Problem 27 - Single Variable Calculus

Question

Determine where the graph of the function is concave upward and where it is concave downward. Also, find all inflection points of the function. $$ h(t)=\sin t+\cos t, \quad 0 \leq t \leq 2 \pi $$

Step-by-Step Solution

Answer

The graph of the function is concave upward for \(0 < t < \frac{3\pi}{4}\) and \(\frac{7\pi}{4} < t < 2\pi\), concave downward for \(\frac{3\pi}{4} < t < \frac{7\pi}{4}\), and the inflection points are located at \(\left(\frac{3\pi}{4},\sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4})\right)\) and \(\left(\frac{7\pi}{4},\sin(\frac{7\pi}{4}) + \cos(\frac{7\pi}{4})\right)\).

Step 1: Find the first derivative

To find the first derivative of h(t), we will apply differentiation rules (sum rule and chain rule) to \(h(t) = \sin t + \cos t\). $$ h'(t) = \frac{d(\sin t)}{dt} + \frac{d(\cos t)}{dt} = \cos t - \sin t $$

Step 2: Find the second derivative

To find the second derivative of h(t), we will differentiate h'(t) again: $$ h''(t) = \frac{d(\cos t - \sin t)}{dt} = -\sin t - \cos t $$

Step 3: Determine the intervals where h''(t) is positive or negative

To know where h''(t) is positive or negative, we first need to find the critical points. These occur when \(h''(t) = 0\). $$ -\sin t - \cos t = 0 \Rightarrow \sin t = -\cos t $$ The critical points occur when \(\tan t = -1\). The values of t in the given interval that satisfy this condition are \(t = \frac{3\pi}{4}\) and \(t = \frac{7\pi}{4}\). Now, we can check the intervals between these critical points to see where the second derivative is positive (concave upward) or negative (concave downward). - For \(0 < t < \frac{3\pi}{4}\), let's pick the test point \(t = \pi\). We have \(h''(\pi) = -\sin(\pi) - \cos(\pi) > 0\). Thus, the graph is concave upward in this interval. - For \(\frac{3\pi}{4} < t < \frac{7\pi}{4}\), let's pick the test point \(t = \frac{5\pi}{2}\). We have \(h''(\frac{5\pi}{2}) = -\sin(\frac{5\pi}{2}) - \cos(\frac{5\pi}{2}) < 0\). Thus, the graph is concave downward in this interval. - For \(\frac{7\pi}{4} < t < 2\pi\), let's pick the test point \(t = \frac{15\pi}{8}\). We have \(h''(\frac{15\pi}{8}) = -\sin(\frac{15\pi}{8}) - \cos(\frac{15\pi}{8}) > 0\). Thus, the graph is concave upward in this interval.

Step 4: Identify inflection points

Inflection points occur when the graph of the function changes its concavity. Based on our findings in the previous step, the inflection points occur at \(t = \frac{3\pi}{4}\) and \(t = \frac{7\pi}{4}\). To find the coordinates of these inflection points, plug the values of t back into the original function h(t): - For \(t = \frac{3\pi}{4}\), the inflection point is \(\left(\frac{3\pi}{4}, h\left(\frac{3\pi}{4}\right)\right) = \left(\frac{3\pi}{4}, \sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4})\right)\). - For \(t = \frac{7\pi}{4}\), the inflection point is \(\left(\frac{7\pi}{4}, h\left(\frac{7\pi}{4}\right)\right) = \left(\frac{7\pi}{4}, \sin(\frac{7\pi}{4}) + \cos(\frac{7\pi}{4})\right)\). In conclusion, the graph of the function is concave upward for \(0 < t < \frac{3\pi}{4}\) and \(\frac{7\pi}{4} < t < 2\pi\), concave downward for \(\frac{3\pi}{4} < t < \frac{7\pi}{4}\), and the inflection points are located at \(\left(\frac{3\pi}{4},\sin(\frac{3\pi}{4}) + \cos(\frac{3\pi}{4})\right)\) and \(\left(\frac{7\pi}{4},\sin(\frac{7\pi}{4}) + \cos(\frac{7\pi}{4})\right)\).

Key Concepts

Inflection Points

Inflection points are key to understanding the behavior of a function's graph. These points occur where a graph changes its direction of concavity. In simpler terms, it's where the curve "bends". If a graph goes from being concave upward (like a cup) to concave downward (like a frown), or vice versa, an inflection point is present. In calculus, to identify these points, you look for changes in the sign of the second derivative, denoted as \( h''(t) \) in this context. If the second derivative changes from positive to negative or from negative to positive as \( t \) crosses over a certain value, that value is an inflection point. In the example provided, the inflection points present at \( t = \frac{3\pi}{4} \) and \( t = \frac{7\pi}{4} \) indicate such changes. By substituting these values back into the original function, you can find their exact coordinates, which helps to further analyze and graph the function's behavior.

Trigonometric Functions

Trigonometric functions such as \( \sin t \) and \( \cos t \) are fundamental in studying periodic behavior in mathematics. In the context of this exercise, the function \( h(t) = \sin t + \cos t \) combines these two basic trigonometric functions. Together, they create a unique wave-like pattern as \( t \) varies.

• Sine Function (\( \sin t \)): It represents the y-coordinate of a point on the unit circle as \( t \) changes. The sine function oscillates smoothly between -1 and 1.
• Cosine Function (\( \cos t \)): Similar to sine, it represents the x-coordinate of the same point. It also oscillates between -1 and 1 but starts at 1 when \( t = 0 \).

When combined, \( \sin t + \cos t \) has a varied oscillation influenced by the sum of these functions. Analyzing how this function behaves involves understanding both the nature of the sine and cosine and how their combination affects concavity and inflection points.

Second Derivative Test

The Second Derivative Test is a valuable tool used to determine the concavity of a function and identify inflection points. When you take the second derivative of a function, you obtain an expression that can tell you where the function is concave upward or downward.

• Concavity: If the second derivative \( h''(t) \) is positive, the function is concave upward in that interval. If \( h''(t) \) is negative, the function is concave downward.
• Identifying Inflection Points: As previously discussed, an inflection point occurs where \( h''(t) = 0 \) and the concavity changes sign. This means the graph shifts from concave upwards to downwards or vice versa.

In this exercise, determining where the function \( h(t) = \sin t + \cos t \) changes concavity results in understanding the nature of its graph. By setting \( h''(t) = -\sin t - \cos t \) to zero, and solving for \( t \), we identify critical points. Checking intervals around these points using the second derivative gives insight into the intervals of concave upward and downward behaviors, thus allowing for a clear understanding of the graph's shape.