Problem 43 - Calculus an Applied Approach

Question

Find an equation of the tangent line to the graph of \(f\) at the point \((2, f(2)) .\) Use a graphing utility to check your result by graphing the original function and the tangent line in the same viewing window. $$ f(x)=\sqrt{4 x^{2}-7} $$

Step-by-Step Solution

Answer

The equation of the tangent line to the graph of \(f(x)= \sqrt{4x^{2}-7}\) at \((2,3)\) is \(y = \frac{8}{3}x - \frac{1}{3}\).

Step 1: Calculate f(2)

We first need to find \(f(2)\). That is obtained by substituting \(x = 2\) in the equation; \(f(2)=\sqrt{4*(2)^{2} -7} = \sqrt{9} = 3. Thus, the point \((2,f(2))\) is \((2,3)\).

Step 2: Differentiate the function

To get the slope at \((2,3)\), we differentiate \(f(x)\), i.e. find \(f'(x)\). The derivative of \(f(x)\) is obtained by the power rule for differentiation. The derivative of \(\sqrt{4x^2 - 7}\) is: \(f'(x) = 1/2 * (4x^2 - 7)^{-1/2} * 2 * 4x = \frac{4x}{\sqrt{4x^2 - 7}}\).

Step 3: Calculate f'(2)

Now, substitute \(x = 2\) in \(f'(x)\) to get the slope at \((2,3)\): \(f'(2)= \frac{4*2}{\sqrt{4*(2)^2 -7}}= \frac{8}{3}\). This is the slope of the line at \((2,3)\).

Step 4: Write down the equation for the tangent line

Use the point-slope form of the equation, \(y - y_1 = m (x - x_1)\), where \((x_1, y_1)\) is the point at which the slope is given and \(m\) is the slope. Then, substituting \(x_1 = 2\), \(y_1 = 3\) and \(m = 8/3\), the equation becomes: \(y - 3 = \frac{8}{3}(x - 2)\). Simplify to obtain: \(y = \frac{8}{3}x - \frac{1}{3}\). This is the equation of the tangent to the curve at \((2,3)\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to find the rate at which a function is changing at any given point. When we differentiate a function like \(f(x)\), we get its derivative, which tells us the slope of the tangent line to the curve of the function at any point \(x\). This is crucial when we want to find the equation of a tangent line.

To differentiate the function \(f(x) = \sqrt{4x^2 - 7}\), we apply the power rule. This involves taking derivatives of powers of \(x\) and using chain rules if complicated expressions are involved. After following these steps, we find that the derivative \(f'(x)\) is \(\frac{4x}{\sqrt{4x^2 - 7}}\).

By calculating \(f'(2)\), we determine the specific slope of the tangent at the point \((2,3)\). Differentiation tells us how steep or flat the tangent line is at the precise point on the curve.

Graphing Utility

A graphing utility is a tool used to visualize mathematical functions and their corresponding graphs. These can be software applications or calculators designed to handle complex equations and graph them accurately. Using a graphing utility helps students see the function \(f(x) = \sqrt{4x^2 - 7}\) and its tangent line on the same graph, allowing for a visual verification of the calculated tangent line's accuracy.

By plotting both the function and the tangent line in a single window, students can visually confirm whether the tangent line correctly touches the curve at the point \((2,3)\) without crossing it, which is characteristic of tangent lines. This reinforces understanding and ensures the solution matches the problem's requirement.

Slope of a Curve

The slope of a curve at a given point refers to the steepness or inclination of the curve at that specific point. This is essentially the same as the derivative at that point. Finding the slope is crucial, especially when trying to draw tangents to curve functions.

In our exercise, the associated slope is found by computing \(f'(2) = \frac{8}{3}\). This tells us that at \(x = 2\), the curve rises 8 units for every 3 units it runs horizontally, indicating a positive and fairly steep slope.

Understanding the slope helps in not only drawing tangent lines but also interpreting the behavior of the function around that particular point. The slope’s sign (positive, negative, or zero) can provide insights into whether the function is increasing, decreasing, or flat at that point.

Point-Slope Form

The point-slope form is a way to represent the equation of a line when you know the slope and a point on the line. The formula is expressed as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a known point and \(m\) is the slope.

This method is especially useful in finding the equation of a tangent line to a curve because, once the slope \(f'(x)\) is known, you can easily plug in the point and slope to get the line's equation. In our example, using \((2,3)\) as the point and \(\frac{8}{3}\) as the slope, the equation becomes \(y - 3 = \frac{8}{3}(x - 2)\).

Simplifying gives \(y = \frac{8}{3}x - \frac{1}{3}\), providing a clear linear equation that represents the tangent to the curve at the specified point. This form is versatile and straightforward for writing equations quickly when dealing with linear approximations.