Math Problem Statement
Consider the vector function: A⃗ =6(−x^y+y^x)/(x2+y2) . The magnitude of ∇×A⃗ at a point with coordinates (x=1,y=2,z=3) is (give your answer to three significant digits)
Solution
To solve for the magnitude of at the given coordinates , let's work step by step:
1. Recall the formula for
The curl of a vector field is given by:
\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix}$$ Here, $$A_x, A_y,$$ and $$A_z$$ are the components of the vector $$\mathbf{A}$$. For $$\mathbf{A} = 6 \frac{-x^y + y^x}{x^2 + y^2}$$, the vector field is two-dimensional (in the $$x$$-$$y$$ plane), and we can write: $$\mathbf{A} = (A_x, A_y, 0),$$ where: $$A_x = 6 \frac{-x^y + y^x}{x^2 + y^2}, \quad A_y = 6 \frac{x^y - y^x}{x^2 + y^2}.$$ ### 2. Compute partial derivatives We compute $$\partial A_y / \partial x$$ and $$\partial A_x / \partial y$$, since the curl in 2D reduces to: $$\nabla \times \mathbf{A} = \left( 0, 0, \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right).$$ #### (a) Partial derivatives 1. Compute $$\frac{\partial A_y}{\partial x}$$: $$A_y = 6 \frac{x^y - y^x}{x^2 + y^2}.$$ Apply the quotient rule and product rule as necessary. 2. Compute $$\frac{\partial A_x}{\partial y}$$: $$A_x = 6 \frac{-x^y + y^x}{x^2 + y^2}.$$ Similarly, apply the quotient rule and product rule. ### 3. Plug in $$x = 1, y = 2$$ After finding the expressions for the partial derivatives, substitute $$x = 1$$ and $$y = 2$$ into the expressions. ### 4. Compute magnitude of $$\nabla \times \mathbf{A}$$ The magnitude is given by: $$|\nabla \times \mathbf{A}| = \sqrt{(0)^2 + (0)^2 + \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right)^2}.$$ Let me calculate these explicitly. sympy as sp # Define variables x, y = sp.symbols('x y') # Define the components of A⃗ A_x = 6 * (-x**y + y**x) / (x**2 + y**2) A_y = 6 * (x**y - y**x) / (x**2 + y**2) # Partial derivatives dA_y_dx = sp.diff(A_y, x) dA_x_dy = sp.diff(A_x, y) # Curl in 2D: (∂A_y/∂x - ∂A_x/∂y) curl = dA_y_dx - dA_x_dy # Substitute values x=1, y=2 curl_at_point = curl.subs({x: 1, y: 2}) # Magnitude of the curl (it's scalar in 2D) magnitude_curl = sp.Abs(curl_at_point) magnitude_curlThe magnitude of $$\nabla \times \mathbf{A}$$ at the point $$(x = 1, y = 2)$$ is: $$\frac{66}{25} - \frac{12 \ln(2)}{5}$$ To get a numerical approximation, let me calculate this to three significant digits. Numerical evaluation to three significant digits magnitude_curl_numeric = magnitude_curl.evalf() magnitude_curl_numericThe magnitude of $$\nabla \times \mathbf{A}$$ at the point $$(x = 1, y = 2)$$ is approximately **0.976** (to three significant digits). Let me know if you would like a detailed breakdown of the steps or further clarification. Here are 5 related questions to deepen understanding: 1. How does the curl of a vector field relate to its rotational properties? 2. Can you explain the physical significance of $$\nabla \times \mathbf{A}$$ in fluid mechanics or electromagnetism? 3. How would the calculation change if $$\mathbf{A}$$ had a nonzero $$z$$-component? 4. What is the divergence of the same vector field $$\mathbf{A}$$? 5. Why does the curl of a 2D vector field have only a $$z$$-component? **Tip:** Always double-check your substitutions and numerical approximations to avoid errors in final results.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Vector Calculus
Curl of a Vector Field
Partial Derivatives
Formulas
Curl of a vector field: ∇×A⃗ = (∂A_y/∂x - ∂A_x/∂y)k̂ in 2D
Quotient Rule: d(u/v)/dx = (v(du/dx) - u(dv/dx))/v^2
Theorems
Stokes' Theorem (conceptually related)
Suitable Grade Level
Grades 11-12 or College Level
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