Problem 26 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

(a) \(y = x\), (b) \(y = -x + \pi\), (c) \(y = \frac{\sqrt{2}}{2}x - \frac{\sqrt{2}\pi}{8} + \frac{\sqrt{2}}{2}\)."

Step 1: Understanding the Tangent Line

The equation of the tangent line to a function \(f(x)\) at a point \(x = a\) is given by \(y = f'(a)(x - a) + f(a)\), where \(f'(a)\) is the derivative of \(f(x)\) evaluated at \(x = a\). We need to find this equation for the function \(f(x) = \sin x\) and the given points.

Step 2: Calculating the Derivative of \(\sin x\)

The derivative of \(f(x) = \sin x\) is \(f'(x) = \cos x\). We will use this derivative to find the slope of the tangent line at the given points \(x = 0\), \(x = \pi\), and \(x = \pi/4\).

Step 3: Finding the Tangent Line at \(x=0\)

Evaluate \(f(0) = \sin 0 = 0\) and \(f'(0) = \cos 0 = 1\). Substitute into the tangent line formula: \[y = 1 \cdot (x - 0) + 0 = x.\] So, the tangent line equation at \(x = 0\) is \(y = x\).

Step 4: Finding the Tangent Line at \(x=\pi\)

Evaluate \(f(\pi) = \sin \pi = 0\) and \(f'(\pi) = \cos \pi = -1\). Substitute into the tangent line formula: \[y = -1 \cdot (x - \pi) + 0 = -x + \pi.\] So, the tangent line equation at \(x = \pi\) is \(y = -x + \pi\).

Step 5: Finding the Tangent Line at \(x=\pi/4\)

Evaluate \(f(\pi/4) = \sin \pi/4 = \frac{\sqrt{2}}{2}\) and \(f'(\pi/4) = \cos \pi/4 = \frac{\sqrt{2}}{2}\). Substitute into the tangent line formula: \[y = \frac{\sqrt{2}}{2} \cdot \left( x - \frac{\pi}{4} \right) + \frac{\sqrt{2}}{2}.\] Thus, the tangent line equation at \(x = \pi/4\) is \(y = \frac{\sqrt{2}}{2}x - \frac{\sqrt{2}\pi}{8} + \frac{\sqrt{2}}{2}\).

Key Concepts

Derivative of Sine

A derivative is a key concept in calculus, used to determine the rate at which a function changes. For the sine function, the derivative is particularly interesting and fundamental in understanding the behavior of trigonometric functions in calculus.

The derivative of \((x) = \sin x\) is \(\cos x\). This means that for any value of \(x\), the slope of the tangent line to the curve of \(\sin x\) is given by \(\cos x\).

If you think of the sine graph as a wave, the derivative tells us how steep the wave is at any point. When you see the derivative \(\cos x\), it's like an instruction manual telling you if the wave is going uphill or downhill at \(x\). This concept is pivotal when solving problems that involve rates of change, like the motion of a pendulum.

Trigonometric Functions

Trigonometric functions are mathematical functions like sine, cosine, and tangent, which relate the angles of triangles to the lengths of their sides. They are periodic, meaning they repeat their values in regular intervals.

In the context of calculus, trigonometric functions are vital for analyzing oscillatory phenomena such as waves and circular motion. For example:

• \(\sin x\) has a period of \(2\pi\), completing a full cycle every \(2\pi\) units.
• \(\cos x\) shares the same period, complementing the sine curve perfectly.

When dealing with these functions, having a clear grasp of their shapes and properties is very helpful in visualizing problems. A solid understanding of how these functions behave will ease the process of solving calculus problems involving them.

Calculus Problem Solving

Problem solving in calculus often involves a clear sequence of steps to tackle seemingly complex problems. For functions like \(\sin x\), this usually entails understanding derivatives and how they transform into tangible solutions.

Let's break it down further:

• **Identify the Function**: First, you need to know what function you're dealing with, like \(\sin x\) in our case.
• **Derivative**: Next, find the derivative, as it helps in understanding how the function behaves at any given point. For \(\sin x\), you take \(\cos x\).
• **Tangent Line**: Use the derivative to find the equation of the tangent line at specific points. Tangents give insights into the function's slope at those points.
• **Substitution**: Insert the points into your equations to get precise solutions, just like when finding tangent lines at \(x = 0, \pi, \pi/4\).

Calmly moving step by step, you will find that calculus problems become much more manageable and even enjoyable! It's about dissecting the problem, which makes solving exercises such as tangent lines of trigonometric functions a walk in the park.