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Chapter 4: Problem 30

Calculate the gradients of the functions and verify that \(\nabla \times \nabla f=\mathbf{0}\). $$f(x, y)=\tan ^{-1}\left(x^{2}+y^{2}\right)$$

Short Answer

Expert verified
The curl of the gradient of the function is zero: \(\nabla \times \nabla f = \mathbf{0}.\)

Step by step solution

01

Calculate the Gradient of f(x, y)

The gradient of a function \(f(x, y)\) is given by the vector of its first partial derivatives. Let's denote \(f(x, y) = \tan^{-1}(x^2 + y^2)\), and calculate its gradient:1. Compute the partial derivative with respect to \(x\): \[\frac{\partial f}{\partial x} = \frac{1}{1+(x^2+y^2)^2} \cdot 2x.\]2. Compute the partial derivative with respect to \(y\): \[\frac{\partial f}{\partial y} = \frac{1}{1+(x^2+y^2)^2} \cdot 2y.\]Therefore, the gradient \(abla f\) is:\[abla f(x, y) = \left( \frac{2x}{1+(x^2+y^2)^2}, \frac{2y}{1+(x^2+y^2)^2} \right).\]
02

Calculate the Curl of the Gradient

The curl operation, denoted \(abla \times \vec{F}\), applies to a vector field \(\vec{F}\) in three dimensions. However, the function \(f(x, y)\) is in two dimensions, so we need to consider an extension in three dimensions where the third component is zero. The function in three dimensions becomes: \(abla f(x, y, z) = \left( \frac{2x}{1+(x^2+y^2)^2}, \frac{2y}{1+(x^2+y^2)^2}, 0 \right).\)To compute the curl of this vector field, use the formula:\[abla \times \vec{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right).\]Since \(F_3 = 0\), the reduced form of the curl operation gives:- \(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = \frac{\partial}{\partial x}\left( \frac{2y}{1+(x^2+y^2)^2} \right) - \frac{\partial}{\partial y}\left( \frac{2x}{1+(x^2+y^2)^2} \right) = 0.\)This confirms that \(abla \times abla f = \mathbf{0}.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gradient
The gradient is a vector that consists of the partial derivatives of a function with respect to each variable in its domain. Written mathematically, the gradient of a function \( f(x, y) \) is represented as \( abla f(x, y) \). The components of this vector are the first partial derivatives with respect to \( x \) and \( y \).
  • The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} \).
  • The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} \).
These derivatives give us the rate of change of the function \( f \) in the direction of each axis. The gradient points in the direction of the steepest increase of the function's value.
For the function \( f(x, y) = \tan^{-1}(x^2 + y^2) \), the gradient was found to be:
\[ abla f(x, y) = \left( \frac{2x}{1+(x^2+y^2)^2}, \frac{2y}{1+(x^2+y^2)^2} \right). \] This means the function changes most rapidly in this direction, depending on the values of \( x \) and \( y \).
Partial Derivatives
Partial derivatives are used when dealing with functions of multiple variables. They measure the rate of change of a function concerning one variable, keeping others constant.
To find a partial derivative of a function \( f(x, y) \), you differentiate the function with respect to one variable, treating the other variable as a constant.
  • With respect to \( x \): This derivative shows how the function changes when only the variable \( x \) changes, given by \( \frac{\partial f}{\partial x} \).
  • With respect to \( y \): Similarly, this represents the change in the function when \( y \) changes, shown as \( \frac{\partial f}{\partial y} \).
Partial derivatives are essential in understanding and interpreting the behavior of multivariable functions. In solving the original exercise, these derivatives form the components of the gradient. They tell us how the function's output value changes in different directions with respect to each input variable.
Curl of a Vector Field
The curl is a measure of the rotational tendency of a vector field. It shows how much and in what manner a vector field tends to "rotate" around any given point.
The formula for the curl in three dimensions is given by:\[ abla \times \vec{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right). \]However, for a two-dimensional field that is extended to three dimensions with a zero third component, calculating the curl can demonstrate something interesting: the curl of the gradient of any scalar function is always zero, \( abla \times abla f = \mathbf{0} \).
  • This property highlights that a scalar field's gradient does not form loops or rotations, which makes it irrotational.
  • For our function \( f(x, y) = \tan^{-1}(x^2 + y^2) \), this principle holds true.
In the given solution, the curl was verified to be zero, as expected, confirming the role of the gradient as depicting purely directional change without rotation.

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Most popular questions from this chapter

Develop some of the classic differential geometry of curves. Let \(c:[a, b] \rightarrow \mathbb{R}^{3}\) be an infinitely differentiable path (derivatives of all orders exist). Assume \(\mathbf{c}^{\prime}(t) \neq \mathbf{0}\) for any \(t .\) The vector \(\mathbf{c}^{\prime}(t) /\left\|\mathbf{c}^{\prime}(t)\right\|=\mathbf{T}(t)\) is tangent to \(\mathbf{c}\) at \(\mathbf{c}(t),\) and, because \(\|\mathbf{T}(t)\|=1, \mathbf{T}\) is called the unit tangent to \(c\) (a) Show that \(\mathbf{T}^{\prime}(t) \cdot \mathbf{T}(t)=0 .\) [HINT: Differentiate \(\mathbf{T}(t) \cdot \mathbf{T}(t)=1 .]\) (b) Write down a formula for \(\mathbf{T}^{\prime}(t)\) in terms of \(\mathbf{c}\)

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