Problem 34 - Calculus with Concepts in Calculus

Question

Use the Second Derivative Test to determine the relative extreme values (if any) of the function. $$ f(t)=t+\cos 2 t $$

Step-by-Step Solution

Answer

Relative maxima at \( t = \frac{\pi}{12} + n\pi \) and minima at \( t = \frac{5\pi}{12} + n\pi \).

Step 1: Find the first derivative

The first step is to find the first derivative of the given function \( f(t) = t + \cos 2t \). We use the derivative rules:\[ f'(t) = \frac{d}{dt}(t) + \frac{d}{dt}(\cos 2t) = 1 - 2\sin 2t \] This is because the derivative of \( \cos x \) is \( -\sin x \) and using the chain rule for \( \cos 2t \).

Step 2: Set the first derivative to zero

To find possible extreme points, set the first derivative to zero:\[ f'(t) = 1 - 2\sin 2t = 0 \] Solving for \( \sin 2t \), we get:\[ \sin 2t = \frac{1}{2} \] The solutions for \( \sin 2t = \frac{1}{2} \) are:\[ 2t = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad 2t = \frac{5\pi}{6} + 2n\pi \] for integer \( n \).

Step 3: Solve for t values

Divide the solutions from Step 2 by 2 to solve for \( t \):\[ t = \frac{\pi}{12} + n\pi \quad \text{and} \quad t = \frac{5\pi}{12} + n\pi \] These \( t \) values are where potential extreme points occur.

Step 4: Find the second derivative

The next step is to find the second derivative for the Second Derivative Test. Starting with:\[ f'(t) = 1 - 2\sin 2t \]The second derivative is:\[ f''(t) = \frac{d}{dt}(-2\sin 2t) = -4\cos 2t \] since the derivative of \( \sin x \) is \( \cos x \).

Step 5: Evaluate the second derivative at the critical points

Evaluate \( f''(t) = -4\cos 2t \) at the critical points \( t = \frac{\pi}{12} + n\pi \) and \( t = \frac{5\pi}{12} + n\pi \):1. For \( t = \frac{\pi}{12} + n\pi \): \[ f''(t) = -4\cos(\frac{\pi}{6} + 2n\pi) = -4 \times \frac{\sqrt{3}}{2} = -2\sqrt{3} \] This value is less than 0, indicating a relative maximum.2. For \( t = \frac{5\pi}{12} + n\pi \): \[ f''(t) = -4\cos(\frac{5\pi}{6} + 2n\pi) = -4 \times -\frac{\sqrt{3}}{2} = 2\sqrt{3} \] This value is greater than 0, indicating a relative minimum.

Key Concepts

Critical Points in Calculus

When working with functions, critical points play a crucial role in analyzing the behavior of these functions. A critical point occurs where the first derivative of a function is zero or undefined.
This means:

• The slope of the tangent line is flat.
• It is a potential location for a maximum, minimum, or saddle point.

For the function \( f(t) = t + \cos 2t \), we find the critical points by setting the first derivative \( f'(t) = 1 - 2\sin 2t \) to zero. Solving \( \sin 2t = \frac{1}{2} \), we find two families of solutions for \( t \): \( t = \frac{\pi}{12} + n\pi \) and \( t = \frac{5\pi}{12} + n\pi \), where \( n \) is an integer.
These points are tested to check if they are places of interest.

Understanding Relative Extrema

Relative extrema refer to the points where a function reaches a local maximum or minimum. To determine if critical points are extrema, we often use the Second Derivative Test:

• If \( f''(t) > 0 \) at a critical point, it indicates a relative minimum - the function curves upwards.
• If \( f''(t) < 0 \) at a critical point, it indicates a relative maximum - the function curves downwards.
• If \( f''(t) = 0 \), the test is inconclusive.

For the function \( f(t) = t + \cos 2t \), evaluating the second derivative \( f''(t) = -4\cos 2t \) at the critical points gives us:- A relative maximum at \( t = \frac{\pi}{12} + n\pi \) since \( f''(t) < 0 \).- A relative minimum at \( t = \frac{5\pi}{12} + n\pi \) since \( f''(t) > 0 \).Understanding how derivatives influence extrema can help predict and sketch the behavior of functions.

Trigonometric Functions in Calculus

Trigonometric functions such as sine and cosine frequently appear in calculus due to their oscillatory nature. They introduce periodic behavior, critical for various real-world applications.
The derivatives of these functions are characteristic and predictable:

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

Their periodic properties can result in multiple critical points when dealing with equations like \( \sin 2t = \frac{1}{2} \). Solutions reflect this periodic nature, producing families of solutions like \( t = \frac{\pi}{12} + n\pi \) and \( t = \frac{5\pi}{12} + n\pi \).
By understanding these properties, students can solve complex trigonometric equations and understand their broader impact on function behavior and calculus.