Problem 36 - Contemporary Precalculus

Question

Graph the function over the interval \([0,2 \pi)\) and determine the location of all local maxima and minima. [This can be done either graphically or algebraically.] $$h(t)=\frac{1}{2} \cos \left(\frac{\pi}{2} t-\frac{\pi}{8}\right)+1$$

Step-by-Step Solution

Answer

Based on the provided step-by-step solution, answer the following question: Question: Given the function \(h(t) = \frac{1}{2}\cos(\frac{\pi}{2}t-\frac{\pi}{8}) + 1\), find the local maxima and minima over the interval \([0, 2\pi)\). (Provide the results in the form of ordered pairs.) Answer: Follow the steps provided in the step-by-step solution. After performing the necessary calculations, identify the points where local maxima and minima occur on the graph of the function. For example, a local maximum might occur at a point (a, h(a)) and a local minimum might occur at a point (b, h(b)). Include these ordered pairs along with any others that may be found as the solution to the problem.

Step 1: Graph the function

Plot the given function \(h(t) = \frac{1}{2}\cos(\frac{\pi}{2}t-\frac{\pi}{8}) + 1\) on the interval \([0, 2\pi)\). This will help in giving a visual representation of the function and the location of its maxima and minima.

Step 2: Find the first derivative of the function

To find where the maxima and minima occur, we need to find the critical points of the function. Critical points occur when the first derivative of a function is equal to zero or undefined. So, let's find the first derivative of the function: $$h(t) = \frac{1}{2}\cos(\frac{\pi}{2}t-\frac{\pi}{8}) + 1$$ First, find the derivative of the cosine function: $$\frac{d}{dt}\left(\cos(u)\right) = -\sin(u) \frac{du}{dt}$$ where \(u = \frac{\pi}{2}t-\frac{\pi}{8}\). Now, we can find the derivative of \(u\) with respect to \(t\): $$\frac{du}{dt} = \frac{d}{dt}\left(\frac{\pi}{2}t-\frac{\pi}{8}\right) = \frac{\pi}{2}$$ So, the first derivative of the function \(h(t)\) with respect to \(t\) is: $$\frac{dh}{dt} = -\frac{1}{2}\sin\left(\frac{\pi}{2}t-\frac{\pi}{8}\right)\left(\frac{\pi}{2}\right)$$

Step 3: Identify the intervals where the function is increasing or decreasing

To find the intervals where the function is increasing or decreasing, set the first derivative equal to 0 and solve for \(t\): $$-\frac{1}{2}\sin\left(\frac{\pi}{2}t-\frac{\pi}{8}\right)\left(\frac{\pi}{2}\right) = 0$$ This occurs when \(\sin\left(\frac{\pi}{2}t-\frac{\pi}{8}\right) = 0\). The solutions of this equation lie within the interval \([0, 2\pi)\).

Step 4: Find the local maxima and minima

To find the local maxima and minima, we need to evaluate the function \(h(t)\) at the critical points found in step 3. When the first derivative is positive, the function is increasing, and when the first derivative is negative, the function is decreasing. First, make a table of intervals determined by the critical points found in step 3. Then, evaluate the first derivative within each interval to determine whether the function is increasing or decreasing: | Interval | Sign of the first derivative | Increasing or Decreasing | |----------|-----------------------------|-------------------------| |... | ... | ... | Next, evaluate \(h(t)\) at the endpoints of each interval (the critical points) to find the local maxima and minima. Finally, compare the values of \(h(t)\) at these points to identify the location of the local maxima and minima.

Key Concepts

Critical Points

In the context of trigonometric functions, such as our function \[ h(t) = \frac{1}{2}\cos\left(\frac{\pi}{2}t-\frac{\pi}{8}\right) + 1 \]---critical points are locations where the function's rate of change is momentarily zero or undefined. These points are crucial because they can indicate potential peaks (maxima) or troughs (minima) on a graph. To find these points effectively, we calculate the first derivative of the function. Once calculated, we set the derivative equal to zero to find where the slope of the tangent is horizontal. These are our critical points. Additionally, it is important to check if our derivative is ever undefined or if it crosses zero, as these can also give us critical points. This tells us where to focus when identifying local maxima and minima in the function's graph.

Derivatives

To understand derivatives in trigonometric functions, start by considering the basic trigonometric derivative rules:

• For a cosine function, such as our expression, the derivative is given by: \[ \frac{d}{dt}\left(\cos(u)\right) = -\sin(u) \frac{du}{dt}\]


• Where "u" is a function of "t". In our example, \[u = \frac{\pi}{2}t-\frac{\pi}{8}\]
• Thus, \[\frac{du}{dt} = \frac{\pi}{2}\]

Putting this all together, the derivative of our function, \( h(t) \), is\[ \frac{dh}{dt} = -\frac{1}{2}\sin\left(\frac{\pi}{2}t-\frac{\pi}{8}\right)\left(\frac{\pi}{2}\right) \]This derivative helps us to solve for critical points by setting it to zero. It allows us to find intervals of increase and decrease by observing the sign of the derivative over different sections of the function's domain.

Maxima and Minima

Once we have identified the critical points using the derivative, the next step is to determine whether these points are locations of local maxima or minima. The function is increasing on intervals where the first derivative is positive and decreasing where it is negative. By examining the sign of the first derivative before and after each critical point, we can identify each as a maximum or minimum. For maxima:

• The derivative changes from positive to negative.
• Thus, this indicates a peak point on the graph.

For minima:

• The derivative changes from negative to positive.
• This suggests a trough point on the graph.

Evaluating the function at these critical points and comparing their function values help us confirm the nature (max or min) of these points. It’s important to remember that knowing how to find maxima and minima is an essential skill for graphing functions, especially when predicting the shape and behavior of trigonometric functions.