Problem 26 - Calculus - AP Edition

Question

Find the equation of the line tangent to the graph of \(\sin x\) at (a) \(x=0\) (b) \(x=\pi\) (c) \(x=\pi / 4\)

Step-by-Step Solution

Answer

(a) \(y=x\), (b) \(y=-x+\pi\), (c) \(y=\frac{1}{\sqrt{2}}x - \frac{\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}}\)."

Step 1: Find the Derivative of the Function

The function given is \( f(x) = \sin x \). The first step in finding the tangent line is to find the derivative of this function, which represents the slope of the tangent line at any point \(x\). The derivative of \(f(x) = \sin x\) is \(f'(x) = \cos x\).

Step 2: Evaluate the Derivative at the Given Points

The slope of the tangent line at a given point is found by evaluating the derivative at that point.(a) For \(x = 0\), the slope is computed as:\[f'(0) = \cos(0) = 1\](b) For \(x = \pi\), the slope is:\[f'(\pi) = \cos(\pi) = -1\](c) For \(x = \pi/4\), the slope is:\[f'(\pi/4) = \cos(\pi/4) = \frac{1}{\sqrt{2}}\]

Step 3: Find the Value of the Function at Given Points

Next, determine the y-coordinate of the tangent line by evaluating the original function at the given points.(a) \(f(0) = \sin(0) = 0\)(b) \(f(\pi) = \sin(\pi) = 0\)(c) \(f(\pi/4) = \sin(\pi/4) = \frac{1}{\sqrt{2}}\)

Step 4: Use the Point-Slope Form to Write the Tangent Line Equations

The point-slope form of a line is given by \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point on the line.(a) For \(x = 0\):The point is \((0,0)\), and the slope is 1:\[y - 0 = 1(x - 0) \Rightarrow y = x\](b) For \(x = \pi\):The point is \((\pi, 0)\), and the slope is -1:\[y - 0 = -1(x - \pi) \Rightarrow y = -x + \pi\](c) For \(x = \pi/4\):The point is \((\pi/4, \frac{1}{\sqrt{2}})\), and the slope is \(\frac{1}{\sqrt{2}}\):\[y - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(x - \frac{\pi}{4})\]Simplifying, we get:\[y = \frac{1}{\sqrt{2}}x - \frac{\pi}{4\sqrt{2}} + \frac{1}{\sqrt{2}}\]

Key Concepts

Understanding the Derivative of the Sine Function

The sine function, written as \( f(x) = \sin x \), is a fundamental trigonometric function. The derivative of this function, with respect to \( x \), is of immense importance when we want to understand how the sine function behaves, especially when we need to find tangent lines to its graph. The derivative is a mathematical tool that gives us the rate at which the function is changing at any point.
For the sine function, its derivative is the cosine function:\( f'(x) = \cos x \). This means that for any specific value of \( x \), the slope of the tangent line to the sine curve is given by the cosine of that angle.
If you remember the unit circle, the cosine relates to the x-coordinate of the unit circle at a given angle, helping us see visually why \( \cos x \) gives us the slope. So, when you find a derivative of the sine function, think about it like finding a slope at a particular tangent point on a wavy sine graph.

Getting to Know the Point-Slope Form

Creating equations for tangent lines often involves a handy formula called the point-slope form. This form is incredibly useful because it allows us to express the equation of a line when we know a point on the line and the slope (which could come from a derivative).
The point-slope formula is:\[ y - y_1 = m(x - x_1) \]Here, \( m \) is the slope of the line, and \( (x_1, y_1) \) is a known point on the line. This formula elegantly ties together the points on the line and its steepness (slope).
To apply this to the tangent line problem, you first find the slope using the derivative, and determine a specific point on the curve using the original sine function, then plug these into the point-slope form to get the tangent line equation.

Evaluating Derivatives to Find Slopes

We evaluate derivatives to find the slope of the tangent line at specific points on the graph. Evaluating derivatives is simply substituting particular values into the derivative function.
In the case of the sine function with a derivative \( \cos x \), compute the cosine of the given \( x \)-values. For example:

• At \( x = 0 \), \( \cos(0) = 1 \). This tells us the slope of the tangent line at this point is 1.
• At \( x = \pi \), \( \cos(\pi) = -1 \), showing a slope of -1 at \( x = \pi \).
• At \( x = \pi/4 \), \( \cos(\pi/4) = \frac{1}{\sqrt{2}} \), giving a slope of \( \frac{1}{\sqrt{2}} \).

These slopes are the values that enter into your point-slope equation, allowing you to construct the equations for the tangent lines at these points on the sine wave.