Problem 28 - Thomas Calculus

Question

In Exercises \(27-30\) , graph the curves over the given intervals, together with their tangents at the given values of \(x\) . Label each curve and tangent with its equation. $$ \begin{array}{l}{y=\sec x, \quad-\pi / 2

Step-by-Step Solution

Answer

Graph \(y = \sec x\) for \( -\frac{\pi}{2} < x < \frac{\pi}{2} \) with tangents at \(x = -\frac{\pi}{3}, \pi/4\). Tangent lines: \(y = -2\sqrt{3}x - \frac{2\sqrt{3}\pi}{3} + 2\) and \(y = \sqrt{2}x - \frac{\sqrt{2}\pi}{4} + \sqrt{2}\).

Step 1: Understand the Function and Interval

The given function is a secant function, \(y = \sec x\). This function is defined as \(\sec x = \frac{1}{\cos x}\). The interval provided is \(-\frac{\pi}{2} < x < \frac{\pi}{2}\). The function \(\sec x\) will have vertical asymptotes where \(\cos x = 0\). On this interval, the secant function is defined except at \(x = \pm \frac{\pi}{2}\).

Step 2: Calculate the Derivative

To find the tangent lines, we need the derivative of the function. The derivative of \(y = \sec x\) is \(y' = \sec x \tan x\). This helps us calculate the slopes of the tangents at the given points.

Step 3: Calculate Values at Given Points

First, at \(x = -\frac{\pi}{3}\), calculate \(y = \sec\left(-\frac{\pi}{3}\right)\) and \(y' = \sec\left(-\frac{\pi}{3}\right) \tan\left(-\frac{\pi}{3}\right)\). The secant of \(-\frac{\pi}{3}\) is \(2\), and the tangent is \(-\sqrt{3}\). Thus, the slope is \(-2\sqrt{3}\).

Step 4: Equation of the Tangent Line at \(x = -\pi/3\)

Using the point \((-\frac{\pi}{3}, 2)\) and slope \(-2\sqrt{3}\), the equation of the tangent line using \(y - y_1 = m(x - x_1)\) is \(y - 2 = -2\sqrt{3}(x + \frac{\pi}{3})\). Simplifying gives \(y = -2\sqrt{3}x - \frac{2\sqrt{3}\pi}{3} + 2\).

Step 5: Repeat for \(x = \pi/4\)

Calculate \(y = \sec(\pi/4)\) and \(y' = \sec(\pi/4) \tan(\pi/4)\). \(\sec(\pi/4) = \sqrt{2}\) and \(\tan(\pi/4) = 1\), so slopes is \(\sqrt{2}\).

Step 6: Equation of the Tangent Line at \(x = \pi/4\)

Using the point \((\pi/4, \sqrt{2})\) and slope \(\sqrt{2}\), the equation is \(y - \sqrt{2} = \sqrt{2}(x - \pi/4)\), which simplifies to \(y = \sqrt{2}x - \frac{\sqrt{2}\pi}{4} + \sqrt{2}\).

Step 7: Plot the Graph

Draw the graph of \(y = \sec x\) over \(-\frac{\pi}{2} < x < \frac{\pi}{2}\) with its characteristic asymmetric shape with vertical asymptotes approaching \(x = \pm \frac{\pi}{2}\). Plot the tangents at \(x = -\frac{\pi}{3}\) and \(x = \pi/4\), using the tangent line equations derived in the previous steps. Label each curve and tangent accordingly.

Key Concepts

Secant Function

The secant function, represented as \(y = \sec x\), is best understood as the reciprocal of the cosine function. Thus, \(\sec x = \frac{1}{\cos x}\). This means that wherever \(\cos x = 0\), \(\sec x\) will have vertical asymptotes because division by zero is undefined. In the interval \(-\frac{\pi}{2} < x < \frac{\pi}{2}\), the secant function behaves in a unique oscillating manner, creating these asymptotes exactly at the points where \(x = \pm \frac{\pi}{2}\). The presence of asymptotes is what gives the secant function its distinctive shape with segments extending upwards or downwards as they approach the lines where \(\cos x\) becomes zero.

Tangent Lines

Tangent lines are straight lines that touch a curve at just one point. At this point, the tangent line has the same slope as the curve, meaning it mimics the curve's immediate direction. When considering the secant function \(y = \sec x\), we want to find tangent lines at specific points.

• The tangent line shows how the secant function is changing exactly at a given point.
• Graphically, it provides an approximation of the function close to the point of tangency.

These linear representations are useful for understanding the instantaneous rate of change at specific points on the curve.

Derivative Calculation

The derivative of a function provides us with the slope of the tangent line at any point on the curve. For the secant function, the derivative can be calculated using the product and chain rules, resulting in \(y' = \sec x \tan x\). This derivative tells us two things:

• The slope of the tangent at each point on the secant curve.
• How fast the function \(\sec x\) is changing at any point within the interval.

For instance, at \(x = -\frac{\pi}{3}\), the derivative gives us a slope of \(-2\sqrt{3}\), indicating the direction and steepness of the curve at that point.

Equation of Tangents

To find the equation of a tangent line, we use the point-slope form: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope from the derivative and \((x_1, y_1)\) is the point of tangency on the curve. Here's a breakdown for the secant function:

• At \(x = -\frac{\pi}{3}\): The point is \((-\frac{\pi}{3}, 2)\) with a slope of \(-2\sqrt{3}\). The tangent's equation is \(y = -2\sqrt{3}x - \frac{2\sqrt{3}\pi}{3} + 2\).
• At \(x = \pi/4\): The point is \((\pi/4, \sqrt{2})\) and the slope is \(\sqrt{2}\). The equation here is \(y = \sqrt{2}x - \frac{\sqrt{2}\pi}{4} + \sqrt{2}\).

These equations allow us to plot the tangent lines on the graph and see how they touch the curve at their respective points of tangency, providing a linear approximation of the curve at these points.