Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find all points on the graph of \(y=\tan ^{2} x\) where the tangent line is horizontal.

Step-by-Step Solution

Answer

Points (nπ, 0) for integer n.

Step 1: Understand the Problem

We need to find points where the tangent to the graph of the function \(y=\tan^{2}x\) is horizontal. A horizontal tangent line occurs where the derivative of the function equals zero.

Step 2: Differentiate the Function

We start by differentiating the function \(y=\tan^{2}x\). Using the chain rule, we have \( \frac{dy}{dx} = 2\tan x \cdot \sec^{2}x = 2\tan x \cdot (1+\tan^{2} x) \).

Step 3: Set the Derivative to Zero

To find where the tangent is horizontal, set the derivative equal to zero: \(2\tan x \cdot (1+\tan^{2} x) = 0\).

Step 4: Solve for \(\tan x = 0\)

For the product \(2\tan x \cdot (1+\tan^{2} x) = 0\), either \(\tan x = 0\) or \((1+\tan^{2} x) = 0\). Since \(1+\tan^{2} x\) cannot be zero for real \(x\), we solve \(\tan x = 0\).

Step 5: Find \(x",

The equation \(\tan x = 0\) has solutions where \(x = n\pi\), where \(n\) is an integer. So the points are at integer multiples of \(\pi\).

Step 6: Find Corresponding \(y",

For \(x = n\pi\), substitute back into \(y = \tan^{2} x\). Since \(\tan(n\pi) = 0\), then \(y = 0\).

Step 7: State the Points

The points where the tangent line is horizontal are \((n\pi, 0)\) where \(n\) is any integer.

Key Concepts

Graph of Trigonometric Functions

The function graph of trigonometric equations, such as the one in the given problem, can provide deep insights into their behavior. For the function \(y=\tan^2 x\), understanding its graph helps us visualize the changes and slopes of the equation.

• \(\tan x\) naturally has vertical asymptotes at each odd multiple of \(\frac{\pi}{2}\), meaning it will not be defined there.
• \(\tan^2 x\) is non-negative because it is the square of \(\tan x\).
• There are repetitive units and periodic behavior, as \(\tan x\) repeats every \(\pi\), leading \(\tan^2 x\) to repeat over this interval.
• The function is symmetrically increasing around these periods, creating noticeable parabolic-like arcs that approach infinity near the undefined points.

Recognizing these properties helps when determining specific behavior such as where the slope of the tangent line is horizontal.

Derivative of Trigonometric Functions

Differentiation is a fundamental concept in calculus that measures how a function changes as its input changes. When you differentiate \(y = \tan^2 x\), you are trying to understand the slope or inclination of the function at various points.

To find the derivative, use the chain rule:

• The derivative of \(\tan x\) is \(\sec^2 x\).
• Applying the chain rule to \(y = \tan^2 x\), you differentiate the outer function \(x^2\) as \(2x\) and multiply it by the derivative of the inner function, \(\sec^2 x\).
• This results in \(\frac{dy}{dx} = 2\tan x \cdot \sec^2 x = 2\tan x \cdot (1+\tan^2 x)\).

Understanding these steps allows us to identify where the tangent line is horizontal by setting this derivative equal to zero.

Equation of Tangent Lines

The tangent line to a curve at a given point represents the instantaneous direction of the curve at that point. When this line is horizontal, the slope of the tangent (derivative) is zero.

From the derivative \(\frac{dy}{dx} = 2\tan x \cdot (1+\tan^2 x)\), set the equation to zero:

• Since \(2\tan x = 0\) leads to \(\tan x = 0\), solving this yields points \(x = n\pi\), where \(n\) is an integer.
• At these values, the function \(y = \tan^2 x\) results in \(y = 0\).
• Thus, horizontal tangent lines occur at points \((n\pi, 0)\) across the graph of the function.

These points show where the function flattens, providing key insights into the function's behavior on its graph.