Problem 55 - Calculus Single Variable

Question

Find the linearization of the given function \(f(x)\) at the given point \(c\). $$ f(x)=\arcsin (x), c=1 / 2 $$

Step-by-Step Solution

Answer

The linearization of \( f(x) = \arcsin(x) \) at \( c = \frac{1}{2} \) is \( L(x) = \frac{\pi}{6} + \frac{2}{\sqrt{3}}x - \frac{1}{\sqrt{3}} \).

Step 1: Understand Linearization

Linearization is the process of approximating a function near a point by a line. The linearization of a function \( f \) at point \( x = c \) is given by the formula:\[ L(x) = f(c) + f'(c)(x-c). \]

Step 2: Evaluate the Function at Point c

First, find \( f(c) \) by evaluating the function at \( c = \frac{1}{2} \):\[ f\left(\frac{1}{2}\right) = \arcsin\left(\frac{1}{2}\right). \]Since \( \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \), we have \( f\left(\frac{1}{2}\right) = \frac{\pi}{6} \).

Step 3: Compute the Derivative of the Function

The derivative of the function \( f(x) = \arcsin(x) \) is \( f'(x) = \frac{1}{\sqrt{1-x^2}} \).

Step 4: Evaluate the Derivative at Point c

Next, substitute \( c = \frac{1}{2} \) into the derivative:\[ f'\left(\frac{1}{2}\right) = \frac{1}{\sqrt{1-\left(\frac{1}{2}\right)^2}} = \frac{1}{\sqrt{\frac{3}{4}}} = \frac{2}{\sqrt{3}}. \]

Step 5: Write the Linearization Formula

Substitute \( f(c) \), \( f'(c) \), and \( c \) into the linearization formula:\[ L(x) = \frac{\pi}{6} + \frac{2}{\sqrt{3}} \left(x - \frac{1}{2}\right). \]

Step 6: Simplify the Linearization Expression

Further simplify the expression to find the linear approximation of \( f(x) \):\[ L(x) = \frac{\pi}{6} + \frac{2}{\sqrt{3}}x - \frac{1}{\sqrt{3}}. \]

Key Concepts

Derivative

A derivative represents the rate at which a function is changing at any given point. For a smooth curve described by a function \( f(x) \), the derivative \( f'(x) \) gives the slope of the tangent line to the curve at a point \( x \).

For example, if you have a position-time graph of a moving object, the derivative of the object's position with respect to time gives its velocity at each instant. In this case, the function \( f(x) = \arcsin(x) \) is an inverse trigonometric function, which describes the angle whose sine is \( x \).

To find the derivative of \( \arcsin(x) \), there is a standard result: \( f'(x) = \frac{1}{\sqrt{1-x^2}} \).

This derivative is particularly useful here because it helps us understand how \( \arcsin(x) \) changes when \( x \) changes in the vicinity of a particular point. Calculating \( f'(1/2) \) allows us to see not only the current value of \( \arcsin(x) \) at \( x = 1/2 \), but also the rate at which it is changing.

Function Approximation

Function approximation is a technique used to represent complex functions with simpler ones, most often linear functions, that are easier to work with. Linearization is a common method of function approximation. The idea behind it is that you can approximate a function near a certain point \( c \) with a straight line, using the point and the slope of the function at that point.

The linearization of a function \( f \) at point \( x = c \) can be represented by the formula:

• \( L(x) = f(c) + f'(c)(x - c) \)

When applying this method, you take the value of the original function at \( c \), \( f(c) \), and add it to the derivative at \( c \), \( f'(c) \), times the deviation \( x - c \) from the point of interest. This provides a line that closely approximates the function around \( c \).

In practice, this is incredibly helpful in scenarios where the exact calculation of \( f(x) \) might be complicated, but a rough estimate is sufficient. This linear model simplifies computation while preserving essential characteristics of the function around the chosen point.

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the effects of the original trigonometric functions. These functions answer questions like, "What angle gives a sine of \( x \)?" For \( \arcsin(x) \), \( x \) is the sine value, and the result is the angle whose sine is \( x \).

Working with \( \arcsin(x) \), you need to recall its range: typically, between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), representing the range of possible angles. This ensures any input value for \( x \) must satisfy \( -1 \leq x \leq 1 \).

These functions are not just mathematically fascinating; they also have practical applications in various fields, such as engineering and physics, where calculating angles based on known lengths or other parameters is necessary.

The derivative of \( \arcsin(x) \), \( \frac{1}{\sqrt{1-x^2}} \), is significant because it allows us to use calculus principles on these functions, enabling linearization and similar analyses. By understanding how \( \arcsin(x) \) behaves at different values, we can use this information in real-world calculations where trigonometric models are applicable.