Problem 46 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ f(t)=\sin t+\cos t $$

Step-by-Step Solution

Answer

Increasing on \((0, \frac{\pi}{4})\) and \((\frac{5\pi}{4}, 2\pi)\); decreasing on \((\frac{\pi}{4}, \frac{5\pi}{4})\).

Step 1: Find the First Derivative

The first step to determine where the function is increasing or decreasing is to find its first derivative. The function is given as \( f(t) = \sin t + \cos t \). The derivative of \( \sin t \) is \( \cos t \), and the derivative of \( \cos t \) is \( -\sin t \). Therefore, the first derivative, \( f'(t) \), is: \[ f'(t) = \cos t - \sin t \]

Step 2: Solve f'(t) = 0

To find critical points where the slope changes sign, set the derivative equal to zero: \[ \cos t - \sin t = 0 \]Rearranging gives \( \cos t = \sin t \). Dividing both sides by \( \cos t \) (assuming \( \cos t eq 0 \)) gives: \[ \tan t = 1 \]The solutions to \( \tan t = 1 \) within one period \( 0 \leq t < 2\pi \) are: \[ t = \frac{\pi}{4}, \frac{5\pi}{4} \]

Step 3: Determine Intervals

The critical points \( \frac{\pi}{4} \) and \( \frac{5\pi}{4} \) divide the interval from \( 0 \) to \( 2\pi \) into three parts: \( \left(0, \frac{\pi}{4}\right) \), \( \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \), and \( \left(\frac{5\pi}{4}, 2\pi\right) \). We need to test the sign of \( f'(t) \) in each interval.

Step 4: Test Intervals for Increasing/Decreasing

Choosing test points in each interval:- For \( t = \frac{\pi}{8} \) (in \( (0, \frac{\pi}{4}) \)), \( f'(t) = \cos\left(\frac{\pi}{8}\right) - \sin\left(\frac{\pi}{8}\right) > 0 \), so the function is increasing.- For \( t = \pi \) (in \( (\frac{\pi}{4}, \frac{5\pi}{4}) \)), \( f'(t) = \cos\pi - \sin\pi = -1 < 0 \), so the function is decreasing.- For \( t = \frac{3\pi}{2} \) (in \( (\frac{5\pi}{4}, 2\pi) \)), \( f'(t) = \cos\left(\frac{3\pi}{2}\right) - \sin\left(\frac{3\pi}{2}\right) = 1 > 0 \), so the function is increasing.

Step 5: State Solution

The function \( f(t) = \sin t + \cos t \) is increasing on the intervals \( (0, \frac{\pi}{4}) \) and \( (\frac{5\pi}{4}, 2\pi) \), and it is decreasing on the interval \( (\frac{\pi}{4}, \frac{5\pi}{4}) \).

Key Concepts

First Derivative

The first derivative of a function is a powerful tool in calculus that helps us determine where a function is increasing or decreasing. To do this, we first need to find the derivative of the given function. For the trigonometric function \( f(t) = \sin t + \cos t \), the derivative involves applying basic rules for derivatives of sine and cosine:

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

Using these rules, the first derivative of the function becomes:
\[ f'(t) = \cos t - \sin t \]
This tells us how the function behaves in terms of increasing or decreasing in different intervals. By inspecting the sign of \( f'(t) \), we can identify these intervals.

Trigonometric Functions

Trigonometric functions like sine and cosine are wave-like functions with a repeating pattern (periodicity). Understanding their behaviors makes it easier to analyze functions involving them. The function \( f(t) = \sin t + \cos t \) combines two basic trigonometric functions:

• Both \( \sin t \) and \( \cos t \) repeat every \( 2\pi \) radians.
• They have local maxima and minima within one period, allowing us to find critical points.
The periodic nature of these functions means they return to the same values after every \( 2\pi \) radians, which is crucial in finding where our function increases or decreases. Each interval can be systematically studied by using these periodic properties of sine and cosine.

Critical Points

Critical points are where the first derivative equals zero, which can indicate a change in the behavior of a function. For the function \( f(t) = \sin t + \cos t \), we set \( f'(t) = 0 \) to locate these points. This results in the equation: \[ \cos t - \sin t = 0 \]
By solving the equation, we find when \( \cos t = \sin t \). Dividing both sides by \( \cos t \) gives:

• \( \tan t = 1 \)

The solutions for this equation within the interval \( 0 \leq t < 2\pi \) are \( t = \frac{\pi}{4} \) and \( t = \frac{5\pi}{4} \). These points help break the function's domain into intervals where we determine whether the function is increasing or decreasing by testing the sign of the derivative at test points around them.Finding these critical points is imperative because they mark where a function changes direction, thus providing essential information for sketching the graph of the function or analyzing its behavior.