At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To solve the given problem, we need to analyze the potential energy function [tex]\( U(x) \)[/tex] and find its critical points, as well as determine the nature of these critical points to identify the maximum and minimum potential energy values. Let's go through this step-by-step.
1. Given Potential Energy Function:
[tex]\[ U(x) = \frac{x^4}{4} - \frac{x^2}{2} \][/tex]
2. Find the First Derivative:
We need to find the first derivative [tex]\( U'(x) \)[/tex] to locate the critical points:
[tex]\[ U'(x) = \frac{d}{dx}\left(\frac{x^4}{4} - \frac{x^2}{2}\right) \][/tex]
[tex]\[ U'(x) = x^3 - x \][/tex]
3. Set the First Derivative to Zero:
To find the critical points, we set [tex]\( U'(x) = 0 \)[/tex]:
[tex]\[ x^3 - x = 0 \][/tex]
Factorizing the equation:
[tex]\[ x(x^2 - 1) = 0 \][/tex]
[tex]\[ x(x-1)(x+1) = 0 \][/tex]
This gives us the critical points:
[tex]\[ x = -1, 0, 1 \][/tex]
4. Evaluate the Potential Energy at the Critical Points:
Next, we substitute each critical point back into the potential energy function to find their corresponding values:
[tex]\[ U(-1) = \frac{(-1)^4}{4} - \frac{(-1)^2}{2} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \][/tex]
[tex]\[ U(0) = \frac{0^4}{4} - \frac{0^2}{2} = 0 \][/tex]
[tex]\[ U(1) = \frac{1^4}{4} - \frac{1^2}{2} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \][/tex]
Therefore, the potential energy values at the critical points are:
[tex]\[ U(-1) = -\frac{1}{4}, \quad U(0) = 0, \quad U(1) = -\frac{1}{4} \][/tex]
5. Find the Second Derivative:
We need the second derivative [tex]\( U''(x) \)[/tex] to determine if each critical point is a maximum or minimum:
[tex]\[ U''(x) = \frac{d^2}{dx^2}\left(\frac{x^4}{4} - \frac{x^2}{2}\right) \][/tex]
[tex]\[ U''(x) = 3x^2 - 1 \][/tex]
6. Evaluate the Second Derivative at the Critical Points:
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ U''(-1) = 3(-1)^2 - 1 = 3(1) - 1 = 2 \quad (\text{positive, minimum}) \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ U''(0) = 3(0)^2 - 1 = -1 \quad (\text{negative, maximum}) \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ U''(1) = 3(1)^2 - 1 = 3(1) - 1 = 2 \quad (\text{positive, minimum}) \][/tex]
7. Identify the Maximum and Minimum Potential Energy:
- The maximum potential energy is at [tex]\( x = 0 \)[/tex] with [tex]\( U(0) = 0 \)[/tex]
- The minimum potential energy is at [tex]\( x = -1 \)[/tex] and [tex]\( x = 1 \)[/tex] with [tex]\( U(-1) = U(1) = -\frac{1}{4} \)[/tex]
Thus, the maximum potential energy is [tex]\( \boxed{0} \)[/tex] and the minimum potential energy is [tex]\( \boxed{-\frac{1}{4}} \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{(3) \ 0, -\frac{1}{4}} \][/tex]
1. Given Potential Energy Function:
[tex]\[ U(x) = \frac{x^4}{4} - \frac{x^2}{2} \][/tex]
2. Find the First Derivative:
We need to find the first derivative [tex]\( U'(x) \)[/tex] to locate the critical points:
[tex]\[ U'(x) = \frac{d}{dx}\left(\frac{x^4}{4} - \frac{x^2}{2}\right) \][/tex]
[tex]\[ U'(x) = x^3 - x \][/tex]
3. Set the First Derivative to Zero:
To find the critical points, we set [tex]\( U'(x) = 0 \)[/tex]:
[tex]\[ x^3 - x = 0 \][/tex]
Factorizing the equation:
[tex]\[ x(x^2 - 1) = 0 \][/tex]
[tex]\[ x(x-1)(x+1) = 0 \][/tex]
This gives us the critical points:
[tex]\[ x = -1, 0, 1 \][/tex]
4. Evaluate the Potential Energy at the Critical Points:
Next, we substitute each critical point back into the potential energy function to find their corresponding values:
[tex]\[ U(-1) = \frac{(-1)^4}{4} - \frac{(-1)^2}{2} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \][/tex]
[tex]\[ U(0) = \frac{0^4}{4} - \frac{0^2}{2} = 0 \][/tex]
[tex]\[ U(1) = \frac{1^4}{4} - \frac{1^2}{2} = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \][/tex]
Therefore, the potential energy values at the critical points are:
[tex]\[ U(-1) = -\frac{1}{4}, \quad U(0) = 0, \quad U(1) = -\frac{1}{4} \][/tex]
5. Find the Second Derivative:
We need the second derivative [tex]\( U''(x) \)[/tex] to determine if each critical point is a maximum or minimum:
[tex]\[ U''(x) = \frac{d^2}{dx^2}\left(\frac{x^4}{4} - \frac{x^2}{2}\right) \][/tex]
[tex]\[ U''(x) = 3x^2 - 1 \][/tex]
6. Evaluate the Second Derivative at the Critical Points:
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ U''(-1) = 3(-1)^2 - 1 = 3(1) - 1 = 2 \quad (\text{positive, minimum}) \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ U''(0) = 3(0)^2 - 1 = -1 \quad (\text{negative, maximum}) \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ U''(1) = 3(1)^2 - 1 = 3(1) - 1 = 2 \quad (\text{positive, minimum}) \][/tex]
7. Identify the Maximum and Minimum Potential Energy:
- The maximum potential energy is at [tex]\( x = 0 \)[/tex] with [tex]\( U(0) = 0 \)[/tex]
- The minimum potential energy is at [tex]\( x = -1 \)[/tex] and [tex]\( x = 1 \)[/tex] with [tex]\( U(-1) = U(1) = -\frac{1}{4} \)[/tex]
Thus, the maximum potential energy is [tex]\( \boxed{0} \)[/tex] and the minimum potential energy is [tex]\( \boxed{-\frac{1}{4}} \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{(3) \ 0, -\frac{1}{4}} \][/tex]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
