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Magazine Chapter 4: Problem 46
\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(t)=t+\cot \left(\frac{1}{2} t\right), \quad[\pi / 4,7 \pi / 4]$$
Short Answer
Expert verified
Maximum at \( t = 3\pi/2 \), Minimum at \( t = \pi/4 \).
Step by step solution
01
Find the derivative
First, we need to find the derivative of the function \( f(t) = t + \cot \left(\frac{1}{2} t \right) \). The derivative of \(t\) is 1. The derivative of \(\cot\left(\frac{1}{2}t\right)\) is \(-\frac{1}{2}\csc^2\left(\frac{1}{2}t\right)\) using the chain rule. Thus, \( f'(t) = 1 - \frac{1}{2} \csc^2\left(\frac{1}{2} t\right) \).
02
Set the derivative to zero
To find critical points, set the derivative equal to zero: \(1 - \frac{1}{2} \csc^2\left(\frac{1}{2}t\right) = 0\). Solving for \( \csc^2\left(\frac{1}{2}t\right) \), we get \( \csc^2\left(\frac{1}{2}t\right) = 2 \). So \( \csc\left(\frac{1}{2}t\right) \) can be \( \sqrt{2} \) or \(-\sqrt{2} \).
03
Solve for t
From \( \csc\left(\frac{1}{2}t\right) = \sqrt{2} \), we have \( \sin\left(\frac{1}{2}t\right) = \frac{1}{\sqrt{2}} \). This leads to \( \frac{1}{2}t = \frac{\pi}{4} + k\pi \). Solving for t gives \( t = \frac{\pi}{2} + 2k\pi \). Similarly, for \( \csc\left(\frac{1}{2}t\right) = -\sqrt{2} \), we get \( \sin\left(\frac{1}{2}t\right) = -\frac{1}{\sqrt{2}} \), leading to \( t = \frac{3\pi}{2} + 2k\pi \).
04
Find critical points in interval
The critical points in the interval \([\frac{\pi}{4}, 7\frac{\pi}{4}]\) occur when \( t = \frac{\pi}{2} \) and \( t = \frac{3\pi}{2} \). With \( k = 0 \), these are the points within our interval.
05
Evaluate the function at critical points and endpoints
Calculate \( f\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + \cot\left(\frac{1}{8}\pi\right) \), \( f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + \cot\left(\frac{1}{4}\pi\right) \), \( f\left(\frac{3\pi}{2}\right) = \frac{3\pi}{2} + \cot\left(\frac{3}{4}\pi\right) \), and \( f\left(7\frac{\pi}{4}\right) = 7\frac{\pi}{4} + \cot\left(\frac{7}{8}\pi\right) \). Evaluate each of these for numerical values.
06
Compare values to find maximum and minimum
Compare the values obtained previously: \( f\left(\frac{\pi}{4}\right) \), \( f\left(\frac{\pi}{2}\right) \), \( f\left(\frac{3\pi}{2}\right) \), and \( f\left(7\frac{\pi}{4}\right) \). \( f\left(\frac{\pi}{4}\right) \) is the smallest value, and \( f\left(\frac{3\pi}{2}\right) \) is the largest value.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
Critical points are the points on a graph where the derivative of a function is zero or undefined.
These points are essential in calculus optimization because they help in identifying potential maximum and minimum values of a function.
These points are essential in calculus optimization because they help in identifying potential maximum and minimum values of a function.
- To find critical points, we set the derivative of the function equal to zero and solve for the variable.
- For example, if a function鈥檚 derivative is given by \( f'(t) = 1 - \frac{1}{2} \csc^2\left(\frac{1}{2} t\right) \), set it to zero: \( 1 - \frac{1}{2} \csc^2\left(\frac{1}{2}t\right) = 0 \).
- Solve for \( \csc^2\left(\frac{1}{2}t\right) \) to find possible values where the function might change direction, indicating a maximum or minimum.
Derivative
The derivative of a function is like its "rate of change" or the "slope" of the function at any given point.
It tells us how the function is changing at a specific point, providing critical insights into the function鈥檚 behavior.
It tells us how the function is changing at a specific point, providing critical insights into the function鈥檚 behavior.
- For a function \( f(t) = t + \cot\left(\frac{1}{2}t\right) \), the derivative is \( f'(t) = 1 - \frac{1}{2} \csc^2\left(\frac{1}{2} t\right) \).
- This is found by taking the derivative of each part separately and combining the results. The derivative of \( t \) is straightforward, \( 1 \), while the derivative of \( \cot\left(\frac{1}{2}t\right) \) involves the trigonometric identity \( -\frac{1}{2}\csc^2\left(\frac{1}{2}t\right) \).
Trigonometric Functions
Trigonometric functions are fundamental in calculus, especially in problems involving periodic behavior.
Functions like \( \sin \), \( \cos \), and in this case, \( \cot \), are used to modeling wave-like patterns or periodic processes.
Functions like \( \sin \), \( \cos \), and in this case, \( \cot \), are used to modeling wave-like patterns or periodic processes.
- In the function \( f(t) = t + \cot\left(\frac{1}{2}t\right) \), \( \cot \) is the trigonometric function used. It is the complement of the tangent and defined as \( \cot\theta = \frac{1}{\tan\theta} \).
- In optimization problems, trigonometric functions can affect where critical points occur due to their periodic nature.
Chain Rule
The chain rule is a fundamental tool in calculus used to differentiate composite functions.
It allows us to break down complex expressions into simpler parts that are easier to work with.
It allows us to break down complex expressions into simpler parts that are easier to work with.
- The chain rule is used when differentiating \( \cot\left(\frac{1}{2}t\right) \). Normally, \( \cot(x) \) differentiates to \( -\csc^2(x) \).
- With the chain rule, we account for the inner function \( \frac{1}{2}t \), getting \( -\frac{1}{2}\csc^2\left(\frac{1}{2}t\right) \).
- The general form of the chain rule is \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \).
