Problem 33 - Calculus with Concepts in Calculus

Question

Use the Second Derivative Test to determine the relative extreme values (if any) of the function. $$ f(t)=\sin t+\cos t $$

Step-by-Step Solution

Answer

The function has relative maxima at \( t = \frac{\pi}{4} + n\pi \) for integer \( n \).

Step 1: Find the First Derivative

First, we need to find the derivative of the function, \[ f(t) = \sin t + \cos t. \] The derivative is \[ f'(t) = \frac{d}{dt}(\sin t) + \frac{d}{dt}(\cos t) = \cos t - \sin t. \]

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find the critical points. \[ \cos t - \sin t = 0. \] Simplify this to \[ \cos t = \sin t. \] Divide both sides by \( \cos t \) (assuming \( \cos t eq 0 \)) to get \[ \tan t = 1. \] The general solutions for \( t \) are \[ t = \frac{\pi}{4} + n\pi, \] where \( n \) is an integer.

Step 3: Find the Second Derivative

Now compute the second derivative \[ f''(t) = \frac{d}{dt}(\cos t - \sin t) = -\sin t - \cos t. \]

Step 4: Evaluate the Second Derivative at Critical Points

We substitute the critical points into the second derivative to determine concavity at these points. At \[ t = \frac{\pi}{4}, \] we have \[ f''\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}. \] Since this is negative, there is a relative maximum at \( t = \frac{\pi}{4} + n\pi \) for odd \( n \). Similarly, since the second derivative is always negative at multiples of \( \pi/4 \) (the quadratic trinomial involved), all these are relative maximum points.

Key Concepts

Critical Points

Critical points are essential values in calculus where the slope of the function is zero or undefined. At these points, potential relative maxima, minima, or saddle points may occur. To find critical points, you set the first derivative of the function equal to zero.

For the function \( f(t) = \sin t + \cos t \), the first derivative is \( f'(t) = \cos t - \sin t \). To find the critical points, solve the equation \( \cos t - \sin t = 0 \). Simplifying this gives \( \tan t = 1 \), meaning angles \( t \) where the tangent is 1 are our critical points.

The general solutions for these critical points are given by \( t = \frac{\pi}{4} + n\pi \), where \( n \) is any integer. At these points, the behavior of the function changes, making them the key areas to check for any extreme values using second derivative methods.

Derivative of Trigonometric Functions

Trigonometric functions, like sine and cosine, have derivatives that are also trigonometric functions. Understanding these derivatives is crucial for solving calculus problems.

For \( f(t) = \sin t + \cos t \):

• The derivative of \( \sin t \) is \( \cos t \)
• The derivative of \( \cos t \) is \( -\sin t \)

Therefore, the first derivative becomes \( f'(t) = \cos t - \sin t \).

Finding the second derivative involves differentiating \( f'(t) \) again, leading to \( f''(t) = -\sin t - \cos t \). This second derivative helps determine the concavity of the function at critical points, which further assists in identifying whether these points are maxima, minima, or neither.

Relative Extrema

Relative extrema refer to the local maximum or minimum points on a function within a specified interval. The Second Derivative Test is a handy tool for determining these points using the second derivative of the function.

After finding critical points from the first derivative, we use the second derivative \( f''(t) = -\sin t - \cos t \) to analyze concavity:

• If \( f''(t) > 0 \) at any critical point, the function has a relative minimum there.
• If \( f''(t) < 0 \) at any critical point, the function has a relative maximum there.

In this scenario, we evaluate \( f''(\frac{\pi}{4}) = -\sqrt{2} \), which is less than zero, indicating a relative maximum. Given the periodic nature of sine and cosine, this pattern repeats for each critical point \( t = \frac{\pi}{4} + n\pi \) with odd \( n \). Thus, these points are consistently relative maxima for the function.