Problem 46 - Calculus Single Variable

Question

Find the tangent line to the graph of the given function at the given point. $$ f(x)=x \cos (x), \quad P=(\pi,-\pi) $$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = -x \).

Step 1: Identify the Function and the Point

We are given the function \( f(x) = x \cos(x) \) and the point \( P = (\pi, -\pi) \) at which we need to find the tangent line. The task is to find the slope of the tangent line at this point and use it to write the equation of the line.

Step 2: Find the Derivative of the Function

To find the slope of the tangent line, we need the derivative of the function, \( f(x) = x \cos(x) \). Use the product rule: if \( u(x) = x \) and \( v(x) = \cos(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \). Compute: \( u'(x) = 1 \) and \( v'(x) = -\sin(x) \), so:\[ f'(x) = 1 \cdot \cos(x) + x \cdot (-\sin(x)) = \cos(x) - x \sin(x). \]

Step 3: Evaluate the Derivative at the Given Point

Now, substitute \( x = \pi \) into \( f'(x) = \cos(x) - x \sin(x) \) to find the slope at point \( P \):\[ f'(\pi) = \cos(\pi) - \pi \sin(\pi) = -1 - 0 = -1. \]

Step 4: Write the Equation of the Tangent Line

The slope of the tangent line is \( -1 \). We use the point-slope form \( y - y_1 = m(x - x_1) \). Here, \( m = -1 \), \( x_1 = \pi \), and \( y_1 = -\pi \):\[ y - (-\pi) = -1(x - \pi). \]Simplify to:\[ y + \pi = -x + \pi. \]Thus, the equation of the tangent line is:\[ y = -x. \]

Step 5: Verification (Optional)

Ensure the tangent line passes through the given point \((\pi, -\pi)\). Substitute \(x = \pi\) into \( y = -x \):\[ y = -\pi, \]which matches the point \((\pi, -\pi)\). Therefore, the equation \( y = -x \) is correct.

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus. It helps us find the rate at which one quantity changes with respect to another. Think of it as a way to measure how the function behaves at a particular point. In the context of a curve, the derivative gives us the slope of the tangent line to the function at any point.
For a function like \( f(x) = x \cos(x) \), finding the derivative is crucial to understand how this function changes around a specific point. Using the rules of differentiation, we can determine the exact instant rate of change, which helps us compute the slope of the tangent line. By determining \( f'(x) \), we know how \( f \) is moving directly at \( x \), providing a stepping stone to sketch a tangent to the curve.

Product Rule

The product rule is a key technique in calculus, used for finding the derivative of a product of two functions. It tells us how to handle derivatives when two functions are multiplied together.
In our function \( f(x) = x \cos(x) \), we have two parts: \( u(x) = x \) and \( v(x) = \cos(x) \). The product rule states:

• First, find the derivative of each function: \( u'(x) = 1 \) and \( v'(x) = -\sin(x) \).
• Then, multiply the first function by the derivative of the second and the second by the derivative of the first: \( f'(x) = u'(x)v(x) + u(x)v'(x) \).
• This gives us \( f'(x) = 1 \cdot \cos(x) + x \cdot (-\sin(x)) = \cos(x) - x \sin(x) \).

By using the product rule, we've computed the derivative, which is essential for finding the slope of the tangent line.

Slope of Tangent Line

The slope of the tangent line at a specific point tells us the direction the curve is heading at that point. In simpler terms, it's like drawing a line that just 'touches' the curve at one spot without crossing it. To find this value, we evaluate the derivative at the given point.
For the point \( P = (\pi, -\pi) \) on our function, we substitute \( x = \pi \) into our derivative \( f'(x) = \cos(x) - x \sin(x) \). This gives us:

• \( f'(\pi) = \cos(\pi) - \pi \sin(\pi) \).
• Since \( \cos(\pi) = -1 \) and \( \sin(\pi) = 0 \), it simplifies to \( -1 - 0 = -1 \).

Thus, the slope of the tangent line at this point is \(-1\), indicating the line is decreasing.

Point-Slope Form

Once we know the slope, the point-slope form allows us to write the equation of the tangent line. This form is \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point and \(m\) is the slope.
For our point \( P = (\pi, -\pi) \) and slope \( m = -1 \), the formula becomes:

• \( y - (-\pi) = -1(x - \pi) \).
• Expanding this, we get: \( y + \pi = -x + \pi \).
• Simplifying gives the equation: \( y = -x \).

The point-slope form is incredibly useful because it clearly incorporates the slope and specific point, providing a direct way to get the equation of the tangent line efficiently.