Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

. Find all points on the graph of \(y=9 \sin x \cos x\) where the tangent line is horizontal.

Step-by-Step Solution

Answer

Solve \(\cos^2 x - \sin^2 x = 0\); solutions are when \(x = n\pi + \frac{\pi}{4}\).

Step 1: Identify the condition for the tangent to be horizontal

A tangent line is horizontal when its slope is 0. The slope of the tangent line is given by the derivative of the function. So, we need to find when the derivative of the function equals 0.

Step 2: Find the derivative of y

The given function is \(y = 9 \sin x \cos x\). To find the derivative, use the product rule. The derivative \(y'\) is given by \(d(9 \sin x \cos x)/dx = 9 [\cos x \cdot \frac{d(\sin x)}{dx} + \sin x \cdot \frac{d(\cos x)}{dx}]\). This simplifies to \(y' = 9 \cos x \cdot \cos x - 9 \sin x \cdot \sin x\). Using the identity \(\sin^2 x + \cos^2 x = 1\), this can be rewritten as \(y' = 9(\cos^2 x - \sin^2 x)\).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in mathematics, especially when dealing with problems involving trigonometric functions. These identities are equations that are true for all values of the variables involved. A key identity used frequently is the Pythagorean identity:

• \( \sin^2 x + \cos^2 x = 1 \)

This identity is instrumental in simplifying expressions, especially ones with squares of sine and cosine. For example, if you are working with an expression involving both \( \sin^2 x \) and \( \cos^2 x \), this identity allows you to express one in terms of the other, greatly simplifying the task. In our exercise, it was used to rewrite the expression for the derivative \( y' \) from \( 9 \cos^2 x - 9 \sin^2 x \) to \( 9(\cos^2 x - \sin^2 x) \), thanks to the so-called double angle identity \( \cos 2x = \cos^2 x - \sin^2 x \). This approach is effective in simplifying and solving derivative problems by reducing complexity.

Horizontal Tangent Line

A horizontal tangent line occurs at points where the slope of the tangent line is zero. In calculus, the slope of the tangent line to a curve at a given point is given by the derivative of the function at that point. When we are tasked with finding where a tangent line is horizontal, we first need to calculate the derivative. Once we have the derivative, we set it equal to zero and solve for the variable(s) involved.For the function \( y=9 \sin x \cos x \), we found the derivative to be \( y' = 9(\cos^2 x - \sin^2 x) \). To find where this derivative is zero, we solve the equation:

• \( 9(\cos^2 x - \sin^2 x) = 0 \)
• This simplifies to \( \cos^2 x = \sin^2 x \)

From here, you can apply the identity \( \cos^2 x = \sin^2 x \) translates to \( \tan^2 x = 1 \), revealing points along the curve where the tangent is indeed horizontal.

Product Rule

The product rule is a fundamental technique for finding the derivative of a function that is the product of two or more functions. When a function \( y = u(x)v(x) \) is given, the product rule states that its derivative, \( y' \), can be found by:

• \( y' = u'(x)v(x) + u(x)v'(x) \)

In our exercise, we have the function \( y = 9 \sin x \cos x \), where the functions \( u(x) = \sin x \) and \( v(x) = \cos x \) are multiplied. Applying the product rule, we have:1. Differentiate \( u(x) = \sin x \), giving \( u'(x) = \cos x \).2. Differentiate \( v(x) = \cos x \), yielding \( v'(x) = -\sin x \).3. Substitute into the product rule: \( y' = 9(\cos x \cdot \cos x + \sin x \cdot (-\sin x)) \).This usage of the product rule simplifies the process of differentiation, making complex products of functions much more manageable to handle.