Problem 22 - The Calculus with Analytic Geometry

Question

Find an equation of the tangent line to the curve \(y=2 /(4-x)^{2}\) at each of the following points: \(\left(0, \frac{1}{8}\right),\left(1, \frac{2}{9}\right),\left(2, \frac{1}{2}\right)\), \((3,2),(5,2),\left(6, \frac{1}{2}\right)\). Draw a sketch of the graph and segments of the tangent lines at the given points.

Step-by-Step Solution

Answer

The equations of the tangent lines are: \( y = \frac{1}{16}x + \frac{1}{8} \), \( y = \frac{4}{27}x + \frac{2}{27} \), \( y = \frac{1}{2}x - \frac{1}{2} \), \( y = 4x - 10 \), \( y = -4x + 22 \), \( y = -\frac{1}{2}x + \frac{7}{2} \).

Step 1 - Find the derivative of the function

The derivative of the function will give the slope of the tangent line. Begin by differentiating the given function: \( y = \frac{2}{(4-x)^2} \) Use the chain rule for differentiation.

Step 2 - Apply the chain rule

Rewrite the function as \( y = 2(4-x)^{-2} \) and then differentiate: \( \frac{dy}{dx} = 2 \cdot -2(4-x)^{-3} \cdot (-1) \). This simplifies to: \( \frac{dy}{dx} = \frac{4}{(4-x)^3} \).

Step 3 - Calculate the slope at each point

Substitute each x-coordinate into the derivative to find the slope at each point: \(m_1 = \frac{4}{(4-0)^3} = \frac{4}{64} = \frac{1}{16}\) \(m_2 = \frac{4}{(4-1)^3} = \frac{4}{27}\) \(m_3 = \frac{4}{(4-2)^3} = \frac{4}{8} = \frac{1}{2}\) \(m_4 = \frac{4}{(4-3)^3} = \frac{4}{1} = 4\) \(m_5 = \frac{4}{(4-5)^3} = \frac{4}{-1} = -4\) \(m_6 = \frac{4}{(4-6)^3} = \frac{4}{-8} = -\frac{1}{2}\).

Step 4 - Write the equation of the tangent line at each point

Using the point-slope form \( y - y_1 = m(x - x_1) \), substitute each point \((x_1, y_1)\) and the respective slope \( m \): For \( (0, \frac{1}{8}) \): \( y - \frac{1}{8} = \frac{1}{16}(x - 0) \rightarrow y = \frac{1}{16}x + \frac{1}{8} \) For \( (1, \frac{2}{9}) \): \( y - \frac{2}{9} = \frac{4}{27}(x - 1) \rightarrow y = \frac{4}{27}x - \frac{4}{27} + \frac{2}{9} \rightarrow y = \frac{4}{27}x + \frac{2}{27} \) For \( (2, \frac{1}{2}) \): \( y - \frac{1}{2} = \frac{1}{2}(x - 2) \rightarrow y = \frac{1}{2}x - 1 + \frac{1}{2} \rightarrow y = \frac{1}{2}x - \frac{1}{2} \) For \( (3, 2) \): \( y - 2 = 4(x - 3) \rightarrow y = 4x - 12 + 2 \rightarrow y = 4x - 10 \) For \( (5, 2) \): \( y - 2 = -4(x - 5) \rightarrow y = -4x + 20 + 2 \rightarrow y = -4x + 22 \) For \( (6, \frac{1}{2}) \): \( y - \frac{1}{2} = -\frac{1}{2}(x - 6) \rightarrow y = -\frac{1}{2}x + 3 + \frac{1}{2} \rightarrow y = -\frac{1}{2}x + \frac{7}{2} \).

Step 5 - Sketch the graph and tangent lines

Plot the function \( y = \frac{2}{(4-x)^2} \) and draw tangent lines at the given points using the equations derived in the previous step to observe the behavior and correctness visually.

Key Concepts

Tangent line

A tangent line is a straight line that touches a curve at one and only one point. This point is called the point of tangency.
The slope of the tangent line reflects the rate of change of the curve at that particular point. Notably, the tangent line does not intersect the curve at any other point near this contact.
For example, when we need to find the equation of a tangent line to the curve given by the function: y = \(\frac{2}{(4-x)^2}\) at specific points, the first step is to determine the slope of the tangent line at each designated point. This involves finding the derivative of the function.

Derivative

In calculus, the derivative of a function measures how a function's value changes as its input changes. In other words, it gives the slope of the function at any point.
Differentiation is the process of finding the derivative. The derivative of the function \( y = \frac{2}{(4-x)^2} \) can be determined by rewriting the function and applying the chain rule.
The derivative, noted as \( \frac{dy}{dx} \), is needed to compute the slope of the tangent line. For our function, after applying the chain rule, the derivative is determined as: \( \frac{dy}{dx} = \frac{4}{(4-x)^3} \).
We then substitute each x-coordinate from the points into this formula to find the slope at the corresponding points. This process consists of multiple steps to unveil the connection between the function's behavior and its slopes.

Chain Rule

The chain rule is a method for differentiating composite functions. Whenever you have a function inside another function, you need this rule.
For the given exercise, you can see the chain rule in action when differentiating \( y = \frac{2}{(4-x)^2} \).
Firstly, rewrite the function as \( y = 2(4-x)^{-2} \) to make the differentiation process easier.Then apply the chain rule: begin by differentiating the outer function and then multiply by the derivative of the inner function. This gives us the steps:

• Differentiating the outer function: \ y = 2 \times -2(4-x)^{-3} \
• Then multiply by the derivative of the inner function (-1): \( 2 \times -2(4-x)^{-3} \times (-1) \)

This streamlines our role in finding the critical slope needed to determine the tangent lines and drawing insights from the curve's behavior.

Point-slope form

To find the equation of the tangent line, we use the point-slope form, which is especially useful for these types of problems.
The point-slope form is written as: \( y - y_1 = m(x - x_1) \).
In our example, \( (x_1, y_1) \) are the coordinates of the given point on the curve, and \( m \) is the slope found from the derivative. By substituting \( x \) into the derivative function, we get the slope of the tangent.
We then substitute \( (x_1, y_1) \) and \( m \) into the point-slope formula to derive the equation of the tangent line at specific points. For instance, for the point \( (0, \frac{1}{8}) \):

• Slope calculation: \( m = \frac{1}{16} \)
• Point-slope form insertion: \( y - \frac{1}{8} = \frac{1}{16}(x - 0) \)
• Simplifying gives: \( y = \frac{1}{16}x + \frac{1}{8} \)

Influencing each tangent line has its own customized equation based on the specific points provided.

Graph sketching

Graph sketching helps bring visualization to mathematical concepts by plotting curves and their tangent lines on a graph.
To effectively sketch the graph of the function \( y=\frac{2}{(4-x)^2} \), we can begin by plotting the curve using its equation. By calculating a few points, particularly the ones listed in the exercise, you can determine the behavior of the function.
Next, leverage the tangent line equations derived from the point-slope form. Plot these tangent lines at the respective points on the curve to observe their relationship visually.

• The curve of the function provides a continuous overview of how y changes with x.
• The tangent lines demonstrate how the slope of the curve behaves at each specific point.
  This visual investigation leads to deeper insights into understanding the core principles of calculus and its applications in curve analysis and slope determination.