Problem 38
Question
Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=\tan (x), c=\pi / 4 $$
Step-by-Step Solution
Verified Answer
The linearization is \(L(x) = 2x + 1 - \frac{\pi}{2}.\)
1Step 1: Evaluate the Function at c
To find the linearization, first evaluate the function at the given value of c. For the function \(f(x) = \tan(x)\), calculate \(f(c)\) by substituting \(c = \frac{\pi}{4}\): \[ f\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1.\]
2Step 2: Find the Derivative of the Function
Next, calculate the derivative of \(f(x) = \tan(x)\). The derivative is \(f'(x) = \sec^2(x)\).
3Step 3: Evaluate the Derivative at c
Find the value of the derivative at \(c = \frac{\pi}{4}\) by substituting into the derivative: \[ f'\left(\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi}{4}\right) = 2.\]
4Step 4: Substitute into the Linearization Formula
Now substitute the values from the previous steps into the linearization formula: \[ L(x) = f(c) + f'(c) \cdot (x - c).\] Substituting \(f(c) = 1\), \(f'(c) = 2\), and \(c = \frac{\pi}{4}\), we get: \[ L(x) = 1 + 2(x - \frac{\pi}{4}).\]
5Step 5: Simplify the Linearization
Finally, simplify the expression for the linearization: \[ L(x) = 1 + 2x - \frac{\pi}{2}.\]
Key Concepts
Understanding LinearizationThe Concept of the DerivativeThe Role of Trigonometric FunctionsUnderstanding the Tangent Function
Understanding Linearization
Linearization is a powerful method in calculus that helps us approximate complex functions with simpler linear functions. It's particularly useful when you want to estimate the value of a function near a certain point. The formula for linearization is given as \(L(x) = f(c) + f'(c) \cdot (x - c)\). Here:
- \(f(c)\) represents the value of the function at a specific point \(c\).
- \(f'(c)\) is the derivative of the function at the same point \(c\), reflecting the slope of the tangent.
- \((x - c)\) indicates the small displacement from the point \(c\).
The Concept of the Derivative
In calculus, the derivative is the measure of how a function changes as its input changes. It is essentially the slope of the function at any point, and it tells us the rate at which the function is increasing or decreasing. For a function \(f(x)\), its derivative is often written as \(f'(x)\).
To find the derivative of a given function, such as \(f(x) = \tan(x)\), we can use differentiation rules. The derivative is \(f'(x) = \sec^2(x)\).
By evaluating this derivative at a specific point \(c\), we get important insight into the function's behavior at that exact point. This evaluation, \(f'(c)\), is a crucial part of calculating the linear approximation, as it reflects the tangent line's slope.
To find the derivative of a given function, such as \(f(x) = \tan(x)\), we can use differentiation rules. The derivative is \(f'(x) = \sec^2(x)\).
By evaluating this derivative at a specific point \(c\), we get important insight into the function's behavior at that exact point. This evaluation, \(f'(c)\), is a crucial part of calculating the linear approximation, as it reflects the tangent line's slope.
The Role of Trigonometric Functions
Trigonometric functions like \(\tan(x)\) are fundamental in both pure and applied mathematics. These functions are based on the ratios of sides in right triangles and have periodic properties. They are inherently related to angles and are extensively used in problems involving periodic or wave-like phenomena.
In the context of this problem, the function \(\tan(x)\) plays the central role and brings unique characteristics like vertical asymptotes and periodicity. Calculating the linearization of such functions helps in approximating their behavior around points of interest, like \(c = \pi/4\), thereby simplifying complex analyses by using linear functions.
In the context of this problem, the function \(\tan(x)\) plays the central role and brings unique characteristics like vertical asymptotes and periodicity. Calculating the linearization of such functions helps in approximating their behavior around points of interest, like \(c = \pi/4\), thereby simplifying complex analyses by using linear functions.
Understanding the Tangent Function
In calculus, the tangent function—here represented as \(\tan(x)\)—is one of the core trigonometric functions. It provides the ratio of the length of the opposite side to the adjacent side in a right-angled triangle. For circular motion or wave analysis, \(\tan(x)\) features quite prominently as it relates angles to ratios varying from zero to infinite, making it a versatile tool in different environments.
For approximation or analysis, considering the tangent of a point, like \(\tan(\pi/4)\), involves understanding that at precisely \(\pi/4\), the secant of the angle squared, \(\sec^2(\pi/4) = 2\), provides the slope of the tangent line derived in derivative calculations. This serves as a foundational element when constructing the linear approximation (or linearization) for the function in question.
For approximation or analysis, considering the tangent of a point, like \(\tan(\pi/4)\), involves understanding that at precisely \(\pi/4\), the secant of the angle squared, \(\sec^2(\pi/4) = 2\), provides the slope of the tangent line derived in derivative calculations. This serves as a foundational element when constructing the linear approximation (or linearization) for the function in question.
Other exercises in this chapter
Problem 37
Find a polynomial function \(p(x)\) of degree 1 such that \(p(2)=6\) and \(p^{\prime}(4)=-5\).
View solution Problem 38
Differentiate the given expression with respect to \(x\). $$ \cosh (x-2) $$
View solution Problem 38
Find the tangent line to the parametric curve \(x=\varphi_{1}(t), y=\varphi_{2}(t)\) at the point corresponding to the given value \(t_{0}\) of the parameter. $
View solution Problem 38
A velocity \(v(t)\) is given. Calculate the acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) ). \(v(t)=t^{2}-5 t \mathrm{~m} / \mathrm{s}\)
View solution