Problem 18 - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

To find the tangent line to the curve \( y = \sin x \) at a given point \( x = a \), we need to find the derivative of the function, \( y' = \frac{dy}{dx} \), and evaluate it at \( x = a \). This gives us the slope of the tangent line. We then use the point-slope form of a linear equation to find the equation of the tangent line.

Step 2: Differentiate the Function

The derivative of \( y = \sin x \) is \( y' = \cos x \). This derivative tells us the slope of the tangent line at any point \( x = a \) on the curve.

Step 3: Calculate the Slope at Each Point

(a) At \( x = 0 \), the slope is \( \cos(0) = 1 \). (b) At \( x = \pi \), the slope is \( \cos(\pi) = -1 \). (c) At \( x = \pi/4 \), the slope is \( \cos(\pi/4) = \frac{\sqrt{2}}{2} \).

Step 4: Determine the Equation of the Tangent Line for Each Case

We use the point-slope form of a line, given by \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency. (a) At \( x = 0 \), the point is \((0, \sin(0)) = (0, 0)\). The equation is: \[ y - 0 = 1 (x - 0) \rightarrow y = x \](b) At \( x = \pi \), the point is \((\pi, \sin(\pi)) = (\pi, 0)\). The equation is: \[ y - 0 = -1(x - \pi) \rightarrow y = -x + \pi \](c) At \( x = \pi/4 \), the point is \((\pi/4, \sin(\pi/4)) = (\pi/4, \frac{\sqrt{2}}{2})\). The equation is: \[ y - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) \rightarrow y = \frac{\sqrt{2}}{2}x + \frac{\sqrt{2}}{2}(1 - \frac{\pi}{4}) \]

Key Concepts

Derivative of Sine

A derivative helps us understand how a function is changing at any point. When we talk about the derivative of the sine function, we're examining how the function \(y = \sin x\) changes as \(x\) changes. The derivative of \(\sin x\) is \(\cos x\). This gives us the slope of the tangent line at any particular point on the curve.

• The derivative is found by differentiating \(\sin x\), which directly yields \(\cos x\).
• For any point \(x\) on the sine curve, \(\cos x\) will tell us how steep the curve is at that point.
• This derivative is crucial when intending to find tangent lines, as it directly tells us the slope of the tangent.

By knowing the derivative, we can predict and understand the behavior of the sine curve at any location, which is a fundamental aspect when dealing with calculus problems involving changes and tangents.

Equation of a Tangent Line

A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. To find the equation of this line, we rely on the slope given by the derivative and the specific point of tangency.

• Once the slope is known, calculated from the derivative, the tangent line can be fully determined.
• The process involves plugging the slope into an equation of a line. For a curve \(y = \sin x\), the slope at any point \(x = a\) is given by \(\cos(a)\).
• Using the slope \(m = \cos a\) and the point \((a, \sin a)\), the tangent line can be expressed.

To illustrate, for example at \(x = 0\), since \(\sin(0) = 0\) and \(\cos(0) = 1\), the tangent line becomes \(y = x\). Similarly, this process is applied for other points to find their respective tangent lines.

Point-Slope Form

In coordinate geometry, the point-slope form is a simple way to write the equation of a line. This form is very handy because it requires just a single point on the line and the slope. The equation is expressed as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope.

• This form is particularly helpful for deriving the equation of a tangent line because once you know the slope (from the derivative) and the point (from the curve), you can easily write the line's equation.
• For instance, to find the tangent line at \(x = \pi/4\) for \(y = \sin x\):
  • Find the slope, which is \(\cos(\pi/4) = \frac{\sqrt{2}}{2}\).
  • Identify the point of tangency: \((\pi/4, \sin(\pi/4)) = (\pi/4, \frac{\sqrt{2}}{2})\).
  • Apply these to the point-slope formula: \(y - \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}(x - \frac{\pi}{4})\).

Understanding the point-slope form enables us to translate our understanding of a curve's behavior at a single point into the precise line that accurately represents its tangent at that location.