Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-26, identify the critical points and find the maximum value and minimum value on the given interval. $$ s(t)=\sin t-\cos t ; I=[0, \pi] $$

Step-by-Step Solution

Answer

Minimum value: -1 at \(t=0\); Maximum value: \(\sqrt{2}\) at \(t=\frac{3\pi}{4}\).

Step 1: Find the Derivative

To find the critical points of the function, we first need to find its derivative. The function given is \( s(t) = \sin t - \cos t \). The derivative \( s'(t) \) is found by differentiating each term:\[s'(t) = \cos t + \sin t.\]

Step 2: Set the Derivative to Zero

Next, we need to find the values of \( t \) where the derivative is zero, as these are possible critical points. Set the derivative \( s'(t) = \cos t + \sin t = 0 \):\[ \cos t = -\sin t. \]

Step 3: Solve the Trigonometric Equation

From \( \cos t = -\sin t \), divide both sides by \( \cos t \), yielding:\[ \tan t = -1. \]This occurs when \( t = \frac{3\pi}{4} \) in the given interval \([0, \pi]\).

Step 4: Evaluate the Function at Boundaries and Critical Points

Evaluate \( s(t) \) at the critical point and at the endpoints of the interval:- At \( t = 0 \), \( s(0) = \sin 0 - \cos 0 = -1 \).- At \( t = \frac{3\pi}{4} \), \( s\left(\frac{3\pi}{4}\right) = \sin\left(\frac{3\pi}{4}\right) - \cos\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \).- At \( t = \pi \), \( s(\pi) = \sin \pi - \cos \pi = 1 \).

Step 5: Identify Maximum and Minimum Values

Compare the values obtained in Step 4:- \( s(0) = -1 \)- \( s\left(\frac{3\pi}{4}\right) = \sqrt{2} \)- \( s(\pi) = 1 \)The minimum value of \( s(t) \) is -1, and the maximum value is \( \sqrt{2} \) on the interval \( [0, \pi] \).

Key Concepts

The Derivative

When analyzing a function like \( s(t) = \sin t - \cos t \), the first step we often take is finding the derivative. This helps to identify critical points, which are essential in understanding the function's behavior. By applying basic differentiation rules for trigonometric functions, we find the derivative:

• The sine function differentiates to cosine, so \( \frac{d}{dt}(\sin t) = \cos t \).
• The cosine function differentiates to negative sine, so \( \frac{d}{dt}(-\cos t) = \sin t \).

Hence, the derivative of \( s(t) = \sin t - \cos t \) becomes \( s'(t) = \cos t + \sin t \).
This derivative is crucial because it tells us how the function \( s(t) \) changes over \( t \). When the derivative is set to zero, it reveals potential critical points where the function could have peaks or valleys.

Maximum Value

Determining a function's maximum value involves comparing values at critical points and endpoints of the interval. Once the critical point is identified from setting the derivative to zero, evaluate the function at:

• The critical point \( t = \frac{3\pi}{4} \).
• The boundaries of the interval, which are \( t = 0 \) and \( t = \pi \) for this problem.

Evaluate the function \( s(t) \) at each of these points:

• \( s(0) = \sin 0 - \cos 0 = -1 \)
• \( s(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \)
• \( s(\pi) = \sin \pi - \cos \pi = 1 \)

From these calculations, the highest value is \( \sqrt{2} \), which occurs at the critical point \( t = \frac{3\pi}{4} \). This is the maximum value of the function on the given interval.

Minimum Value

To find the minimum value of a function, consider the same points assessed for the maximum. For the function \( s(t) \), the values calculated reveal that:

• At \( t = 0 \), \( s(0) = -1 \).
• At \( t = \frac{3\pi}{4} \), \( s(\frac{3\pi}{4}) = \sqrt{2} \).
• At \( t = \pi \), \( s(\pi) = 1 \).

Among these, the lowest value observed is \(-1\) at \( t = 0 \). This is the minimum value for the function on the interval \([0, \pi]\). The minimum value is as vital as the maximum because it helps understand the function's range of output values.

Trigonometric Functions

Trigonometric functions such as sine and cosine play a significant role in this exercise. Given the function \( s(t) = \sin t - \cos t \), understanding these basic trigonometric functions is fundamental.

• Sine Function (\(\sin t\)): This function gives the vertical coordinate of a point on the unit circle as it moves around the circle.
• Cosine Function (\(\cos t\)): This function provides the horizontal coordinate of a point on the unit circle.

One critical aspect of trigonometric functions is their periodicity, meaning they repeat values over regular intervals. This property helps in solving equations like \( \cos t = -\sin t \) because solutions typically recur.
For the problem in question, understanding that both sine and cosine take values from \(-1\) to \(1\) allows for a direct assessment in calculating \( s(t) \) at critical and endpoint values. Mastery of these trigonometric identities and behavior simplifies the process of finding extremum points in functions involving \( \sin \) and \( \cos \).