Discussion
The divergence of the vector field \(\vec{u} = e^{x}\left(\cos y\,\hat{i} + \sin y\,\hat{j}\right)\) is
Answer: C
The divergence is \(2e^{x}\cos y\).
The field has components \(u_{x} = e^{x}\cos y\) and \(u_{y} = e^{x}\sin y\). The divergence is the sum of each component differentiated along its own direction:
\[ \begin{aligned} \nabla\cdot\vec{u} &= \frac{\partial}{\partial x}(e^{x}\cos y) + \frac{\partial}{\partial y}(e^{x}\sin y) \\ &= e^{x}\cos y + e^{x}\cos y \\ &= 2e^{x}\cos y \end{aligned} \]
\(e^{x}\cos y + e^{x}\sin y\) forgets to differentiate the sine with respect to y, and \(2e^{x}\sin y\) mixes up which component is differentiated. The divergence is not zero, since both terms are positive where cos y is.
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