Problem 26 - Calculus Single Variable

Question

A function \(f\) and a value \(c\) are given. Find an equation of the tangent line to the graph of \(f\) at \((c, f(c))\). $$ f(x)=3 x-\cos (x), c=\pi / 2 $$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 4x - 2\pi \).

Step 1: Calculate f(c)

To find the y-coordinate of the point of tangency, we need to evaluate the function at the given point. The function is \( f(x) = 3x - \cos(x) \) and \( c = \frac{\pi}{2} \). We substitute \( x = \frac{\pi}{2} \) into \( f(x) \): \[ f\left(\frac{\pi}{2}\right) = 3\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right) = \frac{3\pi}{2} - 0 = \frac{3\pi}{2} \] Thus, the point of tangency is \( \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \).

Step 2: Find Derivative of f(x)

Next, we need to find the derivative \( f'(x) \) which represents the slope of the tangent line. Differentiate \( f(x) = 3x - \cos(x) \): \[ f'(x) = \frac{d}{dx}(3x) - \frac{d}{dx}(\cos(x)) = 3 - (-\sin(x)) = 3 + \sin(x) \]

Step 3: Evaluate f'(c)

Substitute \( c = \frac{\pi}{2} \) into the derivative to find the slope at the given point: \[ f'\left(\frac{\pi}{2}\right) = 3 + \sin\left(\frac{\pi}{2}\right) = 3 + 1 = 4 \] Therefore, the slope of the tangent line at \( c = \frac{\pi}{2} \) is 4.

Step 4: Write the equation of the tangent line

With the point \( (\frac{\pi}{2}, \frac{3\pi}{2}) \) and the slope 4, we use the point-slope formula for the line: \[ y - y_1 = m(x - x_1) \] Substitute \( m = 4 \), \( x_1 = \frac{\pi}{2} \), and \( y_1 = \frac{3\pi}{2} \):\[ y - \frac{3\pi}{2} = 4\left(x - \frac{\pi}{2}\right) \] Simplify to get the equation in slope-intercept form:\[ y = 4x - 2\pi \]

Key Concepts

Derivative Calculation

To find the equation of a tangent line, we first need its slope. The slope of the tangent line is the derivative of the function at a particular point. When given a function like \( f(x) = 3x - \cos(x) \), finding its derivative involves applying basic differentiation rules.

The derivative \( f'(x) \) tells us how the function's output changes as the input changes. For the function \( f(x) = 3x - \cos(x) \):

• The derivative of \( 3x \) is simply \( 3 \) since the constant rule tells us that the derivative of \( ax \) is \( a \).
• The derivative of \( -\cos(x) \) is \( \sin(x) \). This follows from the rule that states the derivative of \( \cos(x) \) is \( -\sin(x) \), so negating it gives \( \sin(x) \).

Therefore, the derivative \( f'(x) = 3 + \sin(x) \). Calculating this derivative allows us to determine the slope of the tangent line at any point \( x \).

Point-Slope Form

Once we have the slope from the derivative, we can use it in the equation of the tangent line. The point-slope form is an effective way to write the equation of a line when a point on the line and the slope are known. The formula is:

• \( y - y_1 = m(x - x_1) \)

where \( (x_1, y_1) \) is the point of tangency, and \( m \) is the slope.

From the original problem, we found the point of tangency to be \( \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \) and the slope \( m = 4 \). Substituting these into the point-slope form gives:\[ y - \frac{3\pi}{2} = 4 \left(x - \frac{\pi}{2}\right) \] This expression describes the tangent line specific to our function at point \( c = \frac{\pi}{2} \).

Evaluating Functions

To find the coordinates of the point of tangency, we must evaluate the given function at the specified point \( c \). This process involves substituting \( c \) into the function.

Given the function \( f(x) = 3x - \cos(x) \) and \( c = \frac{\pi}{2} \), substitute \( x = \frac{\pi}{2} \) to find \( f(c) \): \[ f\left(\frac{\pi}{2}\right) = 3\left(\frac{\pi}{2}\right) - \cos\left(\frac{\pi}{2}\right) \]

• \( 3\left(\frac{\pi}{2}\right) \) simplifies to \( \frac{3\pi}{2} \).
• Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), the function simplifies to \( \frac{3\pi}{2} \).

Thus, the point of tangency is \( \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \), using which we can proceed to find the tangent line equation.

Trigonometric Functions

Trigonometric functions like \( \cos \) and \( \sin \) appear in many different types of mathematical problems, including the one we are examining here. These functions measure relationships in right triangles and have certain values at specific points.

In our function \( f(x) = 3x - \cos(x) \), we encounter \( \cos(x) \). Important points to know for \( \cos(x) \) are:

• \( \cos(0) = 1 \)
• \( \cos\left(\frac{\pi}{2}\right) = 0 \)
• \( \cos(\pi) = -1 \)

In this exercise, substituting \( \frac{\pi}{2} \) into \( \cos(x) \) gives us 0, which is crucial for simplification steps. Similarly, knowing \( \sin\left(\frac{\pi}{2}\right) = 1 \) helps determine the derivative \( f'(x) = 3 + \sin(x) \) evaluated at \( x = \frac{\pi}{2} \), giving a slope of 4. Understanding these fundamental values allows us to solve problems effortlessly involving trigonometric functions.