Problem 25 - 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 \((c, f(c))\). $$ f(x)=2 \cos (x)-4 \sin (x), c=3 \pi / 4 $$

Step-by-Step Solution

Answer

The tangent line at \(x = \frac{3\pi}{4}\) is \(y + 3\sqrt{2} = \sqrt{2}(x - \frac{3\pi}{4})\).

Step 1: Find the Derivative

To find the equation of the tangent line, we need to know the slope at the point. We obtain this by finding the derivative of the function. Given: \[ f(x) = 2 \cos(x) - 4 \sin(x) \] Taking the derivative, we use the derivatives of sine and cosine: \[ f'(x) = -2 \sin(x) - 4 \cos(x) \]

Step 2: Calculate the Slope at the Point

Substitute \(x = \frac{3\pi}{4}\) into \( f'(x) \) to find the slope of the tangent line at this point: \[ f'\left(\frac{3\pi}{4}\right) = -2 \sin\left(\frac{3\pi}{4}\right) - 4 \cos\left(\frac{3\pi}{4}\right) \]Using trigonometric values: \[ \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} \quad \text{and} \quad \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \] Substitute these values in: \[ f'\left(\frac{3\pi}{4}\right) = -2 \left(\frac{\sqrt{2}}{2}\right) - 4 \left(-\frac{\sqrt{2}}{2}\right) \] \[ = - \sqrt{2} + 2\sqrt{2} = \sqrt{2} \]

Step 3: Find f(c)

Calculate \( f(c) \), or \( f\left(\frac{3\pi}{4}\right) \), using the original function: \[ f\left(\frac{3\pi}{4}\right) = 2 \cos\left(\frac{3\pi}{4}\right) - 4 \sin\left(\frac{3\pi}{4}\right) \] Substitute values: \[ = 2\left(-\frac{\sqrt{2}}{2}\right) - 4\left(\frac{\sqrt{2}}{2}\right) \]\[ = -\sqrt{2} - 2\sqrt{2} = -3\sqrt{2} \]

Step 4: Write the Equation of the Tangent Line

We now use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m = \sqrt{2} \) and \((x_1, y_1) = \left( \frac{3\pi}{4}, -3\sqrt{2} \right)\): \[ y + 3\sqrt{2} = \sqrt{2}(x - \frac{3\pi}{4}) \] This is the equation of the tangent line in point-slope form. You can also rearrange it to the slope-intercept form \(y = mx + b\) if needed.

Key Concepts

Derivatives

Derivatives are a cornerstone in calculus, helping us find how a function changes at any given point. Imagine you're driving a car; the derivative represents the speed of your car at a specific moment. For a function like \[ f(x) = 2 \cos(x) - 4 \sin(x) \]we want to find how the function behaves at any particular point.
To do this, we calculate the derivative, denoted as \( f'(x)\). The derivative of sine is cosine, and the derivative of cosine is negative sine. Therefore, applying these rules, we get:

• Derivative of \( 2 \cos(x) \) is \( -2 \sin(x) \).
• Derivative of \( -4 \sin(x) \) is \( -4 \cos(x) \).

This gives us:\[ f'(x) = -2 \sin(x) - 4 \cos(x) \]This derivative tells us the slope of the tangent line to the function at any point \(x\). Understanding how to find derivatives of trigonometric functions is key to solving such problems.

Trigonometric Functions

Trigonometric functions, like sine and cosine, describe angles and have values between -1 and 1. They are critically important in various fields such as physics, engineering, and even finance. Sine and cosine functions repeatedly occur in periodic phenomena, like sound waves or planetary motion.
Given the function:\[ f(x) = 2 \cos(x) - 4 \sin(x) \]we need specific trigonometric values to solve our original problem at \( x = \frac{3\pi}{4} \). The cosine and sine values for certain angles, such as \( \frac{3\pi}{4} \), are standard:

• \( \sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} \)
• \( \cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \)

Using these values, we compute the function's value and its slope at this angle. These computations give us the needed values to find the tangent line's equation. Mastering trigonometric functions can help unravel various practical and theoretical mathematical problems.

Point-Slope Form

Once you have the slope from the derivative and a point on the curve, you can write the equation of a tangent line. The point-slope form is\[ y - y_1 = m(x - x_1) \] where \(m\) is the slope, and \((x_1, y_1)\) is the point where the tangent touches the curve.

• For our problem, the slope \( m \) is \( \sqrt{2} \), a result from calculating \( f'(\frac{3\pi}{4}) \).
• The point \((x_1, y_1)\) is \( (\frac{3\pi}{4}, -3\sqrt{2}) \), obtained by evaluating \( f \) at \( c \).

Now, using these in the point-slope formula, we get:\[ y + 3\sqrt{2} = \sqrt{2}(x - \frac{3\pi}{4}) \]This equation gives us the tangent line at the specific point on the function's graph. Using the point-slope form is a fundamental skill in calculus to find linear tangents to curves quickly.