A) r
B) \[\frac{1}{r}\]
C) 1
D) 0
Correct Answer: D
Solution :
Given, \[\sqrt{r}=a{{e}^{\theta \cot \alpha }}\] ?...(i) Differentiating w.r.t. \[\theta ,\] \[\frac{1}{2\sqrt{r}}\frac{dr}{d\theta }=a\,\cot \alpha .{{e}^{\theta \cot \alpha }}\] \[\frac{dr}{d\theta }=2a\sqrt{r}\,\cot \alpha \,{{e}^{\theta \cot \alpha }}\] \[\frac{dr}{d\theta }=2a.a{{e}^{\theta \cot \alpha }}.\cot \alpha .{{e}^{\theta \cot \alpha }}\] [form Eq. (i)] \[\frac{dr}{d\theta }=2{{a}^{2}}\cot \alpha .\alpha .{{e}^{2\theta \cot \alpha }}\] Agian r differentiating w.r.t. \[\theta \] \[\frac{{{d}^{2}}r}{d{{\theta }^{2}}}=2{{a}^{2}}\cot \alpha .{{e}^{2\theta \,\cot \alpha }}.2\cot \alpha \] \[\frac{{{d}^{2}}r}{d{{\theta }^{2}}}=4{{a}^{2}}{{\cot }^{2}}\alpha .{{e}^{2\theta \cot \theta }}\] \[\frac{{{d}^{2}}r}{d{{\theta }^{2}}}=4{{\cot }^{2}}\alpha .{{(a{{e}^{\theta \cot \alpha }})}^{2}}\] \[\frac{{{d}^{2}}r}{d{{\theta }^{2}}}=4{{\cot }^{2}}\alpha .{{(\sqrt{r})}^{2}}\] [from Eq. (i)] \[\frac{{{d}^{2}}r}{d{{\theta }^{2}}}=4r\,\,{{\cot }^{2}}\alpha =0\]
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