Problem 55 - Calculus Volume 1

Question

Find the linear approximation \(L(x)\) to \(y=f(x)\) near \(x=a\) for the function. \(f(x)=\sin ^{2} x, a=0\)

Step-by-Step Solution

Answer

The linear approximation \(L(x)\) of \(f(x) = \sin^2 x\) near \(x=0\) is \(L(x) = 0\).

Step 1: Find the Function's Value at x=a

The first step in finding the linear approximation is to evaluate the function at the point of interest, which is where the linear approximation will be based. In this case, evaluate the function at \(x=0\). Given \(f(x)=\sin^{2} x\), we have \(f(0)=\sin^{2}(0) = 0\).

Step 2: Calculate the Derivative of the Function

Next, find the derivative of the function \(f(x)=\sin^{2} x\). Using the chain rule, the derivative is \(f'(x) = 2\sin x \cdot \cos x = \sin(2x)\).

Step 3: Evaluate the Derivative at x=a

Now, evaluate the derivative at \(x=0\) to find the slope of the tangent line at this point. Plugging \(x=0\) into the derivative gives \(f'(0)=\sin(2 \cdot 0) = \sin(0) = 0\).

Step 4: Write the Linear Approximation Formula

The linear approximation of the function near \(x=a\) is given by the formula: \(L(x) = f(a) + f'(a)(x-a)\). Inserting \(f(0) = 0\) and \(f'(0) = 0\), the formula simplifies to \(L(x) = 0 + 0 \cdot (x - 0) = 0\).

Step 5: Confirm the Result

Recheck the calculations to ensure no errors were made. The function value \(f(0)\) and its derivative at \(x=0\) both equal zero, confirming \(L(x) = 0\) as the correct linear approximation.

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of a composition of functions. When you have a function of a function, like \(f(g(x))\), you use the chain rule to take the derivative. This rule is essential when dealing with composite functions because it allows us to "chain" the inner and outer derivatives together.
Here’s the basic idea:

• Find the derivative of the outer function with respect to the inner function.
• Multiply it by the derivative of the inner function with respect to x.

In the exercise, the function \(f(x) = \sin^{2} x\) is a composition where the outer function is \(u^2\) and the inner function is \(\sin x\). Using the chain rule, the derivative is \(2 \sin x \cdot \cos x\), which simplifies to \(\sin(2x)\). It’s like peeling an orange; you first peel the outer layer (outer function) and then get to the segments inside (inner function) with their own derivatives to handle.

Derivative

Derivatives play a central role in calculus and describe how a function changes at any given point.
The derivative of a function gives us the slope of the tangent line to the function’s graph at any point, describing how steep the graph is at that point. It’s essentially the "rate of change" of the function with respect to its variable.
In our exercise, finding the derivative of \(f(x) = \sin^{2} x\) was key to determining the slope for the linear approximation. We applied the chain rule, resulting in \(f'(x) = \sin(2x)\), which shows how \(f(x)\) changes as \(x\) changes. For example, at \(x=0\), the derivative \(f'(0) = 0\) indicates that the function is constant (or flat) at this point, and explains why the linear approximation is simply \(L(x) = 0\).

Sine Function

The sine function is a fundamental trigonometric function that generates waves. Represented as \(\sin x\), it gives the y-coordinate of a point on the unit circle corresponding to an angle \(x\). It’s characterized by its periodic nature, repeating every \(2\pi\) radians, and by its smooth, wave-like graph that oscillates between -1 and 1.
In this exercise, we squared the sine function: \(f(x) = \sin^2 x\). When we evaluate this function at \(x = 0\), we find \(f(0) = \sin^2(0) = 0\). Despite being transformed (squared), the function retains the periodic properties of the sine function, helping to determine the behavior near \(x = 0\) for the linear approximation.

Tangent Line

A tangent line to a curve at a given point is the straight line that "just touches" the curve at that point, sharing the same slope as the curve there.
You can think of it as the "best linear approximation" to the curve at that specific point.
In calculus, this idea is important when we want to approximate the value of a function near a point. The slope of this tangent line is given by the derivative of the function at the point of interest.

• For the function \(f(x) = \sin^2 x\) in our exercise, the tangent line at \(x = 0\) has a slope of \(f'(0) = 0\).
• Thus, the tangent line, which is the linear approximation, is horizontal at \(x = 0\), and expressed as \(L(x) = 0\).

This concept beautifully showcases how derivatives can simplify the complex shapes of curves into simple linear representations at points, making calculations and predictions much easier.