Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

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

Step 1: Find the derivative of the function

To find critical points, we first need to know where the derivative is zero or undefined. The function given is \( s(t) = \sin t - \cos t \). The first step is to differentiate this function with respect to \( t \). The derivative \( s'(t) \) is computed as follows:\[ s'(t) = \cos t + \sin t \].

Step 2: Set the derivative equal to zero

Now, to find critical points, we set the derivative equal to zero and solve for \( t \):\[ \cos t + \sin t = 0 \].This simplifies to \( \cos t = -\sin t \). Dividing both sides by \( \cos t \), we get \( \tan t = -1 \).

Step 3: Solve for critical points

We solve for \( t \) in the interval \([0, \pi]\) where \( \tan t = -1 \). The equation \( \tan t = -1 \) holds true when \( t = \frac{3\pi}{4} \), which lies within the interval \([0, \pi]\).

Step 4: Evaluate the function at the boundaries and critical points

We need to evaluate \( s(t) \) at the endpoints of the interval and at any critical points found. - At \( t = 0 \): \( s(0) = \sin 0 - \cos 0 = 0 - 1 = -1 \).- 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 \( t = \pi \): \( s(\pi) = \sin(\pi) - \cos(\pi) = 0 - (-1) = 1 \).

Step 5: Determine maximum and minimum values

The maximum and minimum values within the interval \([0, \pi]\) are determined from the evaluations:- Maximum value is \( \sqrt{2} \) at \( t = \frac{3\pi}{4} \).- Minimum value is \(-1\) at \( t = 0 \).

Key Concepts

Understanding Critical Points in Calculus Optimization

To optimize a function, we need to determine its critical points. Critical points occur where the function's derivative is zero or undefined. They are pivotal because they can indicate potential locations for maxima or minima.
For the function in this exercise, we consider critical points because they help pinpoint extreme values within a defined interval.

• First, differentiate the function to find its derivative.
• Identify values of the variable where the derivative equals zero or does not exist.

In the example given, our function is not undefined anywhere, which simplifies our task to finding where the derivative equals zero. This direct approach helps streamline the optimization process!

Differentiating the Function

The derivative is the foundation of calculus optimization. It represents the function's rate of change and is essential for finding critical points.
In this example, our function is \( s(t) = \sin t - \cos t \). To obtain the derivative, use simple rules of differentiation for trigonometric functions:

• The derivative of \( \sin t \) is \( \cos t \).
• The derivative of \( -\cos t \) is \( \sin t \).

So, the derivative becomes \( s'(t) = \cos t + \sin t \).
This expression will facilitate locating where changes in rate occur and help identify critical points.

Working with Trigonometric Functions

Trigonometric functions often appear in optimization problems, especially those constrained to specific intervals. They possess familiar periodic properties making solution finding predictable.
Let's focus on how these functions behave:

• \( \sin t \) and \( \cos t \) cycle between -1 and 1, showing maxima, minima, and zeros.
• Tangent relationships, like \( \tan t = -1 \), describe specific angles.

Here, we solve the equation \( \cos t + \sin t = 0 \), simplified to \( \tan t = -1 \). The solution is a known angle, \( t = \frac{3\pi}{4} \).
Trigonometric identities and angle properties allow these solutions to be straightforwardly recognized.

Interval Evaluation for Optimization

Once critical points are found, evaluate the function at these points, and at the interval's endpoints, to determine extrema. This completes the optimization process.
For our function, interval \( [0, \pi] \) requires checking:

• \( t = 0 \)
• \( t = \frac{3\pi}{4} \)
• \( t = \pi \)

Evaluate \( s(t) \) at these points to compare values.

Observing the values obtained:
- At \( t = 0 \, : \ s(0) = -1 \)
- At \( t = \frac{3\pi}{4} \, : \ s\left(\frac{3\pi}{4}\right) = \sqrt{2} \)
- At \( t = \pi : s(\pi) = 1 \)
The maximum value here is \( \sqrt{2} \) and the minimum is \(-1\), a result of interval evaluation guiding the determination of optimal points.