Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find, if possible, the (global) maximum and minimum values of the given function on the indicated interval. $$ h(t)=\sin t^{2} \text { on }[0, \pi] $$

Step-by-Step Solution

Answer

The maximum value is 1 at \( t = \sqrt{\frac{\pi}{2}} \); minimum is 0 at \( t = 0 \).

Step 1: Determine Critical Points

First, find the derivative of the function \( h(t) = \sin(t^2) \). Using the chain rule, the derivative is \( h'(t) = 2t \cos(t^2) \). Set \( h'(t) = 0 \) to find critical points. Solving \( 2t \cos(t^2) = 0 \), we get either \( t = 0 \) or \( \cos(t^2) = 0 \). For \( \cos(t^2) = 0 \), solve for \( t^2 = (2n+1)\frac{\pi}{2} \). In the interval \([0, \pi]\), this gives the feasible critical points.

Step 2: Identify Candidates for Critical Points

From \( \cos(t^2) = 0 \), we find \( t^2 = \frac{\pi}{2} \) or \( t = \sqrt{\frac{\pi}{2}} \). Therefore, the critical points are \( t = 0 \), \( t = \sqrt{\frac{\pi}{2}} \), and the end point \( t = \pi \) must also be checked. These are the candidates for evaluating the maximum and minimum values.

Step 3: Evaluate Function at Critical Points and Interval Endpoints

Evaluate the function \( h(t) = \sin(t^2) \) at the critical points and endpoints: \( t = 0 \), \( t = \sqrt{\frac{\pi}{2}} \), and \( t = \pi \). Compute \:- \( h(0) = \sin(0^2) = 0 \)- \( h(\sqrt{\frac{\pi}{2}}) = \sin(\frac{\pi}{2}) = 1 \)- \( h(\pi) = \sin(\pi^2) \).

Step 4: Check Values and Determine Maximum and Minimum

Comparing the calculated values, \( h(0) = 0 \), \( h(\sqrt{\frac{\pi}{2}}) = 1 \), and since \( \pi^2 \) is more than \( 2\pi \), the sine value would be in its periodic decline, possibly negative, making \( h(\pi) \) smaller than \( 1 \). Therefore, the global maximum value is \( 1 \) at \( t = \sqrt{\frac{\pi}{2}} \) and the global minimum value is \( 0 \) at \( t = 0 \).

Key Concepts

Critical Points

Critical points are essential in analyzing functions. They help us identify where the function may have maximum or minimum values, or where the function changes its behavior. To find critical points, we need to look for values where the derivative is zero or undefined. In the given function, we set the derivative equal to zero, which helps us find points where the slope of the tangent to the curve is horizontal. This is often where local peaks or valleys occur.
In the example, when solving the derivative equation \( 2t \cos(t^2) = 0 \), we find that either \( t = 0 \) or \( \cos(t^2) = 0 \). These conditions guide us to specific values possible within the interval. Finding these critical points on the given interval is key to evaluating the global behavior of the given trigonometric function.

Derivative

The derivative is a fundamental concept in calculus that describes how a function changes at any given point. It essentially measures the "rate of change" of the function. When we're concerned with finding maximum and minimum values of a function, the first derivative tells us where the slope is zero or changes, indicating potential peaks (maxima) or troughs (minima).
For the function \( h(t) = \sin(t^2) \), the derivative is calculated using the chain rule. The derivative we found is \( h'(t) = 2t \cos(t^2) \). Solving this equation helps identify points where the function does not increase or decrease momentarily, giving hints of critical points for further evaluation.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent are periodic and have specific ranges that make analyzing them interesting in calculus. In the function \( h(t) = \sin(t^2) \), the inner term \( t^2 \) causes the angle to increase rapidly, affecting the behavior of the sine function in somewhat unexpected ways.
We know that the sine function oscillates between -1 and 1. This behavior means that when evaluating critical points, you may find multiple intervals within a given range where the sine's periodic nature results in the same values. Understanding the trigonometric aspects is crucial to properly analyze and predict function behavior, especially when considering endpoints as well.

Interval

An interval in calculus represents a specific range of values for which a function is analyzed. In our example, the interval \([0, \pi]\), signifies that we're only interested in how the function behaves between 0 and 𝜋. This boundary plays a critical role in our analysis of the trigonometric function.
It's essential to evaluate not only at the critical points found within the interval but also at the endpoints of the interval. This helps in determining the global maximum and minimum values across the entire range. For instance, in this problem, you checked \( t = 0 \) and \( t = \pi \) alongside critical points to ensure no potential extreme values are missed.