the curve at the point \( (2, 2) \).

The tangent line visually depicts the immediate direction of the curve exactly at that point, making it a powerful concept in understanding instantaneous rate of change.