Question
If $ \sqrt{r} = a e^{\theta \cot \alpha} $ where $ a $ and $ \alpha $ are real numbers, then $ \frac{d^{2}r}{d\theta^{2}} - 4r \cot^{2}\alpha $ is ...
If $ \sqrt{r} = a e^{\theta \cot \alpha} $ where $ a $ and $ \alpha $ are real numbers, then $ \frac{d^{2}r}{d\theta^{2}} - 4r \cot^{2}\alpha $ is ...