Problem 33 - Calculus: Early Transcendentals

Question

Slope of tangent line Given the function \(f(x)=1-\cos x\) and the points \(A(\pi / 2, f(\pi / 2)), B(\pi / 2+0.05, f(\pi / 2+0.05))\) \(C(\pi / 2+0.5, f(\pi / 2+0.5)),\) and \(D(\pi, f(\pi))\) (see figure), find the slopes of the secant lines through \(A\) and \(D, A\) and \(C,\) and \(A\) and B. Use your calculations to make a conjecture about the slope of the line tangent to the graph of \(f\) at \(x=\pi / 2\) (graph cannot copy)

Step-by-Step Solution

Answer

Answer: We can conjecture that the slope of the tangent line to the graph of the function \(f(x) = 1 - \cos x\) at \(x = \frac{\pi}{2}\) would be slightly larger than the slope of the secant line AB, i.e., slightly larger than \(\frac{-\cos(\frac{\pi}{2} + 0.05)}{0.05}\).

Step 1: Finding coordinates of points A, B, C and D

The given points are: - A(\(\frac{\pi}{2}\), \(f(\frac{\pi}{2})\)) - B(\(\frac{\pi}{2}+0.05\), \(f(\frac{\pi}{2}+0.05)\)) - C(\(\frac{\pi}{2}+0.5\), \(f(\frac{\pi}{2}+0.5)\)) - D(\(\pi\), \(f(\pi)\)) To find the coordinates of each point, substitute the corresponding x-values in \(f(x)=1-\cos x\). A\((\frac{\pi}{2}, 1 - \cos(\frac{\pi}{2})) \Rightarrow (\frac{\pi}{2}, 1)\) B\((\frac{\pi}{2}+0.05, 1 - \cos(\frac{\pi}{2}+0.05)) \Rightarrow (\frac{\pi}{2}+0.05, 1-\cos(\frac{\pi}{2}+0.05))\) C\((\frac{\pi}{2}+0.5, 1 - \cos(\frac{\pi}{2}+0.5)) \Rightarrow (\frac{\pi}{2}+0.5, 1-\cos(\frac{\pi}{2}+0.5))\) D\((\pi, 1-\cos(\pi)) \Rightarrow (\pi, 1 - (-1)) \Rightarrow (\pi, 2)\)

Step 2: Finding slopes of the secant lines

To find the slopes of the secant lines, use the formula \(m = \frac{f(x_2) - f(x_1)}{x_2 - x_1}\) for each pair of points A & D, A & C, and A & B. For Secant line AD: \(m_{AD} = \frac{f(\pi) - f(\frac{\pi}{2})}{\pi - \frac{\pi}{2}} = \frac{2 - 1}{\pi - \frac{\pi}{2}} = \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi}\) For Secant line AC: \(m_{AC} = \frac{f(\frac{\pi}{2} + 0.5) - f(\frac{\pi}{2})}{\frac{\pi}{2} + 0.5 - \frac{\pi}{2}} = \frac{1 - \cos(\frac{\pi}{2} + 0.5) - 1}{0.5} = \frac{-\cos(\frac{\pi}{2} + 0.5)}{0.5}\) For Secant line AB: \(m_{AB} = \frac{f(\frac{\pi}{2} + 0.05) - f(\frac{\pi}{2})}{\frac{\pi}{2} + 0.05 - \frac{\pi}{2}} = \frac{1 - \cos(\frac{\pi}{2} + 0.05) - 1}{0.05} = \frac{-\cos(\frac{\pi}{2} + 0.05)}{0.05}\)

Step 3: Make conjecture about the slope of the tangent line

Based on the slopes of the secant lines, as the second point (B or C) gets closer to the first point A, the slope goes from \(\frac{2}{\pi}\) and seems to increase. The tangent line is the limit of the secant line when the distance between the points tends to zero. Therefore, we can conjecture that the slope of the tangent line to the graph of the function \(f(x) = 1 - \cos x\) at \(x = \frac{\pi}{2}\) would be slightly larger than the slope of the secant line AB, i.e., slightly larger than \(\frac{-\cos(\frac{\pi}{2} + 0.05)}{0.05}\).

Key Concepts

Tangent Line

A tangent line to a function at a particular point is a straight line that just touches the curve at that point. It forms a perfect first derivative of the function at that point, representing the slope at exactly that moment. Unlike a secant line, which passes through two points on the curve, the tangent only "leans against" the function at one single point.
The concept of a tangent line is fundamental in calculus because it provides the basis for differentiating functions. Differentiation allows us to calculate the slope of a tangent line at any point on a function's graph.
In simple terms, if you imagine skating along the curve of a hill, your tangent line is like the direction or path you'd continue to move if you went straight ahead at that point, without turning your skates.

Secant Line

The secant line is an important concept in calculus, representing a line that intersects a curve at two distinct points. It essentially "cuts" through the curve, providing an average rate of change over an interval between these two points.
To calculate the slope of a secant line between two points, you would use the formula:

• \[ m = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \]

This formula calculates the difference in the "y"-values of the function at these two points, divided by the difference in "x"-values.
By measuring how the secant line changes as you move the second point closer to the first point, you begin to approximate the slope of the tangent line.

Trigonometric Functions

Trigonometric functions are a group of functions that relate angles of a triangle to the ratios of the triangle's sides. In the realm of calculus, these functions help in modeling periodic phenomena such as sound waves, light waves, and alternating current.
Some common trigonometric functions include sine, cosine, and tangent. In this exercise, we particularly dealt with the cosine function. The function under examination was given by:

• \( f(x) = 1 - \cos x \)

Here, the subtraction of cosine from 1 modifies the standard cosine wave, affecting its amplitude and vertical shift.
These trigonometric identities are crucial for students to understand as they enable the modeling of real-world processes and the application of calculus in different fields such as physics and engineering.

Slope Calculation

Slope calculation is a fundamental tool in calculus used to determine the steepness or gradient of a line on a graph. But, it's not just restricted to straight lines; the concept extends to curves via differentiation.
The slope is calculated using the formula for straight lines:

• \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]

When working with curves, as is common in calculus, we take a derivative of the function to find the slope of the tangent line at any given point.
This concept is crucial in finding maximum and minimum points of functions, as well as analyzing rates of change, which are essential in fields like economics, engineering, and the natural sciences. Understanding how to effectively calculate slope equips students with analytical skills necessary for advanced studies.