Q101 / 180 · KCET 2011 · Mathematics
If $\sqrt{r}=a e^{\theta \cot \alpha}$ where $a$ and $\alpha$ are real numbers, then $\frac{d^{2} r}{d \theta^{2}}-4 r \cot ^{2} \alpha$ is
- A$r$
- B$\frac{1}{r}$
- C1
- Correct answerD0
Solution
Given, $\quad \sqrt{r}=a e^{\theta \cot \alpha}$
Differentiating w.r.t. $\theta$,
$$
\begin{aligned}
\frac{1}{2 \sqrt{r}} \frac{d r}{d \theta} &=a \cot \alpha \cdot e^{\theta \cot \alpha} \\
\frac{d r}{d \theta} &=2 a \sqrt{r} \cot \alpha e^{\theta \cot \alpha} \\
\frac{d r}{d \theta} &=2 a \cdot a e^{\theta \cot \alpha} \cdot \cot \alpha \cdot e^{\theta \cot \alpha}
\end{aligned}
$$
[from Eq. (i)]
$$
\frac{d r}{d \theta}=2 a^{2} \cot \alpha \cdot \alpha \cdot e^{2000 t \alpha}
$$
Again $r$ differentiating w.r.t. $\theta$
$\frac{d^{2} r}{d \theta^{2}}=2 a^{2} \cot \alpha \cdot e^{2 \theta \cot \alpha} \cdot 2 \cot \alpha$ $\frac{d^{2} r}{d \theta^{2}}=4 a^{2} \cot ^{2} \alpha \cdot e^{20 \cot \theta}$ $\frac{d^{2} r}{d \theta^{2}}=4 \cot ^{2} \alpha \cdot\left(a e^{\theta \cot \alpha}\right)^{2}$ $\frac{d^{2} r}{d \theta^{2}}=4 \cot ^{2} \alpha \cdot(\sqrt{r})^{2} \quad$ [from Eq. (i)] $\frac{d^{2} r}{d \theta^{2}}-4 r \cot ^{2} \alpha=0$
Hence, the correct answer is option: 'D'.