Problem 45 - Calculus with Concepts in Calculus

Question

Find the intervals on which the given Ir function is increasing and those on which it is decreasing. $$ f(t)=2 \cos t-t $$

Step-by-Step Solution

Answer

Increasing on \((\frac{11\pi}{6}, 2\pi)\), decreasing on \([0, \frac{7\pi}{6})\) and \((\frac{7\pi}{6}, \frac{11\pi}{6})\)."

Step 1: Understand the Problem

We need to determine where the function \( f(t) = 2 \cos t - t \) is increasing or decreasing. A function is increasing when its derivative is positive, and decreasing when its derivative is negative.

Step 2: Find the Derivative

Compute the derivative of \( f(t) = 2 \cos t - t \) with respect to \( t \). The derivative is \( f'(t) = -2 \sin t - 1 \).

Step 3: Set the Derivative Equal to Zero

To find critical points where the function changes from increasing to decreasing or vice-versa, set \( f'(t) = -2 \sin t - 1 = 0 \). Solve for \( t \): \( -2 \sin t = 1 \), which gives \( \sin t = -\frac{1}{2} \).

Step 4: Solve for Critical Points

Solve \( \sin t = -\frac{1}{2} \), which corresponds to angles \( t = \frac{7\pi}{6} + 2k\pi \) and \( t = \frac{11\pi}{6} + 2k\pi \) where \( k \) is an integer, within the interval \( [0, 2\pi] \). These points are \( t = \frac{7\pi}{6}, \frac{11\pi}{6} \).

Step 5: Determine the Sign of the Derivative

Analyze the intervals based on critical points, testing values in the derivative. Select test points in the intervals \( (0, \frac{7\pi}{6}) \), \( (\frac{7\pi}{6}, \frac{11\pi}{6}) \), and \( (\frac{11\pi}{6}, 2\pi) \). For \( t \) in these intervals, check if \( f'(t) \) is positive or negative.

Step 6: Interval Analysis

- For \( t \in (0, \frac{7\pi}{6}) \), choose \( t = \pi \), \( f'(\pi) = -2 \sin(\pi) - 1 = -1 < 0 \), function is decreasing. - For \( t \in (\frac{7\pi}{6}, \frac{11\pi}{6}) \), choose \( t = 2\pi \), \( f'(2\pi) = -2 \sin(2\pi) - 1 = -1 < 0 \), function is decreasing. - For \( t \in (\frac{11\pi}{6}, 2\pi) \), choose \( t = \frac{3\pi}{2} \), \( f'(\frac{3\pi}{2}) = -2(-1) - 1 = 1 > 0 \), function is increasing.

Step 7: Conclusion

The function \( f(t) = 2 \cos t - t \) is increasing on \( (\frac{11\pi}{6}, 2\pi) \) and decreasing on intervals \( (0, \frac{7\pi}{6}) \) and \( (\frac{7\pi}{6}, \frac{11\pi}{6}) \).

Key Concepts

Derivatives

In calculus, derivatives are a fundamental tool that allows us to analyze how a function changes at any given point. Derivatives represent the rate of change or the slope of the function at a particular point on its curve. For our function \( f(t) = 2 \cos t - t \), the derivative is found through differentiation. This process helps to identify how each component of the function contributes to its overall behavior.
\[ f'(t) = -2 \sin t - 1 \]
Here, \(-2 \sin t\) is derived from the \(2 \cos t\) term, and \(-1\) comes from the \(-t\) term. Understanding derivatives involves knowing rules like the chain rule and the derivative of standard functions, such as trigonometric functions. It's helpful to remember these rules:

• The derivative of \( \cos x \) is \(-\sin x\)
• The derivative of \( x \) is \(1\)

By applying these rules, we gain insight into the behavior of the function and can proceed to analyze how it behaves with respect to \( t \), identifying whether it's increasing or decreasing.

Increasing and Decreasing Functions

Once we have the derivative of a function, we can use it to determine where the function itself is increasing or decreasing. A function is said to be increasing on an interval if, as the input variable \( t \) increases, the output of the function also increases. Similarly, it is decreasing if the output decreases as \( t \) increases.
To find these intervals:

• Determine when the derivative \( f'(t) \) is positive. This indicates increasing behavior.
• Determine when \( f'(t) \) is negative. This indicates decreasing behavior.

For \( f'(t) = -2 \sin t - 1 \), we analyze intervals determined by examining critical points and testing sub-intervals:
- On \((0, \frac{7\pi}{6})\), choose \( t = \pi\), yielding \( f'(\pi) < 0 \), indicating a decreasing function.
- On \((\frac{7\pi}{6}, \frac{11\pi}{6})\), use test point \( t = 2\pi\), giving \( f'(2\pi) < 0 \), showing decreasing behavior.
- On \((\frac{11\pi}{6}, 2\pi)\), use the point \( t = \frac{3\pi}{2}\), resulting in \( f'(\frac{3\pi}{2}) > 0 \), indicating the function is increasing.

Critical Points

Critical points are particular values of \( t \) where the derivative of a function is zero or undefined. These points are essential in determining the function's changes in direction, from increasing to decreasing, or vice versa. For \( f(t) = 2 \cos t - t \), finding critical points involves solving \(-2 \sin t - 1 = 0\).
Solving the equation \(-2 \sin t = 1\) leads to:

• \( \sin t = -\frac{1}{2} \)
• This occurs at angles \( t = \frac{7\pi}{6} \) and \( t = \frac{11\pi}{6} \) within \([0, 2\pi]\).

These values are where the function changes its rate of increase or decrease.
Critical points help us divide the function curve into distinct segments. Each segment can then be analyzed to determine its behavior—whether it is rising or falling along the curve. Understanding this can explain a lot about the overall nature and shape of the function's graph.