Problem 14 - Calculus of a Single Variable

Question

Determining Concavity In Exercises \(3-14\) , determine the open intervals on which the graph is concave upward or concave downward. $$ y=x+\frac{2}{\sin x}, \quad(-\pi, \pi) $$

Step-by-Step Solution

Answer

The graph of \(y=x+\frac{2}{\sin x}\) is concave down on \(-\pi < x < 0\) and concave up on \(0 < x < \pi\).

Step 1: Find the first derivative

The first derivative of the function \(y=x+\frac{2}{\sin x}\) can be calculated using the sum rule and the chain rule of differentiation. The sum rule states that the derivative of a sum of two functions is the sum of their respective derivatives, and the chain rule states that the derivative of a composite function is the derivative of the outside function times the derivative of the inside function. In this case the first derivative \(y'\) can be computed as \(y'=1-\frac{2\cos x}{\sin^2 x}\).

Step 2: Find the second derivative

The second derivative is found by taking the derivative of the first derivative. Use the quotient rule and the chain rule again to differentiate \(y'\). The quotient rule states that the derivative of a ratio of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. This gives the second derivative \(y''\) as \(y''=\frac{2\cos^2 x}{\sin^3 x}+\frac{2}{\sin^2 x}\).

Step 3: Determine the intervals of concavity

To find where the graph is concave upwards or downwards, we set \(y''=0\) and solve for \(x\). However, in this case, it is not possible to solve this equation for \(x\). Instead, we examine the denominator and the numerator of \(y''\). We find that the denominator \(\sin^3 x\) is negative on the interval \(-\pi < x < 0\) and positive on \(0< x < \pi\). The numerator \(2\cos^2 x + 2\sin^2 x\) is always nonnegative since it is a sum of squares. Therefore \(y''\) is negative on the interval \(-\pi < x < 0\) and positive on \(0< x < \pi\). This means that \(y=x+\frac{2}{\sin x}\) is concave down on the interval \(-\pi < x < 0\) and concave up on \(0< x < \pi\).

Key Concepts

Concave Upward

When we discuss the shape of a curve on a graph, we use the term 'concave upward' to describe a section where the curve opens upward like a cup. To envision this, imagine how a bowl or spoon rests with its open side facing up; a graph that is concave upward has a similar curvature.

To determine mathematically if a function is concave upward on an interval, we look at its second derivative. If the second derivative is greater than zero throughout an interval, the graph will be concave upward in that region. This happens because when the second derivative is positive, the slope of the tangent line to the curve is increasing. In simple terms, as you move left to right across the graph, the line is bending upwards, just as the slope of a hill increases as it becomes steeper.

Concave Downward

In contrast to concave upward, a graph that is 'concave downward' bows downwards, resembling the shape of an arch or the way a hill descends.

To determine if a function's graph is concave downward over an interval, we examine the second derivative of the function. If the second derivative is less than zero for that stretch, the function's graph will display a downward concavity. This indicates that the slope of the function's tangent line is decreasing across that interval, which can be thought of as going down a hill where the descent becomes less steep as you continue downwards.

Second Derivative Test

Identifying Concavity

The 'second derivative test' is a valuable tool for analyzing the concavity of a function. By taking the derivative of the derivative (hence 'second' derivative), we obtain a measure of how the rate of change of the function is changing.

If this value is positive, the curve is bending upwards and the graph is concave upward. Conversely, if the second derivative is negative, the graph is concave downward. This test not only tells us about the concavity but can also help pinpoint inflection points, where the graph changes from concave upward to concave downward or vice versa.

Application in Optimization

Moreover, the second derivative test can serve another purpose in optimization problems. It can help determine whether a critical point (found using the first derivative) is a maximum or minimum. If the second derivative is positive at a critical point, it suggests a local minimum; if it's negative, a local maximum.

First Derivative

The 'first derivative' of a function gives us the slope of the tangent line at any point on a function's graph. It allows us to analyze the rate at which the function's output is changing with respect to its input. This is immensely useful, as it directly leads to the identification of a function's increasing and decreasing intervals.

In the context of concavity, the first derivative also lays the groundwork for finding the second derivative, which, as explained in previous sections, is necessary to determine the concavity of the function. It is important to remember that by itself, the first derivative does not give information about concavity; it primarily tells us about the slope and the rate of change.