Question

If $\sf\displaystyle \sqrt{r}=ae^{\displaystyle\theta cot \alpha }$ where $\sf \theta$ and $\sf\alpha$ are real numbers, then $\sf\displaystyle \dfrac{d^{2}r}{d \theta^{2}}-4rCot^{2} \alpha$ is ?

Answer 1

Given,

$\displaystyle \sf \: \sqrt{r}=ae^{\displaystyle\theta cot \alpha }$

Differentiating w.r.t theta (Notice that the variables under operation r and ∅ are in parametric form),

$\implies \sf \: \dfrac{d( \sqrt{r} )}{d \theta} = \dfrac{d(ae {}^{ \theta \: \cot( \alpha ) }) }{d \theta} \\ \\ \implies \sf \: \dfrac{d( \sqrt{r}) }{dr} \dfrac{dr}{d \theta} = ae {}^{ \theta \: \cot( \alpha ) } \times \dfrac{ d(\theta \: \cot( \alpha )) }{d \theta} \\ \\ \implies \sf \: \dfrac{1}{2 \sqrt{r} } \dfrac{dr}{d \theta} = a \cot( \alpha ) e {}^{ \theta \: \cot( \alpha ) } \\ \\ \implies \sf \: \dfrac{dr}{d \theta} = \cot( \alpha ) \bigg \{ ae {}^{ \theta \: \cot( \alpha ) } \bigg \} \times 2 \sqrt{r} \\ \\ \implies \sf \: \dfrac{dr}{d \theta} = \cot( \alpha ) \times \sqrt{r} \times 2 \sqrt{r} \\ \\ \implies \sf \: \dfrac{dr}{d \theta} =2r \cot( \alpha ) - - - - - - - (1)$

Differentiating w.r.t theta again,

$\implies \sf \: \dfrac{ {d}^{2}r }{ {d \theta}^{2} } = 2 \cot( \alpha ) \dfrac{dr}{d \theta} \\ \\ \implies \sf \: \dfrac{ {d}^{2}r }{ {d \theta}^{2} } =4r \cot {}^{2} ( \alpha ) \ \ \ \ \ \ \ |Using (1)| \\ \\ \implies \boxed{ \boxed{ \sf \: \dfrac{ {d}^{2}r }{ {d \theta}^{2} } - 4r \cot {}^{2} ( \alpha ) = 0}}$

Answer 2

Given:

$\sf\sqrt{r}=ae^{\displaystyle\theta cot \alpha }$

To find:

$\sf\dfrac{d^{2}r}{d \theta^{2}}-4rCot^{2} \alpha = ?$

Calculation:

• First, let's differentiate w.r.t $\theta$ and then apply Chain Rule.

$\rm \: \sqrt{r} = a {e}^{ \theta \cot( \alpha ) }$

$\rm \implies\: \dfrac{d(\sqrt{r})}{d \theta} = \dfrac{d \bigg(a {e}^{ \theta \cot( \alpha ) } \bigg)}{d \theta}$

$\rm \implies\: \dfrac{d(\sqrt{r})}{d r} \times \dfrac{dr}{d \theta} = a {e}^{ \theta \cot( \alpha ) } \times \dfrac{d \bigg \{ \theta \cot( \alpha ) \bigg \}}{d \theta}$

$\rm \implies\: \dfrac{1}{2 \sqrt{r} } \times \dfrac{dr}{d \theta} = a {e}^{ \theta \cot( \alpha ) } \times \cot( \alpha )$

$\rm \implies\: \dfrac{dr}{d \theta} = a {e}^{ \theta \cot( \alpha ) } \times \cot( \alpha ) \times 2 \sqrt{r}$

$\rm \implies\: \dfrac{dr}{d \theta} = \sqrt{r} \times \cot( \alpha ) \times 2 \sqrt{r}$

$\rm \implies\: \dfrac{dr}{d \theta} = 2r \cot( \alpha )$

• Second order differentiation w.r.t $\theta$.

$\rm \implies\: \dfrac{ {d}^{2} r}{d {\theta}^{2} } = 2 \cot( \alpha ) \times \dfrac{dr}{d \theta}$

$\rm \implies\: \dfrac{ {d}^{2} r}{d {\theta}^{2} } = 2 \cot( \alpha ) \times 2r \cot( \alpha )$

$\rm \implies\: \dfrac{ {d}^{2} r}{d {\theta}^{2} } =4r {\cot}^{2} ( \alpha )$

$\rm \implies\: \dfrac{ {d}^{2} r}{d {\theta}^{2} } - 4r {\cot}^{2} ( \alpha ) = 0$

So, final answer is :

$\boxed{ \bf\: \dfrac{ {d}^{2} r}{d {\theta}^{2} } - 4r {\cot}^{2} ( \alpha ) = 0}$