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Chapter 4: Problem 61

Upper Bounds Show that for any numbers \(a\) and \(b\) \(|\sin b-\sin a| \leq|b-a|\)

Short Answer

Expert verified
The inequality \(|\sin b - \sin a| \leq |b-a|\) can be proved by expressing the left hand side as the difference of two sine functions and applying Mean Value Theorem(MVT). According to MVT, there exists a number c in the interval (a, b) such that the derivative of the function at c is equal to the average rate of change of function over the interval. Using this theorem and the fact that the absolute value of the cosine function is always less than or equal to 1, the inequality is proved.

Step by step solution

01

Express the problem as a difference of two functions

We can express the left hand side of the inequality as the difference of two sine functions: \(f(b) - f(a)\) where \(f(x) = \sin x\). Now, we should think about how we can use the Mean Value Theorem (MVT) to prove the inequality.
02

Apply the Mean Value Theorem (MVT)

According to the MVT, there exists a number \(c\) in the interval \((a, b)\) such that \(f'(c) = (f(b) - f(a)) / (b - a)\) where \(f'(x)\) is the derivative of \(f(x)\). So, \(f'(c) = \cos c\) and \(f'(c) = (\sin b - \sin a) / (b - a)\).
03

Find Absolute values and Apply Inequality

We know that the absolute value of the derivative of f(x), which is \(\cos x\) is always less than or equal to 1. So, \(|\cos c| \leq 1\). From the previous step, we know that \(|\cos c| = |(\sin b - \sin a) / (b - a)|\). So, we get, \(|(\sin b - \sin a) / (b - a)| \leq 1\).
04

Multiply both sides by |b-a|

Multiply both sides by \(|b-a|\) to get \(|\sin b - \sin a| \leq |b - a|\). This completes the proof

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Key Concepts

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

Understanding Trigonometric Functions and Their Properties
Trigonometric functions, like sine and cosine, are essential in mathematics, particularly in geometry and calculus. They relate angles of a triangle to the lengths of its sides. These functions are periodic, meaning they repeat their values in regular intervals. For example, the sine function achieves a full wave from 0 to 2蟺 radians. Understanding this periodical nature is crucial when solving equations and inequalities.

Every trigonometric function is defined using a unit circle. The sine of an angle corresponds to the y-coordinate of the point on this circle at the given angle. This interpretation helps us understand why the sine and cosine functions never exceed 1 or drop below -1, as they represent lengths or distances from the origin. Knowing these properties supports understanding how these functions behave over intervals, and it plays a significant role in applying concepts like the Mean Value Theorem.
Exploring Inequalities with Trigonometric Functions in Calculus
Inequalities in calculus involve finding conditions where one value is greater or less than another. They're essential for determining bounds and behavior of functions. When analyzing trigonometric functions, inequalities help establish limits on their values over intervals.

The given inequality, \(|\sin b - \sin a| \leq |b - a|\), means the absolute difference between the sine of two angles is not greater than the difference between the angles themselves. This results from understanding how change occurs in the sine function over a small interval. The Mean Value Theorem proves this by linking the average rate of change to an instantaneous rate of change within the interval.
  • Inequalities reveal behaviors within functions, helping to anticipate and verify function outputs.
  • Understanding how functions are bounded can simplify solving many calculus problems, as they establish a set of constraints for evaluating other properties.
Role of Derivatives in Calculus and Their Connection to the Mean Value Theorem
Derivatives are a cornerstone of calculus, measuring how functions change. They represent the slope of a function's graph at any point, showing how the function's value shifts in response to changes in its input.

When applying the Mean Value Theorem (MVT) involving derivatives, it gives us insight about a function's behavior on a specific interval. For the sine function, the derivative, \(\cos x\), represents how quickly the sine function rises or falls. The MVT provides that there is some point, \( c \), in an interval \( (a, b) \) where the function's average rate of change equals the instantaneous rate of change at \( c \). This connection helps to bridge derivatives with tangible inequalities, as shown in the problem's solution.
  • Derivatives help to map out function behavior and growth trends, essential for advanced calculus operations like integration and series.
  • They also aid in finding optimizations and can detect critical points where functions attain local minima or maxima.

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Most popular questions from this chapter

Multiple Choice A cylindrical rubber cord is stretched at a constant rate of 2 \(\mathrm{cm}\) per second. Assuming its volume does no change, how fast is its radius shrinking when its length is 100 \(\mathrm{c}\) and its radius is 1 \(\mathrm{cm} ?\) $$\begin{array}{ll}{\text { (A) } 0 \mathrm{cm} / \mathrm{sec}} & {\text { (B) } 0.01 \mathrm{cm} / \mathrm{sec}} 67 {\text{ (C) } 0.02 \mathrm{cm} / \mathrm{sec}}$\\\ {\text { (D) } 2 \mathrm{cm} / \mathrm{sec}} & {\text { (E) } 3.979 \mathrm{cm} / \mathrm{sec}}\end{array}

Draining Conical Reservoir Water is flowing at the rate of 50 \(\mathrm{m}^{3} / \mathrm{min}\) from a concrete conical reservoir (vertex down) of base radius 45 \(\mathrm{m}\) and height 6 \(\mathrm{m} .\) (a) How fast is the water level falling when the water is 5 \(\mathrm{m}\) deep? (b) How fast is the radius of the water's surface changing at that moment? Give your answer in \(\mathrm{cm} / \mathrm{min.}\)

The domain of f^{\prime}\( is \)[0,1) \cup(1,2) \cup(2,3]

Measuring Acceleration of Gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\) . The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g\) . By keeping track of \(\Delta T\) , we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L .\) (a) With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts \((b)\) and \((c)\) . (b) Writing to Learn If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. (c) A clock with a 100 -cm pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001 \mathrm{sec} .\) Find \(d g\) and estimate the value of \(g\) at the new location.

The Linearization is the Best Linear Approximation Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c(m\) and \(c\) constants). If the error \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions i. \(E(a)=0\) ii. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \((x-a)\) .
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