Max & Min Of G(θ) = 2θ - 3sin(θ) On [0, Π]: A Guide
Hey guys! Today, we're diving into a fun problem in calculus: finding the maximum and minimum values of the function g(θ) = 2θ - 3sin(θ) within the interval [0, π]. This type of problem is a classic application of derivatives, and mastering it will seriously boost your calculus skills. So, grab your thinking caps, and let's get started!
Understanding the Problem
Before we jump into the solution, let's break down what we're trying to achieve. We have a function, g(θ) = 2θ - 3sin(θ), which takes an angle θ as input and spits out a real number. Our mission is to find the largest (maximum) and smallest (minimum) values this function can take when θ is restricted to the interval [0, π]. This interval simply means we're only considering angles between 0 and π radians (or 0 and 180 degrees, if you prefer).
Why do we care about maximum and minimum values? Well, these points are super important in many real-world applications. Think about optimizing profit in business, finding the shortest path in navigation, or designing structures that can withstand maximum stress. Calculus gives us the tools to tackle these kinds of optimization problems, and finding the maximum and minimum of a function is a fundamental step.
In this particular case, our function g(θ) combines a linear term (2θ) with a trigonometric term (-3sin(θ)). This combination creates a function that increases overall due to the 2θ term but also oscillates due to the -3sin(θ) term. Finding the exact points where the maximum and minimum occur requires a bit of calculus magic, which we'll explore in the next sections.
Step 1: Finding Critical Points
Our first step in finding the maximum and minimum values is to identify the critical points of the function. Critical points are the points where the derivative of the function is either equal to zero or undefined. These points are crucial because they are the potential locations where the function changes direction – from increasing to decreasing (maximum) or from decreasing to increasing (minimum).
To find the critical points, we need to calculate the derivative of g(θ). Remember, the derivative of a function tells us its rate of change at any given point. Using basic differentiation rules:
- The derivative of 2θ with respect to θ is 2.
- The derivative of -3sin(θ) with respect to θ is -3cos(θ).
Therefore, the derivative of g(θ), denoted as g'(θ), is:
g'(θ) = 2 - 3cos(θ)
Now, we need to find the values of θ for which g'(θ) = 0. This means solving the equation:
2 - 3cos(θ) = 0
Let's isolate cos(θ):
3cos(θ) = 2 cos(θ) = 2/3
To find the values of θ that satisfy this equation within the interval [0, π], we need to use the inverse cosine function (also known as arccos or cos⁻¹):
θ = arccos(2/3)
Using a calculator, we find that:
θ ≈ 0.841 radians
Since the cosine function is positive in the first and fourth quadrants, but our interval is [0, π] (which covers the first and second quadrants), we only have one critical point within our interval: θ ≈ 0.841 radians. This is a key point to remember as we move to the next step.
Step 2: Evaluating the Function at Critical Points and Endpoints
Now that we've found our critical point (θ ≈ 0.841 radians), the next step is to evaluate the original function g(θ) at this point. But we're not done yet! We also need to evaluate g(θ) at the endpoints of our interval, which are θ = 0 and θ = π. This is because the maximum or minimum value could occur at an endpoint, even if the derivative isn't zero there.
So, let's calculate g(θ) at these three points:
-
At θ = 0:
g(0) = 2(0) - 3sin(0) = 0 - 3(0) = 0
-
At θ ≈ 0.841 radians:
g(0.841) = 2(0.841) - 3sin(0.841) ≈ 1.682 - 3(0.745) ≈ 1.682 - 2.235 ≈ -0.553
-
At θ = π:
g(π) = 2(π) - 3sin(π) ≈ 2(3.1416) - 3(0) ≈ 6.283
We now have the values of the function at our critical point and the endpoints:
- g(0) = 0
- g(0.841) ≈ -0.553
- g(π) ≈ 6.283
Step 3: Determining the Maximum and Minimum Values
With the function values calculated at the critical point and endpoints, we can now easily identify the maximum and minimum values. We simply compare the values we obtained in the previous step.
Looking at the values:
- g(0) = 0
- g(0.841) ≈ -0.553
- g(π) ≈ 6.283
It's clear that the smallest value is approximately -0.553, which occurs at θ ≈ 0.841 radians. This is our minimum value.
Similarly, the largest value is approximately 6.283, which occurs at θ = π. This is our maximum value.
Therefore, we can conclude that:
- The minimum value of g(θ) on the interval [0, π] is approximately -0.553, occurring at θ ≈ 0.841 radians.
- The maximum value of g(θ) on the interval [0, π] is approximately 6.283, occurring at θ = π.
Visualizing the Solution
To solidify our understanding, it's helpful to visualize the function and its maximum and minimum points. Imagine the graph of g(θ) = 2θ - 3sin(θ) on the interval [0, π]. The graph would start at the origin (0, 0), dip down to a minimum point around θ ≈ 0.841 radians, and then rise steadily to a maximum point at θ = π.
The minimum point we found represents the lowest point on the graph within the interval, and the maximum point represents the highest point. This visual confirmation reinforces our analytical solution and helps us grasp the behavior of the function.
Key Takeaways
Let's recap the key steps we took to solve this problem:
- Find the derivative: We calculated g'(θ) to find the rate of change of the function.
- Find critical points: We solved g'(θ) = 0 to identify potential locations of maximum and minimum values.
- Evaluate at critical points and endpoints: We calculated g(θ) at the critical points and the endpoints of the interval.
- Determine maximum and minimum: We compared the values to find the largest and smallest values of the function.
This process is a standard approach for finding maximum and minimum values of functions, and it's a valuable tool in many areas of mathematics and its applications.
Conclusion
And there you have it! We've successfully found the maximum and minimum values of the function g(θ) = 2θ - 3sin(θ) on the interval [0, π]. By using calculus techniques like differentiation and critical point analysis, we were able to pinpoint the exact locations and values of these extreme points. Remember, practice makes perfect, so keep tackling these types of problems to sharpen your calculus skills. Keep exploring, keep learning, and I'll catch you in the next one! This process is a standard approach for finding maximum and minimum values of functions, and it's a valuable tool in many areas of mathematics and its applications. Remember, practice makes perfect, so keep tackling these types of problems to sharpen your calculus skills. Keep exploring, keep learning, and I'll catch you in the next one! This process is a standard approach for finding maximum and minimum values of functions, and it's a valuable tool in many areas of mathematics and its applications. Remember, practice makes perfect, so keep tackling these types of problems to sharpen your calculus skills. Keep exploring, keep learning, and I'll catch you in the next one!