Question

Identify the critical points. Then use (a) the First Derivative Test and (if possible) (b) the Second Derivative Test to decide which of the critical points give a local maximum and which give a local minimum. $$ \Psi(\theta)=\sin ^{2} \theta,-\pi / 2<\theta<\pi / 2 $$

Short answer

The critical point is at \( \theta = 0 \), which is a local minimum.

Step-by-step solution

Find the First Derivative

We need to find the derivative of the function \( \Psi(\theta) = \sin^2 \theta \). To do this, use the chain rule. Start by letting \( u = \sin \theta \), making \( \Psi(\theta) = u^2 \). The derivative of \( u^2 \) with respect to \( u \) is \( 2u \), and the derivative of \( u \) with respect to \( \theta \) is \( \cos \theta \). Therefore, \( \Psi'(\theta) = 2\sin \theta \cdot \cos \theta = \sin(2\theta) \).

Find Critical Points

Critical points arise when the first derivative is equal to zero. So, set \( \Psi'(\theta) = \sin(2\theta) = 0 \). This implies \( 2\theta = n\pi \), where \( n \) is an integer. Since \( -\pi/2 < \theta < \pi/2 \), possible solutions for \( \theta \) are \( \theta = 0 \).

Apply the First Derivative Test

To determine the nature of the critical point, examine the sign of the first derivative \( \Psi'(\theta) = \sin(2\theta) \) around \( \theta = 0 \). The function \( \sin(2\theta) > 0 \) for \( 0 < \theta < \pi/2 \) and \( \sin(2\theta) < 0 \) for \( -\pi/2 < \theta < 0 \). Thus, the sign changes from negative to positive at \( \theta = 0 \), indicating a local minimum.

Find the Second Derivative

Now, find the second derivative \( \Psi''(\theta) \) of the function. Start by differentiating \( \Psi'(\theta) = \sin(2\theta) \). The derivative is \( \Psi''(\theta) = 2\cos(2\theta) \).

Apply the Second Derivative Test

Assess the value of the second derivative at \( \theta = 0 \). We find \( \Psi''(0) = 2\cos(0) = 2 > 0 \). Since \( \Psi''(0) > 0 \), the critical point \( \theta = 0 \) is a local minimum.

Key concepts

First Derivative Test

The First Derivative Test is a method used to identify whether a critical point of a function is a local maximum or minimum. Let's break it down to understand how it applies to our example. First, we need the critical points. These occur where the function's first derivative is zero or undefined. In the given problem, we determined that the first derivative is \( \Psi'(\theta) = \sin(2\theta) \). Setting this derivative to zero, we found \( \theta = 0 \) as a critical point. Now, to apply the First Derivative Test, you examine the sign of the derivative just before and after the critical point:- If the derivative changes from positive to negative, the critical point is a local maximum.- If it changes from negative to positive, it indicates a local minimum.In our exercise, \( \sin(2\theta) < 0 \) for \( -\pi/2 < \theta < 0 \) and \( \sin(2\theta) > 0 \) for \( 0 < \theta < \pi/2 \). This sign change from negative to positive at \( \theta = 0 \) implies a local minimum at this point. Through the First Derivative Test, we see that examining the behavior of the derivative around critical points gives significant insights into the nature of these points.

Second Derivative Test

The Second Derivative Test is another insightful tool to determine the nature of a critical point, provided the second derivative exists and is continuous.To use this test, calculate the second derivative of the function at the critical point. For our example, let’s consider the derivative \( \Psi'(\theta) = \sin(2\theta) \) again and find its second derivative. Differentiating, we get \( \Psi''(\theta) = 2\cos(2\theta) \). The Second Derivative Test then goes as follows:- If \( \Psi''(\theta) > 0 \), the function is concave up at the critical point, denoting a local minimum.- If \( \Psi''(\theta) < 0 \), the function is concave down at the critical point, which means a local maximum.Evaluating at the critical point \( \theta = 0 \): \[ \Psi''(0) = 2\cos(0) = 2 \]Since \( 2 > 0 \), the second derivative indicates that \( \theta = 0 \) is indeed a local minimum. This test complements the First Derivative Test by further confirming the nature of the critical point through concavity.

Chain Rule

The Chain Rule is a fundamental tool in calculus used to differentiate compositions of functions. In the context of this problem, it helped us find the derivative of the function \( \Psi(\theta) = \sin^2 \theta \).To apply the chain rule effectively, you break down the function into outer and inner parts. Here’s how it played out:- Consider \( u = \sin \theta \), transforming the expression to \( \Psi(\theta) = u^2 \).- Differentiate the outer function \( u^2 \) with respect to \( u \), resulting in \( 2u \).- Also, find the derivative of the inner function, \( \sin \theta \), which is \( \cos \theta \).With these derivatives, the Chain Rule allows us to multiply them together to find the derivative of the entire function:\[ \Psi'\theta = (2\sin\theta)(\cos\theta) = \sin(2\theta). \]This rule is indispensable in tackling derivatives of complicated functions, particularly when dealing with compositions, as seen in this problem.