Problem 23 - 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 \((c, f(c))\). $$ f(x)=\sin (x), c=\pi / 3 $$

Step-by-Step Solution

Answer

The tangent line equation is \( y - \frac{\sqrt{3}}{2} = \frac{1}{2}\left(x - \frac{\pi}{3}\right) \).

Step 1: Find the Derivative of the Function

The equation of the tangent line requires the slope, which is given by the derivative of the function. Start by finding the derivative of the function \(f(x) = \sin(x)\). The derivative is \(f'(x) = \cos(x)\).

Step 2: Evaluate the Derivative at c

Substitute \(c = \frac{\pi}{3}\) into the derivative to find the slope of the tangent line at this point. Calculate \(f'\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right)\). Since \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\), the slope is \(\frac{1}{2}\).

Step 3: Calculate the Function Value at c

Find the function value at \(x=c\) by calculating \(f\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)\). We know that \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), so \(f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\).

Step 4: Write the Equation of the Tangent Line

Use the point-slope form of a line equation, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is \(\left(\frac{\pi}{3}, \frac{\sqrt{3}}{2}\right)\) and \(m = \frac{1}{2}\). Substituting these values gives: \[ y - \frac{\sqrt{3}}{2} = \frac{1}{2}\left(x - \frac{\pi}{3}\right) \] This is the equation of the tangent line.

Key Concepts

Derivative

A derivative helps us understand how a function changes at a particular point, acting as a foundation for finding the slope of a tangent line to a curve. For our function, we have \( f(x) = \sin(x) \). Taking the derivative means calculating \( f'(x) \), which in this case becomes \( \cos(x) \). This derivative value at any point on the function will tell us how sharply or gently the function is increasing or decreasing at that location. Without the derivative, finding the slope at a single point is akin to shooting in the dark.

Slope of a Function

The slope of a function indicates the rate at which the function value y changes as x changes. It's the steepness or incline of a function at any given point. When dealing with curves, such as \( \sin(x) \), the slope of the tangent line is the derivative evaluated at the specific point of interest. In our problem, substituting \( x = \frac{\pi}{3} \) into our derivative \( f'(x) = \cos(x) \) gives us the value \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \). This tells us that at \( x = \frac{\pi}{3} \), the function rises by half a unit of y for every unit increase in x.

Trigonometric Functions

Trigonometric functions like \( \sin(x) \) and \( \cos(x) \) are fundamental to understanding periodic phenomena. They help us model waves, oscillations, and other cyclical patterns. Specifically, \( \sin(x) \) represents the y-coordinate of a point on the unit circle corresponding to an angle \( x \). When we evaluated \( \sin\left(\frac{\pi}{3}\right) \), which equaled \( \frac{\sqrt{3}}{2} \), we found the y-value where our tangent line touches the function. Recognizing these values using trigonometric identities simplifies solving problems involving angles and curves.

Point-Slope Form

The point-slope form provides a straightforward way to write the equation of a line. It is particularly useful when you know a point on the line and its slope. The formula is \( y - y_1 = m(x - x_1) \). For our exercise, the known point is \( \left(\frac{\pi}{3}, \frac{\sqrt{3}}{2}\right) \) and the slope is \( \frac{1}{2} \). By substituting these into the point-slope form, we create our tangent line equation: \[ y - \frac{\sqrt{3}}{2} = \frac{1}{2}\left(x - \frac{\pi}{3}\right) \]. This provides a tidy and efficient way to represent the line that just skims the curve at our specified point.