Problem 68 - Calculus an Applied Approach

Question

Find an equation of the tangent line to the graph of the function at the given point. Then use a graphing utility to graph the function and the tangent line in the same viewing window. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\\ {s(x)=\frac{1}{\sqrt{x^{2}-3 x+4}}} & {\left(3, \frac{1}{2}\right)}\end{array}$$

Step-by-Step Solution

Answer

The equation of the tangent line to the function \(s(x)=\frac{1}{\sqrt{x^{2}-3 x+4}}\) at the point (3, 1/2) is \(y = \frac{1}{2}x + \frac{1}{2}\).

Step 1: Calculate the Derivative

Let's start by finding the derivative of \(s(x)=\frac{1}{\sqrt{x^{2}-3 x+4}}\). This will involve using the chain rule. The derivative will be \(s'(x)=\frac{-x+3}{2(x^{2}-3x+4)^{\frac{3}{2}}}\).

Step 2: Evaluate Derivative at Given Point

Next, evaluate the derivative at the x-coordinate of the given point. This will give the slope of the tangent line at that point. So, \(s'(3)=\frac{1}{2}\).

Step 3: Use the slope-point form of a line to find the equation of the tangent line

Since the slope of the tangent line at the point (3, 1/2) is 1/2, and the line passes through the point (3, 1/2), we can use the slope-point form of a line, \(y - y1 = m(x - x1)\), where m is the slope and (x1, y1) is the given point, to find the equation of the tangent line. Substituting the slope and the given point, we get:\(y - \frac{1}{2} = \frac{1}{2}(x - 3)\)Solving this equation gives \(y = \frac{1}{2}x + \frac{1}{2}\). This is the equation of the tangent line.

Step 4: Graph the function and the tangent line

To complete the problem, graph the function \(s(x)=\frac{1}{\sqrt{x^{2}-3 x+4}}\) and the tangent line \(y = \frac{1}{2}x + \frac{1}{2}\). This will visually confirm the correctness of the solution.

Key Concepts

Chain Rule

Understanding the process of differentiation is crucial when finding tangent lines, especially for compound functions like the one given in the exercise. The Chain Rule is a differentiation technique used when a function is composed of two or more functions. In simpler terms, it's used to differentiate composite functions - those that apply one function to the result of another. For example, for the function \( s(x) = \frac{1}{\sqrt{x^2 - 3x + 4}} \), the outer function is \( f(u) = u^{-rac{1}{2}} \) and the inner function is \( g(x) = x^2 - 3x + 4 \). The Chain Rule states that to find the derivative of such a function, we take the derivative of the outer function, and multiply it by the derivative of the inner function. Using this rule, the derivative \( s'(x) \) was calculated as \( \frac{-x + 3}{2(x^2 - 3x + 4)^{\frac{3}{2}}} \). This result gives us a new function that can be used to find the slope of the tangent line at any given point on \( s(x) \).

Derivative

The derivative concept is central to calculus and helps us understand rates of change. In the context of our exercise, the derivative \( s'(x) \) represents the rate at which \( s(x) \) changes with respect to \( x \). This concept is crucial when we want to find a tangent line to a curve at a specific point. The derivative function tells us the slope of this tangent line.

• We compute the derivative once we have the formula for it by substituting the x-coordinate of the point of tangency.
• For this exercise, that specific calculation was \( s'(3) = \frac{1}{2} \), revealing that the slope of the tangent line at the point (3, \( \frac{1}{2} \)) is \( \frac{1}{2} \).

Understanding derivatives not only helps in finding slopes but also in optimization problems and analyzing the behavior of functions.

Slope-Point Form

The slope-point form is a straightforward approach to writing the equation of a line when you know the slope and a specific point on the line. The formula is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line. In our exercise:

• We use the slope \( \frac{1}{2} \) found from the derivative.
• The point \( (3, \frac{1}{2}) \) is provided as part of the problem.
• Substituting these values into the slope-point formula gives \( y - \frac{1}{2} = \frac{1}{2}(x - 3) \), which simplifies to \( y = \frac{1}{2}x + \frac{1}{2} \).

This is the equation of the tangent line. The slope-point form is advantageous because it requires minimal information to create a whole linear equation.

Graphing Utility

A graphing utility, like a graphing calculator or software, is a valuable tool that allows us to visualize mathematical problems and solutions. For this exercise, graphing the function \( s(x) = \frac{1}{\sqrt{x^2 - 3x + 4}} \) along with its tangent line \( y = \frac{1}{2}x + \frac{1}{2} \) serves a few purposes:

• It visually confirms that the tangent line calculated truly touches the curve at exactly one point.
• Provides a deeper understanding of how steep the slope is and how the tangent line interacts with the curve.
• Helps in cross-verifying that the algebraic solution aligns with the visual representation.

By utilizing graphing utilities, students can gain insight and intuition behind complex equations and their graphical interpretations.