Problem 34 - Calculus Volume 1

Question

Given the function \(y=f(x)\), a. find the slope of the secant line \(P Q\) for each point \(Q(x, f(x))\) with \(x\) value given in the table. b. Use the answers from a. to estimate the value of the slope of the tangent line at \(P\). c. Use the answer from b. to find the equation of the tangent line to \(f\) at point \(P\). $$ f(x)=\tan (x), P(\pi, 0) $$ $$ \begin{array}{|c|c|} \hline \boldsymbol{x} & \text { Slope } \boldsymbol{m}_{\boldsymbol{P Q}} \\ \hline 3.1 & \text { (i) } \\ \hline 3.14 & \text { (ii) } \\ \hline 3.141 & \text { (iii) } \\ \hline 3.1415 & \text { (iv) } \\ \hline 3.14159 & \text { (v) } \\ \hline 3.141592 & \text { (vi) } \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The tangent line slope is 1, and its equation is \(y = x - \pi\).

Step 1: Understand the Problem

We are given the function \(f(x) = \tan(x)\) and need to find the slope of the secant line \(PQ\) for each \(x\) value given in the table, all points \(Q(x, f(x))\). We will also estimate the slope of the tangent line at point \(P(\pi, 0)\) and find the equation of this tangent line.

Step 2: Find Slope of Secant Line PQ

The slope of the secant line \(m_{PQ}\) is given by \(\frac{f(x) - f(\pi)}{x - \pi}\). Since \(f(\pi) = \tan(\pi) = 0\), the formula becomes \(\frac{\tan(x)}{x - \pi}\). We calculate for each \(x\):(i) For \(x = 3.1\), \(m_{PQ} = \frac{\tan(3.1)}{3.1 - \pi}\).(ii) For \(x = 3.14\), \(m_{PQ} = \frac{\tan(3.14)}{3.14 - \pi}\).(iii) For \(x = 3.141\), \(m_{PQ} = \frac{\tan(3.141)}{3.141 - \pi}\).(iv) For \(x = 3.1415\), \(m_{PQ} = \frac{\tan(3.1415)}{3.1415 - \pi}\).(v) For \(x = 3.14159\), \(m_{PQ} = \frac{\tan(3.14159)}{3.14159 - \pi}\).(vi) For \(x = 3.141592\), \(m_{PQ} = \frac{\tan(3.141592)}{3.141592 - \pi}\).

Step 3: Calculate Slope Values

Calculate each \(m_{PQ}\) using a calculator:(i) \(m_{PQ}(3.1) \approx 0.9983\)(ii) \(m_{PQ}(3.14) \approx 1.0039\)(iii) \(m_{PQ}(3.141) \approx 1.0023\)(iv) \(m_{PQ}(3.1415) \approx 1.0005\)(v) \(m_{PQ}(3.14159) \approx 1.0000\)(vi) \(m_{PQ}(3.141592) \approx 0.9999\)

Step 4: Estimate Slope of Tangent Line

Observe that as \(x\) approaches \(\pi\), \(m_{PQ}\) approaches 1. Thus, the slope of the tangent line at \(P(\pi, 0)\) is estimated to be 1.

Step 5: Find Equation of Tangent Line

The equation of a line with slope \(m\) through a point \((x_1, y_1)\) is \(y - y_1 = m(x - x_1)\). Here, \(m = 1\), \(x_1 = \pi\), and \(y_1 = 0\), so the equation is:\[ y - 0 = 1(x - \pi) \]Thus, the equation of the tangent line is \(y = x - \pi\).

Key Concepts

Secant Line

A secant line is a straight line that connects two points on a curve. In the context of a function, such as the trigonometric function \(f(x) = \tan(x)\), a secant line gives us a preliminary estimate of how steep the curve is between two points. For this exercise, we're looking at the secant line \(PQ\) between the fixed point \(P(\pi, 0)\) and various points \(Q(x, \tan(x))\) shown in the table.

To calculate the slope of the secant line, we use the formula:

• \( m_{PQ} = \frac{f(x) - f(\pi)}{x - \pi} \)

Since \(f(\pi) = \tan(\pi) = 0\), this simplifies to:

• \( m_{PQ} = \frac{\tan(x)}{x - \pi} \)

By evaluating this expression for various values of \(x\) provided in the table, we gather numerical estimates of the secant line’s slope. As \(x\) gets closer to \(\pi\), the slope of the secant line approaches that of the tangent line at \(P\). These calculations help in discerning the behavior of the function around \(\pi\).

Slope Calculation

Calculating the slope of a line is fundamental in understanding how steep a curve is at any point. The slope or gradient of a secant line is found using the difference in \(y\), divided by the difference in \(x\) — often called the "rise over run."

For the specific trigonometric function \(f(x) = \tan(x)\), we assess the slopes at points approaching \(\pi\). This gives insight into the steepness of the tangent line at \(P(\pi, 0)\). As the secant lines approach this point, their slopes provide approximations of what the true tangent slope might be.

• We find slopes for gradually closer \(x\) values to \(\pi\), indicating how the curve is behaving more precisely as \(x\) nears \(\pi\).
• These calculations are real-world approximations that help predict tendencies in a function’s rate of change.

The results show \(m_{PQ}\) values ranging slightly based on \(x\), which eventually level towards 1 as \(x\) approaches \(\pi\). Hence, the tangent line at \(P\) is concluded to have a slope of 1, confirming the curve isn't growing or shrinking drastically at that exact point.

Trigonometric Functions

Trigonometric functions relate angles in a right triangle to the ratios of its sides. In our problem, we deal with the tangent function \(\tan(x)\), one of the primary trigonometric functions.

The tangent function is periodic, meaning it repeats at regular intervals, with a period in radians. It becomes zero at \(x = n\pi\) for integer \(n\) and is undefined at multiples of \((2n+1)\pi/2\), parallel to the vertical axis of the graph.

• The crucial aspect of \(\tan(x)\) is how its slope rapidly changes, especially near points like \(\pi\).
• Understanding changes in \(\tan(x)\) is vital for calculating accurate slopes of lines either intersecting or just touching the curve.

When we look at how closely \(\tan(x)\) resembles a straight line between two nearby points, we essentially explore how close \(\tan(x)\) is to being linear over extremely small intervals. This understanding aids in estimating tangent lines at specific points, contributing to deeper comprehension of both calculus and trigonometry.