Chapter 3: Problem 6
Show that $$ \nabla \times \frac{\mathbf{A} \times \mathbf{r}}{2}=\mathbf{A} $$ where \(\mathbf{r}=\mathbf{i} x+\mathbf{j} y+\mathbf{k} z\) and \(\mathbf{A}\) is a constant vector.
Short Answer
Expert verified
The curl \( \nabla \times \frac{\mathbf{A} \times \mathbf{r}}{2} \) is equal to \( \mathbf{A} \).
Step by step solution
01
Define the Cross Product
Start by defining the cross product \( \mathbf{A} \times \mathbf{r} \). Given that \( \mathbf{r} = \mathbf{i}x + \mathbf{j}y + \mathbf{k}z \) and \( \mathbf{A} = A_x\mathbf{i} + A_y\mathbf{j} + A_z\mathbf{k} \), the cross product \( \mathbf{A} \times \mathbf{r} \) can be determined using the determinant method:\[\mathbf{A} \times \mathbf{r} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ A_x & A_y & A_z \ x & y & z \end{vmatrix}\]After expanding the determinant, we obtain:\[\mathbf{A} \times \mathbf{r} = (A_y z - A_z y)\mathbf{i} - (A_x z - A_z x)\mathbf{j} + (A_x y - A_y x)\mathbf{k}\]
02
Simplify the Expression with Division
The original problem requires investigating the expression \( abla \times \left( \frac{\mathbf{A} \times \mathbf{r}}{2} \right) \). We first rewrite the cross product as:\[ \frac{\mathbf{A} \times \mathbf{r}}{2} = \frac{1}{2}((A_y z - A_z y)\mathbf{i} - (A_x z - A_z x)\mathbf{j} + (A_x y - A_y x)\mathbf{k}) \]
03
Calculate the Curl
Now, we calculate the curl of the expression \( abla \times \left( \frac{\mathbf{A} \times \mathbf{r}}{2} \right) \). Here is the formula for the curl:\[ abla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ F_x & F_y & F_z \end{vmatrix} \]Substitute \( F_x = \frac{1}{2}(A_y z - A_z y), \ F_y = \frac{1}{2}(-(A_x z - A_z x)), \ F_z = \frac{1}{2}(A_x y - A_y x) \) into this formula, and then calculate the partial derivatives and determinants.
04
Evaluate the Determinant
After substituting the components of \( \mathbf{F} \) into the determinant for the curl, calculate each term of the determinant. Evaluate the following:- \( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \)- \( \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \)- \( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \)Each of these expressions simplifies to the components of \( \mathbf{A} \). Therefore, the result becomes \( \mathbf{A} \).
05
Conclude the Result
After completing all the calculations, you will find that:\[ abla \times \left( \frac{\mathbf{A} \times \mathbf{r}}{2} \right) = \mathbf{A} \]This confirms the given statement. Thus, the curl of half the cross product is indeed equal to the constant vector \( \mathbf{A} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Curl
The concept of a curl in vector calculus is crucial for understanding the rotation or swirling strength of a vector field. When you see the notation \( abla \times \mathbf{F} \), it refers to the curl of a vector field \( \mathbf{F} \). It gives us a new vector that represents the rotation at each point in the vector field.
To find the curl of a vector field \( \mathbf{F} = \langle F_x, F_y, F_z \rangle \), use the formula:
To find the curl of a vector field \( \mathbf{F} = \langle F_x, F_y, F_z \rangle \), use the formula:
- \( \mathbf{i} \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \right) \)
- \( -\mathbf{j} \left( \frac{\partial F_z}{\partial x} - \frac{\partial F_x}{\partial z} \right) \)
- \( \mathbf{k} \left( \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \)
Cross Product
The cross product is a fundamental operation in vector calculus. It takes two vectors and produces a third vector perpendicular to both. This is especially useful in physics and engineering where perpendicular vectors are needed, such as in torque calculations.
Given two vectors \( \mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k} \) and \( \mathbf{B} = B_x \mathbf{i} + B_y \mathbf{j} + B_z \mathbf{k} \), their cross product \( \mathbf{A} \times \mathbf{B} \) can be evaluated using the determinant method:
Given two vectors \( \mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k} \) and \( \mathbf{B} = B_x \mathbf{i} + B_y \mathbf{j} + B_z \mathbf{k} \), their cross product \( \mathbf{A} \times \mathbf{B} \) can be evaluated using the determinant method:
- \( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix} \)
Partial Derivatives
Partial derivatives are an extension of the concept of derivatives applied to multi-variable functions. In multivariable calculus, they allow us to understand how a function changes as each variable changes independently while keeping the others constant.
Consider a function \( f(x, y, z) \). The partial derivative of \( f \) with respect to \( x \), written as \( \frac{\partial f}{\partial x} \), measures the rate at which \( f \) changes as \( x \) increases while \( y \) and \( z \) are held constant. Similarly, you calculate \( \frac{\partial f}{\partial y} \) and \( \frac{\partial f}{\partial z} \) for changes with respect to \( y \) and \( z \) respectively.
Consider a function \( f(x, y, z) \). The partial derivative of \( f \) with respect to \( x \), written as \( \frac{\partial f}{\partial x} \), measures the rate at which \( f \) changes as \( x \) increases while \( y \) and \( z \) are held constant. Similarly, you calculate \( \frac{\partial f}{\partial y} \) and \( \frac{\partial f}{\partial z} \) for changes with respect to \( y \) and \( z \) respectively.
- This concept is crucial for multi-dimensional analysis in the fields of physics, engineering, and economics where multiple factors influence outcomes.
- They appear in gradient vectors, describing how functions rapidly change, and are foundational in optimization problems and fields like data analysis.