Question

Find the linear approximation to the given functions at the specified points. Plot the function and its linear approximation over the indicated interval. $$ f(x)=\sqrt{1-x^{2}} \text { at } a=0,[-1,1] $$

Short answer

The linear approximation is \( L(x) = 1 \) and remains constant over \([-1, 1]\).

Step-by-step solution

Find the derivative of the function

We need to find the derivative of the function \( f(x) = \sqrt{1 - x^2} \). Start by applying the chain rule. Let \( u = 1-x^2 \), so \( f(x) = \sqrt{u} = u^{1/2} \). The derivative \( \frac{du}{dx} = -2x \). Apply the chain rule: \( \frac{df}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{1- x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}} \).

Calculate the value of the function and its derivative at \(a=0\)

Substitute \( x = 0 \) into \( f(x) \) to find \( f(0) = \sqrt{1-0^2} = 1 \). Now, find the value of the derivative at \( x = 0 \): \( f'(x) = \frac{-x}{\sqrt{1-x^2}} \). Substituting \( x = 0 \) gives \( f'(0) = 0 \).

Write the linear approximation formula

The linear approximation, or tangent line, at \( x=a \) can be written as \( L(x) = f(a) + f'(a)(x-a) \). Since \( a = 0 \), \( f(0) = 1 \), and \( f'(0) = 0 \), we have: \[ L(x) = 1 + 0 \cdot (x-0) = 1. \]

Plot the function and its linear approximation over the interval

Plot the original function \( f(x) = \sqrt{1-x^2} \) and the linear approximation \( L(x) = 1 \) over the interval \([-1, 1]\). The function is a semicircle's upper half centered at the origin with radius 1, and the linear approximation graphed as a horizontal line at \( y = 1 \).

Key concepts

Derivative Calculation

The process of finding a derivative is fundamental in calculus, as it helps measure how a function changes at any point. For the function \( f(x) = \sqrt{1-x^2} \), the derivative calculation involves differentiating a square root, which can be challenging without applying the right rules.

One efficient method for this task is the chain rule. We start by introducing a substitution. Let \( u = 1-x^2 \), so we can rewrite \( f(x) \) as \( u^{1/2} \).

The chain rule allows us to differentiate a composite function. It involves calculating \( \frac{du}{dx} \) first, which is \(-2x\). The derivative of \( u^{1/2} \) is \( \frac{1}{2\sqrt{u}} \), so combining these using the chain rule gives the final derivative \( f'(x) = \frac{-x}{\sqrt{1-x^2}} \). This derivative tells us the slope of the tangent line to the curve at any point \( x \).

Chain Rule

The chain rule is a powerful tool in calculus for computing the derivative of composite functions. Imagine a function inside another function; the chain rule helps us peel back the outer layer to get to the derivative.

For the function \( f(x) = \sqrt{1 - x^2} \), we see it consists of an inner function \( u = 1-x^2 \) and the outer function \( \sqrt{u} \). The chain rule formula is: \( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \).

In this exercise, we calculated \( \frac{du}{dx} = -2x \) and \( \frac{df}{du} = \frac{1}{2\sqrt{u}} \). So, applying the chain rule gives: \( \frac{df}{dx} = \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) \), leading us to \( \frac{-x}{\sqrt{1-x^2}} \). This technique simplifies the differentiation process significantly, especially for nested functions.

Tangent Line

The concept of a tangent line is crucial for understanding linear approximation. A tangent line touches a curve at exactly one point and represents the best linear approximation of the curve at that point.

The linear approximation formula for a function \( f \) at a point \( a \) is given by \( L(x) = f(a) + f'(a)(x-a) \).

In our problem, we found \( f(0) = 1 \) and \( f'(0) = 0 \). Therefore, the tangent line at \( a = 0 \) is simply \( L(x) = 1 \). This horizontal line indicates no change in \( y \) with respect to \( x \) when \( x \) is around 0 in this case. The tangent line provides a straightforward linear model for approximating complex functions near the point of tangency.

Plotting Functions

Visualizing functions by plotting is an essential step to grasp their behavior. It helps to better understand the relationship between the function and its linear approximation.

For the original function \( f(x) = \sqrt{1-x^2} \), you plot the upper half of a circle with radius 1 over the interval \([-1, 1]\). This semicircle touches the x-axis at -1 and 1 and reaches its peak at 0 (origin).

The linear approximation, represented by the function \( L(x) = 1 \), is a horizontal line across the whole interval. Plotting these together allows you to see how well the line approximates \( f(x) \) near \( a = 0 \).

• Circle with radius: visualizes symmetry and limit points.
• Linear approximation relevance: shows proximity to actual values around \( x=0 \).

Graphing these helps to conceptually connect algebraic results to geometric representations, strengthening understanding of linear approximations.