Question

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

• A. $r$
• B. $1/r$
• C. $1$
• D. $0$

Answer 1

Answer: (D)

Given, $\sqrt{r}=a e^{\theta \cot \alpha}$...(i)
Differentiating w.r.t. $\theta$,
$\frac{1}{2 \sqrt{r}} \frac{d r}{d \theta} =a \cot \alpha \cdot e^{\theta \cot \alpha}$
$\frac{d r}{d \theta} =2 a \sqrt{r} \cot \alpha e^{\theta \cot \alpha}$
$\frac{d r}{d \theta} =2 a \cdot a e^{\theta \cot \alpha} \cdot \cot \alpha \cdot e^{\theta \cot \alpha}$
[form Eq.(i)]
$\frac{d r}{d \theta}=2 a^{2} \cot \alpha \cdot \alpha \cdot e^{2 \theta \cot \alpha}$
Agian $r$ differentiating w.r.t. $\theta$
$\frac{d^{2} r}{d \theta^{2}}=2 a^{2} \cot \alpha \cdot e^{2 \theta \cot \alpha} \cdot 2 \cot \alpha$
$\frac{d^{2} r}{d \theta^{2}}=4 a^{2} \cot ^{2} \alpha \cdot e^{2 \theta \cot \theta}$
$\frac{d^{2} r}{d \theta^{2}}=4 \cot ^{2} \alpha \cdot\left(a e^{\theta \cot \alpha}\right)^{2}$
$\frac{d^{2} r}{d \theta^{2}}=4 \cot ^{2} \alpha \cdot(\sqrt{r})^{2}$ [from Eq. (i)]
$\frac{d^{2} r}{d \theta^{2}}-4 r \cot ^{2} \alpha=0$