Problem 53 - Calculus Volume 1

Question

Find the linear approximation \(L(x)\) to \(y=f(x)\) near \(x=a\) for the function. \(f(x)=\sin x, a=\frac{\pi}{2}\)

Step-by-Step Solution

Answer

The linear approximation is \(L(x) = 1\).

Step 1: Calculate the Derivative

The first step is to find the derivative of the function \(f(x) = \sin x\). The derivative of \(\sin x\) with respect to \(x\) is \(\cos x\). Therefore, \(f'(x) = \cos x\).

Step 2: Evaluate Function and Derivative at a.

Evaluate both the function \(f(x)\) and its derivative \(f'(x)\) at \(x = a\), where \(a = \frac{\pi}{2}\). First, find \(f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1\). Next, calculate \(f'\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0\).

Step 3: Write the Formula for Linear Approximation

The linear approximation \(L(x)\) of the function near \(x = a\) is given by: \(L(x) = f(a) + f'(a)(x - a)\).

Step 4: Substitute Values into Linear Approximation Formula

Substitute \(f(a) = 1\), \(f'(a) = 0\), and \(a = \frac{\pi}{2}\) into the linear approximation formula: \(L(x) = 1 + 0 \cdot (x - \frac{\pi}{2})\).

Step 5: Simplify the Expression

Simplify the expression \(L(x) = 1 + 0 \cdot (x - \frac{\pi}{2})\) to get \(L(x) = 1\).

Key Concepts

Derivative of Sine Function

To understand the linear approximation of a function, we must first know how to find its derivative. For a sine function, like \( f(x) = \sin x \), the derivative tells us how the function changes at any point along its curve. The derivative of \( \sin x \) is \( \cos x \).

• The derivative \( \cos x \) is a result of differentiating \( \sin x \).
• This derivative is essential as it helps in finding the slope of the function at any point.

By understanding the derivative, we can start to see how small changes in \(x\) affect \(y\). This is a critical part of finding the linear approximation.

Evaluate Derivative at a Point

Once we have a derivative, we need to evaluate it at a specific point. Evaluating the derivative at a given point tells us the exact rate of change of the function at that point. For example, if we have \( f(x) = \sin x \) and we want to evaluate at \( a = \frac{\pi}{2} \):

• First, we calculate \( f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1\).
• Next, we find \( f'\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0\).

Knowing both the function value and its derivative at \( a = \frac{\pi}{2} \) is vital for forming the linear approximation.

Linearization Formula

With both the function value and the derivative evaluated at a point, we apply the linearization formula. The linear approximation formula is a perfect tool to understand how the function behaves near a specific point. The formula is:
\[ L(x) = f(a) + f'(a)(x - a) \]

• \(L(x)\) is the linear approximation of \(f(x)\) near \(x = a\).
• In our case, substitute \( f(a) = 1 \), \( f'(a) = 0 \), and \(a = \frac{\pi}{2}\).

This substitution yields the linear approximation that gives an approximate value of the function near \(x = a\). The formula represents the line tangent to the function curve at \(a\).

Simplifying Expressions

The last step in finding the linear approximation is to simplify the expression derived from the linearization formula. Often, simplifying can turn a complex-looking expression into a simple one, making it easier to interpret. Based on our calculation:

• Substitute the values: \(L(x) = 1 + 0 \cdot (x - \frac{\pi}{2})\).
• Simplify to \(L(x) = 1\).

Simplifying involved removing the zero multiplier, which didn't affect the function's value. As a result, the linear approximation, \(L(x) = 1\), shows that near \(x = \frac{\pi}{2}\), \(\sin x\) is approximately equal to 1. Simplification helps clarify the result and makes the approximation easier to use.