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Chapter 3: Problem 12

In Exercises \(11-14\) , find the linearization of \(f\) at \(x=a\) . Then graph the linearization and \(f\) together. $$ f(x)=\cos x \text { at } \quad(a) x=0, \quad \text { (b) } x=-\pi / 2 $$

Short Answer

Expert verified
(a) Linearization at 0 is \(L(x)=1\); (b) at \(-\frac{\pi}{2}\) is \(L(x)=x+\frac{\pi}{2}\).

Step by step solution

01

Determine the Linearization Formula

The linearization of a function \(f\) at a point \(x = a\) is given by the formula \(L(x) = f(a) + f'(a)(x-a)\). This formula approximates the function near the point \(a\).
02

Compute Function Values

Calculate \( f(x) = \cos(x) \) for each specified point:(a) \(a = 0\); therefore, \(f(0) = \cos(0) = 1\).(b) \(a = -\frac{\pi}{2}\); therefore, \(f(-\frac{\pi}{2}) = \cos(-\frac{\pi}{2}) = 0\).
03

Calculate the Derivative of the Function

The derivative of \(f(x) = \cos(x)\) is \(f'(x) = -\sin(x)\). We'll evaluate this at each of the points.(a) \(x = 0\): \(f'(0) = -\sin(0) = 0\).(b) \(x = -\frac{\pi}{2}\): \(f'\left(-\frac{\pi}{2}\right) = -\sin(-\frac{\pi}{2}) = 1\).
04

Formulate the Linearization Equation

Using the formula \(L(x) = f(a) + f'(a)(x-a)\) and the values from steps 2 and 3:(a) At \(x = 0\): \(L(x) = 1 + 0 \cdot (x - 0) = 1\).(b) At \(x = -\frac{\pi}{2}\): \(L(x) = 0 + 1(x + \frac{\pi}{2}) = x + \frac{\pi}{2}\).
05

Graph the Linearization with the Function

Plot the function \(f(x) = \cos(x)\) and its linearization \(L(x)\) on the same graph for visibility.(a) For \(x=0\): \(f(x) = \cos(x)\) and \(L(x) = 1\).(b) For \(x=-\frac{\pi}{2}\): \(f(x) = \cos(x)\) and \(L(x) = x + \frac{\pi}{2}\). Observe how \(L(x)\) approximates \(f(x)\) near each point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linearization
Linearization is a mathematical tool used to approximate a function using its tangent line. When we focus only on scenarios near a specific point, linearization provides a simpler, highly practical method for function approximation. The linearization of a function at a point is expressed by the formula:\[ L(x) = f(a) + f'(a)(x-a) \]where \(f(a)\) is the function value at point \(a\) and \(f'(a)\) is the derivative of the function at the same point. Here are key points to remember:
  • The purpose of linearization is to create a straight-line approximation of a function around a specific value \(a\). It is especially useful for estimating the behavior of complex functions near this point.
  • Linearization can make complicated functions easier to work with, especially in calculus, engineering, and physics.
  • In our exercise, this technique is applied to the cosine function at two points: 0 and \(-\frac{\pi}{2}\).
Understanding linearization provides the base for tackling complex calculus problems, by focusing on how functions behave precisely at or near chosen points.
Derivative
Derivatives are fundamental concepts in calculus, representing the rate at which a function is changing at any point. To find the derivative of a function is to determine its rate of change or its slope at a particular point.
Here are the basics:
  • The derivative of a function \(f(x)\) is often denoted as \(f'(x)\) or \(\frac{df}{dx}\).
  • For trigonometric functions, specific derivatives are well-known: for example, the derivative of \(\cos(x)\) is \(-\sin(x)\).
In the exercise, the derivative \(f'(x) = -\sin(x)\) was computed at two points:
  • At \(x = 0\), the derivative evaluates to 0 since \(-\sin(0) = 0\).
  • At \(x = -\frac{\pi}{2}\), the derivative becomes 1 because \(-\sin(-\frac{\pi}{2}) = 1\).
By understanding derivatives, we gain insights into how graphs behave, how we formulate linear approximations, and how we predict the behavior of functions in various mathematical and real-world applications.
Trigonometric Functions
Trigonometric functions are a type of function that relate angles to the ratios of sides of a right triangle. Some familiar trigonometric functions include sine, cosine, and tangent.
Here are essential points about trigonometric functions:
  • These functions are periodic, meaning they repeat their values in regular intervals. The cosine function, for example, has a period of \(2\pi\).
  • Trigonometric functions are critical in modeling real-world phenomena, such as waves and oscillations.
  • In the problem, we focus on the cosine function \(f(x) = \cos(x)\). Its values at specific points are pivotal in forming linear approximations.
For our exercise:
  • At \(x = 0\), the cosine function outputs a value of 1.
  • At \(x = -\frac{\pi}{2}\), \(\cos(x)\) equals 0.
Understanding trigonometric functions, and their derivatives, aids in forming the correct linearizations. Learning how these functions behave can greatly help when working with many calculus applications.

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Most popular questions from this chapter

Chain Rule Suppose that \(f(x)=x^{2}\) and \(g(x)=|x| .\) Then the composites $$ \begin{aligned}(f \circ g)(x)=|x|^{2}=x^{2} & \text { and } \quad(g \circ f)(x)=\left|x^{2}\right|=x^{2} \\ \text { are both differentiable at } x=0 \text { even though } g \text { itself is not differ- } \\ \text { entiable at } x=0 . \text { Does this contradict the Chain Rule? Explain. } \end{aligned} $$

In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ 6 x^{2}+3 x y+2 y^{2}+17 y-6=0, \quad(-1,0) $$

In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ y=2 \sin (\pi x-y), \quad(1,0) $$

Assuming that the equations in Exercises \(63-66\) define \(x\) and \(y\) implicitly as differentiable functions \(x=f(t), y=g(t),\) find the slope of the curve \(x=f(t), y=g(t)\) at the given value of \(t .\) $$ x^{2}-2 t x+2 t^{2}=4, \quad 2 y^{3}-3 t^{2}=4, \quad t=2 $$

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