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Chapter 7: Problem 60

. Given the components of a vector fleld \(A=A_{1} \mathbf{e}_{1}+A_{2} \mathbf{e}_{2}+A_{3} \mathbf{e}_{3}\), find the components of its curl.

Short Answer

Expert verified
#tag_title#Question Given a vector field A with components \(A_{1} = x^{2}y\), \(A_{2} = 3xz\), and \(A_{3} = y^{2}z\), find the components of its curl. #tag_answer#Curl of A is (-2yz, x^2 - 2yz, 3z).

Step by step solution

01

Recall the definition and formula of the curl of a vector field

The curl of a vector field is a measure of the field's tendency to swirl around a particular point. Mathematically, the curl is defined as the cross product of the del (nabla) operator and the vector field. If we have a vector field A with components \(A_{1}\), \(A_{2}\), and \(A_{3}\), its curl, denoted as \(\nabla \times \mathbf{A}\), can be computed as the following determinant: $\nabla \times \mathbf{A} = \begin{vmatrix} \hat{\mathbf{e}_1} & \hat{\mathbf{e}_2} & \hat{\mathbf{e}_3} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_{1} & A_{2} & A_{3} \end{vmatrix}$
02

Compute the components of the curl

Now, we need to calculate the determinant. We can do this by applying the cofactor expansion method, resulting in: $\nabla \times \mathbf{A} = \begin{pmatrix} \frac{\partial A_{3}}{\partial y} - \frac{\partial A_{2}}{\partial z} \\ \frac{\partial A_{1}}{\partial z} - \frac{\partial A_{3}}{\partial x} \\ \frac{\partial A_{2}}{\partial x} - \frac{\partial A_{1}}{\partial y} \end{pmatrix}$ The components of the curl are given by the elements of the resulting vector. To find the components of the curl of A, take the partial derivatives of the components \(A_{1}\), \(A_{2}\), and \(A_{3}\) with respect to x, y, and z as appropriate, and then substitute them into the formula above.

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Most popular questions from this chapter

Verify that \(f^{*}\) preserves exterior multiplication: \(f^{*}\left(\omega^{*} \wedge \omega^{l}\right)=\left(f^{*} \omega^{k}\right) \wedge\left(f^{*} \omega^{f}\right)\).

Prove the formulas for differentiating a sum and a product: $$ d\left(\omega_{1}+\omega_{2}\right)=d \omega_{1}+d \omega_{2} . $$ and $$ d\left(\omega^{k} \wedge \omega^{\prime}\right)=d \omega^{k} \wedge \omega^{\prime}+(-1)^{k} \omega^{k} \wedge d \omega^{l} . $$

Show that the mapping $$ \left(w_{1}, \omega_{2}\right) \rightarrow w_{1} \wedge w_{2} $$ is bilinear and skew symmetric: $$ \begin{gathered} \omega_{1} \wedge \omega_{2}=-\omega_{2} \wedge \omega_{1} \\ \left(\lambda^{\prime}\left(\omega_{1}^{\prime}+\lambda^{\prime \prime} \omega_{1}^{\prime \prime}\right) \wedge \omega_{2}=\lambda^{\prime} \omega_{1}^{\prime} \wedge \omega_{2}+\lambda^{N} \omega_{1}^{\prime \prime} \wedge \omega_{2}\right. \end{gathered} $$

Show that the product of monomials is associative: $$ \left(\omega^{k} \times \omega^{r}\right) \wedge \omega^{m}-\omega^{2} \times\left(\omega^{2} \times \omega \omega^{m}\right) $$ and skew-commutative: $$ \omega^{k} \wedge \omega^{l}=(-1)^{k} \omega^{l} \wedge \omega^{k} . $$

Show that, under the isomorphisms established above, the exterior product of 1 -forms becomes the vector product in \(\mathbb{R}^{3}\), i.e., that $$ \omega_{\mathbf{A}}^{1} \wedge \omega_{\mathrm{B}}^{\mathrm{H}}=\omega_{\left|A_{,} \mathbf{B}\right|}^{2} \text { for any } \mathbf{A}, \mathbf{B} \in \mathbb{R}^{3} \text {. } $$ In this way the exterior product of 1 -forms can be considered as an extension of the vector product in \(\mathbb{R}^{3}\) to higher dimensions. However, in the \(n\)-dimensional case, the product is not a vector in the same space; the space of 2-forms on \(R^{n}\) is isomorphic to \(R^{n}\) only for \(n=3\).
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