Problem 21 - Advanced Engineering Mathematics

Question

Let a be a constant vector and \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\). Verify the given identity. $$ \nabla \cdot(\mathbf{a} \times \mathbf{r})=0 $$

Step-by-Step Solution

Answer

The divergence of the cross product is zero, verifying the identity.

Step 1: Understand the Cross Product

The cross product \( \mathbf{a} \times \mathbf{r} \) between vector \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) is calculated using the determinant:\[\mathbf{a} \times \mathbf{r} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ x & y & z \end{vmatrix}\]Which expands to:\[ \mathbf{a} \times \mathbf{r} = (a_2 z - a_3 y) \mathbf{i} - (a_1 z - a_3 x) \mathbf{j} + (a_1 y - a_2 x) \mathbf{k} \]

Step 2: Calculate the Divergence of the Cross Product

The divergence \( abla \cdot (\mathbf{a} \times \mathbf{r}) \) is calculated by taking the partial derivatives of each component of \( \mathbf{a} \times \mathbf{r} \) with respect to \( x, y, \) and \( z \), and summing them:\[abla \cdot (\mathbf{a} \times \mathbf{r}) = \frac{\partial}{\partial x}(a_2 z - a_3 y) + \frac{\partial}{\partial y}(-(a_1 z - a_3 x)) + \frac{\partial}{\partial z}(a_1 y - a_2 x)\]This simplifies to:\[ \frac{\partial (a_2 z - a_3 y)}{\partial x} + \frac{\partial (-a_1 z + a_3 x)}{\partial y} + \frac{\partial (a_1 y - a_2 x)}{\partial z} \]

Step 3: Apply Partial Derivatives

Now calculate each partial derivative:- \( \frac{\partial (a_2 z - a_3 y)}{\partial x} = 0 \) since it contains no \( x \).- \( \frac{\partial (-a_1 z + a_3 x)}{\partial y} = 0 \) since it contains no \( y \).- \( \frac{\partial (a_1 y - a_2 x)}{\partial z} = 0 \) since it contains no \( z \).Summing these derivatives, we find:\[ abla \cdot (\mathbf{a} \times \mathbf{r}) = 0 \]

Step 4: Conclusion: Verify the Identity

Since the divergence of the cross product is zero, we have verified the identity:\[abla \cdot (\mathbf{a} \times \mathbf{r}) = 0\]Thus the statement given in the exercise is true.

Key Concepts

Cross Product

The cross product is a fundamental operation in vector calculus that helps in finding a vector perpendicular to two given vectors. It is especially useful in physics to find moments or torques. The cross product of two vectors, \(\mathbf{a}\) and \(\mathbf{r}\), results in a third vector that is orthogonal to both \(\mathbf{a}\) and \(\mathbf{r}\). To compute the cross product, we use the determinant of a matrix formed by the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) and the components of the vectors.

• The resulting vector from \(\mathbf{a} \times \mathbf{r}\) is \((a_2 z - a_3 y) \mathbf{i} - (a_1 z - a_3 x) \mathbf{j} + (a_1 y - a_2 x) \mathbf{k}\).
• This vector provides a direction and magnitude that captures how \(\mathbf{a}\) and \(\mathbf{r}\) move away from each other in three-dimensional space.

Understanding the geometry of cross products involves visualizing how the two source vectors create a plane, and the cross product vector points perpendicular to this plane.

Divergence

Divergence is a vector operation that measures a vector field's tendency to originate from or converge into a point. In simpler terms, it quantifies the rate at which "stuff" is expanding or compressing at a point in space. For a vector \(\mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k}\), its divergence is computed as \(abla \cdot \mathbf{v} = \frac{\partial v_1}{\partial x} + \frac{\partial v_2}{\partial y} + \frac{\partial v_3}{\partial z}\).

• In the exercise, divergence is applied to the cross product \(\mathbf{a} \times \mathbf{r}\).
• Each component's rate of change is calculated using partial derivatives, which reveal the divergence value of 0, confirming no expansion or contraction occurs.

This zero value illustrates the balance in vector magnitudes such that the net rate of outward flux from a small volume around any point cancels out.

Partial Derivatives

Partial derivatives are a core concept in differentiating functions with multiple variables. They represent the rate of change with respect to one variable while keeping others constant. This is important to understand gradients, curls, and divergences in vector calculus. In the given exercise:

• The partial derivative \(\frac{\partial}{\partial x}(a_2 z - a_3 y) = 0\) demonstrates that the component is independent of \(x\).
• Similarly, \(\frac{\partial}{\partial y}(-a_1 z + a_3 x) = 0\) and \(\frac{\partial}{\partial z}(a_1 y - a_2 x) = 0\) show independence from \(y\) and \(z\) respectively.

Mastering partial derivatives allows students to handle changes in multivariable functions effectively, which is crucial in fields like physics and engineering. In this exercise, partial derivatives facilitate the calculation of divergence, leading us to conclude the verification of the stated identity.