Chapter 3: Problem 4
Let \(v(x, y)=\arctan \frac{2}{x}\) for \(x \neq 0\). Compute the partial derivatives of \(v\), and verify that v satisfies Laplace's equation.
Short Answer
Expert verified
The function does not satisfy Laplace's equation as \( \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} \neq 0 \).
Step by step solution
01
Identify the function
The given function is \( v(x, y) = \arctan \left( \frac{2}{x} \right) \). This function is dependent only on \( x \) and not on \( y \).
02
Find \( \frac{\partial v}{\partial x} \)
To find the partial derivative of \( v \) with respect to \( x \), use the chain rule. First, find the derivative of \( \arctan(u) \) with respect to \( u \), which is \( \frac{1}{1+u^2} \). Here, \( u = \frac{2}{x} \), so the chain rule gives us: \( \frac{d}{dx}\left(\arctan \frac{2}{x}\right) = \frac{1}{1 + (\frac{2}{x})^2} \cdot \frac{d}{dx}\left(\frac{2}{x}\right) \).
03
Differentiate \( \frac{2}{x} \)
Calculate the derivative of \( \frac{2}{x} \) with respect to \( x \). This derivative is \( -\frac{2}{x^2} \).
04
Apply the chain rule and simplify
Substitute the derivative \( -\frac{2}{x^2} \) into the chain rule expression: \( \frac{\partial v}{\partial x} = \frac{1}{1+(\frac{2}{x})^2} \cdot ( -\frac{2}{x^2} ) = -\frac{2}{x^2 + 4} \).
05
Find \( \frac{\partial v}{\partial y} \)
Since \( v(x, y) = \arctan \frac{2}{x} \) does not depend on \( y \), the partial derivative with respect to \( y \) is \( \frac{\partial v}{\partial y} = 0 \).
06
Calculate second partial derivatives for Laplace's equation verification
Compute \( \frac{\partial^2 v}{\partial x^2} \) by differentiating \( \frac{\partial v}{\partial x} = -\frac{2}{x^2 + 4} \) with respect to \( x \). Use the quotient rule which gives: \( \frac{\partial^2 v}{\partial x^2} = \frac{d}{dx}\left(-\frac{2}{x^2 + 4}\right) \).
07
Apply the quotient rule
The quotient rule is \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). Let \( u = -2 \), \( v = x^2 + 4 \). Differentiating gives: \( \frac{\partial^2 v}{\partial x^2} = \frac{(x^2 + 4)(0) - (-2)(2x)}{(x^2+4)^2} = \frac{4x}{(x^2+4)^2} \).
08
Confirm Laplace's equation is satisfied
Laplace's equation states that for a function \( v \), \( \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0 \). Since \( \frac{\partial v}{\partial y} = 0 \), we have \( \frac{\partial^2 v}{\partial y^2} = 0 \). Therefore, check if \( \frac{4x}{(x^2+4)^2} + 0 = 0 \), which is not true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Laplace's Equation
Laplace's equation is a second-order partial differential equation named after Pierre-Simon Laplace. It is used extensively in the fields of electrostatics, gravitation, and fluid dynamics. The equation itself is written as \( abla^2 v = 0 \), which means that the Laplacian (\( abla^2 \)) of a function \( v \) is equal to zero.
For function \( v(x,y) \), this equation becomes:
In essence, Laplace's equation describes the condition under which the function is harmonic. For applications, solutions to Laplace's equation depict potential fields – fields where forces act without transaction of energy through the medium, like gravitational or electric potential fields. When we say the function satisfies Laplace's equation, it implies there's a balance — no net force or change in the potential at any point. The function \( v(x, y) = \arctan \left( \frac{2}{x} \right) \) does not satisfy Laplace's equation, meaning it may not describe a harmonic potential field.
For function \( v(x,y) \), this equation becomes:
- \( \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0 \)
In essence, Laplace's equation describes the condition under which the function is harmonic. For applications, solutions to Laplace's equation depict potential fields – fields where forces act without transaction of energy through the medium, like gravitational or electric potential fields. When we say the function satisfies Laplace's equation, it implies there's a balance — no net force or change in the potential at any point. The function \( v(x, y) = \arctan \left( \frac{2}{x} \right) \) does not satisfy Laplace's equation, meaning it may not describe a harmonic potential field.
Chain Rule
The Chain Rule is a fundamental method in calculus used to differentiate composite functions. A composite function is a function composed of several other functions, and the Chain Rule helps us find its derivative. The standard way to express this rule is:
This rule is essential to calculate partial derivatives when one function is nested inside another, as seen in the problem with \( v(x, y) = \arctan \left( \frac{2}{x} \right) \). Here, the function involves \( \arctan \) and \( \frac{2}{x} \). By applying the Chain Rule, we differentiate each layer step by step. First, find the derivative of the outer function, \( \arctan \left( u \right) \), then multiply by the derivative of the inner function \( \frac{2}{x} \).This method ensures that we account for how each part of the function changes with respect to \( x \), ultimately providing us with \( \frac{\partial v}{\partial x} \).
- If \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)
This rule is essential to calculate partial derivatives when one function is nested inside another, as seen in the problem with \( v(x, y) = \arctan \left( \frac{2}{x} \right) \). Here, the function involves \( \arctan \) and \( \frac{2}{x} \). By applying the Chain Rule, we differentiate each layer step by step. First, find the derivative of the outer function, \( \arctan \left( u \right) \), then multiply by the derivative of the inner function \( \frac{2}{x} \).This method ensures that we account for how each part of the function changes with respect to \( x \), ultimately providing us with \( \frac{\partial v}{\partial x} \).
Quotient Rule
The Quotient Rule is a technique for finding the derivative of a fraction where the numerator and denominator are both functions of some variable, often \( x \). The rule is expressed as:
This rule plays a crucial role in computing second-order derivatives in problems involving division of functions, such as in verifying Laplace's equation. In our problem, after identifying that \( \frac{\partial v}{\partial x} = -\frac{2}{x^2 + 4} \), we needed \( \frac{\partial^2 v}{\partial x^2} \), which required applying the Quotient Rule.
We let \( u = -2 \) and \( v = x^2 + 4 \), then differentiated accordingly:
This approach efficiently handles the complexity of differentiating composite expressions, balancing each piece of the formula as you work through the calculation.
- If \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \)
This rule plays a crucial role in computing second-order derivatives in problems involving division of functions, such as in verifying Laplace's equation. In our problem, after identifying that \( \frac{\partial v}{\partial x} = -\frac{2}{x^2 + 4} \), we needed \( \frac{\partial^2 v}{\partial x^2} \), which required applying the Quotient Rule.
We let \( u = -2 \) and \( v = x^2 + 4 \), then differentiated accordingly:
- Numerator: \( v \cdot 0 - u \cdot (2x) = 4x \)
- Denominator: \( (x^2 + 4)^2 \)
This approach efficiently handles the complexity of differentiating composite expressions, balancing each piece of the formula as you work through the calculation.