Problem 62 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent line to the graph of \(y=1+x \sin 3 x\) at \(\left(\frac{\pi}{3}, 1\right) .\) Where does this line cross the \(x\) -axis?

Step-by-Step Solution

Answer

The tangent line's equation is \( y = -\pi x + \frac{\pi^2}{3} + 1 \); it crosses the x-axis at \( x = \frac{\pi}{3} + \frac{1}{\pi} \).

Step 1: Find the Derivative of the Function

To find the equation of the tangent line, we first need the derivative of the function, which will give us the slope at any point. The function given is \( y = 1 + x \sin(3x) \). We will use the product rule and chain rule to find the derivative. The derivative of \( y \) with respect to \( x \) is:\[ y' = \frac{d}{dx}(x \sin(3x)) = \sin(3x) + x \frac{d}{dx}(\sin(3x)) \]The chain rule tells us that \( \frac{d}{dx}(\sin(3x)) = 3 \cos(3x) \), so:\[ y' = \sin(3x) + 3x \cos(3x) \]

Step 2: Evaluate the Derivative at the Given Point

Now we need to find the slope of the tangent line at the point \( \left( \frac{\pi}{3}, 1 \right) \). Plug \( x = \frac{\pi}{3} \) into the derivative:\[ y'\left( \frac{\pi}{3} \right) = \sin(3 \cdot \frac{\pi}{3}) + 3 \cdot \frac{\pi}{3} \cos(3 \cdot \frac{\pi}{3}) \]Calculating the trigonometric values:\[ y'\left( \frac{\pi}{3} \right) = \sin(\pi) + \pi \cos(\pi) = 0 + (-\pi) = -\pi \]Thus, the slope \( m \) of the tangent line is \( -\pi \).

Step 3: Form the Equation of the Tangent Line

We use the point-slope form of the equation of a line: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point. Here, \( m = -\pi \), \( x_1 = \frac{\pi}{3} \), and \( y_1 = 1 \).Plug in these values:\[ y - 1 = -\pi \left(x - \frac{\pi}{3}\right) \]Simplify the equation:\[ y = -\pi x + \frac{\pi^2}{3} + 1 \]

Step 4: Find the x-intercept of the Tangent Line

The x-intercept is where the line crosses the x-axis (where \( y = 0 \)). Set the equation of the tangent line to zero and solve for \( x \):\[ 0 = -\pi x + \frac{\pi^2}{3} + 1 \]Solve for \( x \):\[ \pi x = \frac{\pi^2}{3} + 1 \]\[ x = \frac{\pi^2}{3\pi} + \frac{1}{\pi} \]\[ x = \frac{\pi}{3} + \frac{1}{\pi} \]

Step 5: Final Solution

The equation of the tangent line is \( y = -\pi x + \frac{\pi^2}{3} + 1 \). The x-intercept, where the tangent line crosses the x-axis, is at \( x = \frac{\pi}{3} + \frac{1}{\pi} \).

Key Concepts

Derivative Calculation

To find the tangent line to a graph, you need to calculate the derivative of the function. Derivatives tell us about the rate at which a function is changing at any given point. In essence, they provide the slope of the tangent line at a specific point. Consider the function \( y = 1 + x \sin(3x) \). Our goal is to find the derivative \( y' \).
In basic terms, a derivative \( y' \) is the limit of the function's average rate of change as the interval approaches zero. For more complex functions like \( x \sin(3x) \), we apply rules like the product rule and chain rule to find this. Once calculated, this derivative will allow us to determine the exact slope of the tangent line at any point on the curve.
Specifically, the derivative is used to find the slope of the tangent line at the point \( \left(\frac{\pi}{3}, 1\right) \). This step is crucial because it gives us the necessary slope \( m \), which is used in the point-slope form for the equation of a line.

Product Rule in Calculus

The product rule is a key concept in calculus that helps calculate the derivative of functions that are products of two or more simpler functions. When we have a function like \( x \sin(3x) \), which is the product of \( x \) and \( \sin(3x) \), the product rule comes into play. It states that for two functions \( u(x) \) and \( v(x) \), the derivative of their product \( u \cdot v \) is \( u'v + uv' \).
Applying this, we start with our functions:

• \( u(x) = x \)
• \( v(x) = \sin(3x) \)

Their derivatives are:

• \( u'(x) = 1 \) because the derivative of \( x \) is simple
• \( v'(x) = 3 \cos(3x) \) as determined through the chain rule in this context

Putting it all together, the derivative of \( x \sin(3x) \) using the product rule becomes \( \sin(3x) + 3x \cos(3x) \). This showcases the product rule's power in handling complex derivatives made up of simpler parts.

Chain Rule Application

The chain rule is another essential rule in calculus for differentiating compositions of functions. A composition of functions is like nesting one function inside another, such as \( \sin(3x) \). To handle this, the chain rule assists us by breaking down the different parts effectively.
The chain rule can be expressed as \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). Here, \( \sin(3x) \) is indeed such a composition where \( f(u) = \sin(u) \) and \( g(x) = 3x \).
Using this rule, you differentiate the inside function \( g(x) = 3x \) to get \( g'(x) = 3 \). For the outside function \( \sin(u) \), the derivative is \( \cos(u) \). Now apply the chain rule:

• Find \( f'(g(x)) = \cos(3x) \)
• Multiply by \( g'(x) = 3 \)

So, \( \frac{d}{dx}[\sin(3x)] = 3 \cos(3x) \). The chain rule thus simplifies the process of differentiating nested functions, making it possible to solve problems involving composite expressions.