Problem 15 - Calculus for Biology and Medicine

Question

Determine where each function is increasing, decreasing, concave up, and concave down. With the help of a graphing calculator, sketch the graph of each function and label the intervals where it is increasing, decreasing, concave up, and concave down. Make sure that your graphs and your calculations agree. $$ y=\sin x, 0 \leq x \leq 2 \pi $$

Step-by-Step Solution

Answer

The function is increasing on \( (0, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, 2\pi) \), decreasing on \( (\frac{\pi}{2}, \frac{3\pi}{2}) \). It is concave up on \( (0, \pi) \) and concave down on \( (\pi, 2\pi) \).

Step 1: Analyze Critical Points

The function given is \( y = \sin x \) with an interval from \( 0 \) to \( 2\pi \). First, find the derivative \( y' = \cos x \). The critical points are found by setting \( y' = 0 \), which gives us \( \cos x = 0 \). Hence, the critical points are \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

Step 2: Determine Increasing and Decreasing Intervals

Use the first derivative test. The derivative \( \cos x \) is positive in the intervals \( (0, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, 2\pi) \), indicating \( y \) is increasing there. It is negative in \( (\frac{\pi}{2}, \frac{3\pi}{2}) \), so \( y \) is decreasing in this interval.

Step 3: Find Inflection Points

Calculate the second derivative \( y'' = -\sin x \). Set \( y'' = 0 \) to find inflection points. This gives \( \sin x = 0 \), leading to potential inflection points at \( x = 0, \pi, 2\pi \).

Step 4: Determine Concavity Intervals

The second derivative \( -\sin x \) is positive (concave up) in the intervals \( (0, \pi) \) and negative (concave down) in \( (\pi, 2\pi) \). The points \( x = 0, \pi, 2\pi \) are where concavity changes.

Step 5: Graph the Function

Using a graphing calculator, plot \( y = \sin x \) over the interval \([0, 2\pi]\). On the graph, label \( (0, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, 2\pi) \) as increasing, and \( (\frac{\pi}{2}, \frac{3\pi}{2}) \) as decreasing. Mark \( (0, \pi) \) as concave up and \( (\pi, 2\pi) \) as concave down. Verify these intervals agree with your calculations.

Key Concepts

Derivative

In calculus, the derivative of a function represents the rate at which the function's value changes with respect to changes in its input. For the function \( y = \sin x \), the derivative is \( y' = \cos x \). This derivative tells us about the slopes of the tangent lines at various points on the sine curve.

• When \( \cos x > 0 \), the tangent lines have a positive slope, and thus, the function is increasing.
• When \( \cos x < 0 \), the tangent lines have a negative slope, and the function is decreasing.

To find the critical points, which are where the function changes from increasing to decreasing or vice versa, we set the derivative to zero: \( \cos x = 0 \). This happens at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \). These points divide the interval \( [0, 2\pi] \) into sections where the function either consistently increases or decreases.

Critical Points

Critical points are the places on a graph where the derivative equals zero or does not exist. They are essential for understanding the behavior of a function because they indicate where the function might change direction or have a peak or valley. For \( y = \sin x \), the critical points were calculated by setting \( y' = \cos x = 0 \), resulting in the points \( x = \frac{\pi}{2}, \frac{3\pi}{2} \).

At these points, the function is neither increasing nor decreasing, but potentially changes direction. By using these points, alongside the sign of the derivative in nearby intervals, we understand where \( y \) is increasing—specifically in \( (0, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, 2\pi) \)—and where it is decreasing—in \( (\frac{\pi}{2}, \frac{3\pi}{2}) \). Analyzing critical points helps to sketch an accurate graph of the function's behavior.

Concavity

Concavity describes the nature of the curve of a function—whether it opens upwards or downwards. This can be determined by the second derivative of the function. For \( y = \sin x \), the second derivative is \( y'' = -\sin x \).

• When \( y'' > 0 \), the function is concave up, which means it has a U-shape, like a cup that holds water.
• When \( y'' < 0 \), the function is concave down, resembling an upside-down U.

Setting the second derivative equal to zero identifies potential inflection points, which is where the concavity changes. The inflection points for this function occur at \( x = 0, \pi, 2\pi \), dividing the interval into sections of different concavity: \( (0, \pi) \) where it's concave up, and \( (\pi, 2\pi) \) where it's concave down.

Graphing Calculator

Using a graphing calculator is a great way to visualize theoretical concepts by producing accurate graphs. To analyze \( y = \sin x \) within \( [0, 2\pi] \), a graphing calculator can help seamlessly plot this sine wave.

When you examine the graph:

• Identify the intervals of increase \( (0, \frac{\pi}{2}) \) and \( (\frac{3\pi}{2}, 2\pi) \), and decrease \( (\frac{\pi}{2}, \frac{3\pi}{2}) \) based on the slope of the graph.
• Mark the sections of concavity—concave up in \( (0, \pi) \), and concave down in \( (\pi, 2\pi) \).

The graphing calculator helps confirm our calculations visually. It ensures the sign changes in the derivative align with increases and decreases, while changes in the second derivative are evident in the concavity. This visual approach also reinforces the underlying math concepts by representing them dynamically.