Question:
For a two-dimensional plane, the unit vectors \((\hat{e}_r,\hat{e}_\theta)\) of the polar coordinate system and \((\hat{i},\hat{j})\) of the cartesian coordinate system are related by
\[
\hat{e}_r=\cos\theta\,\hat{i}+\sin\theta\,\hat{j}, \qquad
\hat{e}_\theta=-\sin\theta\,\hat{i}+\cos\theta\,\hat{j}.
\]
Which one of the following is the CORRECT value of \(\dfrac{\partial(\hat{e}_r+\hat{e}_\theta){\partial\theta}\)?}
For a two-dimensional plane, the unit vectors \((\hat{e}_r,\hat{e}_\theta)\) of the polar coordinate system and \((\hat{i},\hat{j})\) of the cartesian coordinate system are related by
\[
\hat{e}_r=\cos\theta\,\hat{i}+\sin\theta\,\hat{j}, \qquad
\hat{e}_\theta=-\sin\theta\,\hat{i}+\cos\theta\,\hat{j}.
\]
Which one of the following is the CORRECT value of \(\dfrac{\partial(\hat{e}_r+\hat{e}_\theta){\partial\theta}\)?}
Show Hint
- In polar coordinates: \(\dfrac{\partial \hat{e}_r}{\partial\theta}=\hat{e}_\theta\) and \(\dfrac{\partial \hat{e}_\theta}{\partial\theta}=-\hat{e}_r\).
- Memorizing these two relations helps in curvilinear coordinate calculus and vector differentiation problems.
- Memorizing these two relations helps in curvilinear coordinate calculus and vector differentiation problems.
Updated On: Aug 26, 2025
- \(1\)
- \(\hat{e}_\theta\)
- \(\hat{e}_r+\hat{e}_\theta\)
- \(-\hat{e}_r+\hat{e}_\theta\)
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The Correct Option is D
Solution and Explanation
Step 1: Differentiate the basis vectors with respect to \(\theta\):
\[
\frac{\partial\hat{e}_r}{\partial\theta}
= -\sin\theta\,\hat{i}+\cos\theta\,\hat{j}
= \hat{e}_\theta,\qquad
\frac{\partial\hat{e}_\theta}{\partial\theta}
= -\cos\theta\,\hat{i}-\sin\theta\,\hat{j}
= -\hat{e}_r.
\]
Step 2: Therefore,
\[
\frac{\partial(\hat{e}_r+\hat{e}_\theta)}{\partial\theta}
= \frac{\partial\hat{e}_r}{\partial\theta}
+ \frac{\partial\hat{e}_\theta}{\partial\theta}
= \hat{e}_\theta - \hat{e}_r
= -\hat{e}_r+\hat{e}_\theta.
\]
\[
\boxed{-\hat{e}_r+\hat{e}_\theta}
\]
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