Discussion
Divergence of the vector field :
V(x, y, z) = −(x cos xy + y)i + (y cos xy)j + (sin z2 + x2 + y2)k is :
Answer: D
The divergence is \(2z \cos z^{2}\).
Take the partial derivative of each component with respect to its own variable and add them.
\[ \begin{aligned} \frac{\partial V_{x}}{\partial x} &= -\cos xy + xy\sin xy \\ \frac{\partial V_{y}}{\partial y} &= \cos xy - xy\sin xy \\ \frac{\partial V_{z}}{\partial z} &= 2z\cos z^{2} \end{aligned} \]
The first two cancel exactly, leaving \(\nabla \cdot \mathbf{V} = 2z \cos z^{2}\). The \(x^{2} + y^{2}\) inside the k-component does not depend on \(z\), so it drops out.
The options with \(\sin xy\) or \(\cos z\) keep terms that cancel or differentiate wrongly, and \(2x \cos x^{2}\) differentiates the wrong component.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development