At any point (r, $ heta$) on the curve $r=ae^{ heta cot\alpha},$ the radius of curvature is equal to:

Q: At any point (r, $\theta$) on the curve $r=ae^{\theta cot\alpha},$ the radius of curvature is equal to:

At any point (r, $\theta$) on the curve $r=ae^{\theta cot\alpha},$ the radius of curvature is equal to:

Question

At any point (r, $	heta$) on the curve $r=ae^{	heta cot\alpha},$ the radius of curvature is equal to:

Accepted Answer

To determine why Option 2 ($r~cosec~\alpha$) is the correct answer for the radius of curvature at any point on the curve given by $ r = ae^{\theta \cot \alpha} $, we need to analyze the properties of the curve and the formula for the radius of curvature in polar coordinates.

### Explanation for Option 2:

1. **Curve Analysis**: The given curve is $ r = ae^{\theta \cot \alpha} $. This is a type of exponential spiral, where $ a $ is a constant and $ \alpha $ is a parameter that affects the shape of the spiral.

2. **Radius of Curvature in Polar Coordinates**: The formula for the radius of curvature $ \rho $ in polar coordinates $(r, \theta)$ is given by:
   $$
   \rho = \frac{(r^2 + \left(\frac{dr}{d\theta}\right)^2)^{3/2}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}
   $$

3. **Derivatives**:
   - First, find $\frac{dr}{d\theta}$:
     $$
     \frac{dr}{d\theta} = \frac{d}{d\theta}(ae^{\theta \cot \alpha}) = ae^{\theta \cot \alpha} \cdot \cot \alpha = r \cot \alpha
     $$
   - Next, find $\frac{d^2r}{d\theta^2}$:
     $$
     \frac{d^2r}{d\theta^2} = \frac{d}{d\theta}(r \cot \alpha) = \frac{d}{d\theta}(ae^{\theta \cot \alpha} \cot \alpha) = r \cot^2 \alpha
     $$

4. **Substitute into the Radius of Curvature Formula**:
   - Substitute $ r $, $\frac{dr}{d\theta}$, and $\frac{d^2r}{d\theta^2}$ into the formula for $\rho$:
     $$
     \rho = \frac{(r^2 + (r \cot \alpha)^2)^{3/2}}{r^2 + 2(r \cot \alpha)^2 - r(r \cot^2 \alpha)}
     $$
   - Simplify the expression:
     $$
     \rho = \frac{(r^2 + r^2 \cot^2 \alpha)^{3/2}}{r^2 + 2r^2 \cot^2 \alpha - r^2 \cot^2 \alpha}
     $$
     $$
     \rho = \frac{(r^2(1 + \cot^2 \alpha))^{3/2}}{r^2(1 + \cot^2 \alpha)}
     $$
     $$
     \rho = \frac{r^3 \csc^3 \alpha}{r^2 \csc^2 \alpha}
     $$
     $$
     \rho = r \csc \alpha
     $$

5. **Conclusion**: The radius of curvature simplifies to $ r \csc \alpha $, which matches Option 2.

### Explanation for Why Other Options are Incorrect:

- **Option 1 ($r \sin \alpha$)**: This option suggests the radius of curvature is directly proportional to $ r $ and $ \sin \alpha $. However, the derivation shows that the radius of curvature involves $ \csc \alpha $, not $ \sin \alpha $.

- **Option 3 ($\frac{1}{r} \sin \alpha$)**: This option implies an inverse relationship with $ r $ and involves $ \sin \alpha $. The correct derivation shows that the radius of curvature is directly proportional to $ r $ and involves $ \csc \alpha $, not $ \sin \alpha $.

- **Option 4 ($\frac{1}{r} \cosec \alpha$)**: This option suggests an inverse relationship with $ r $ and involves $ \csc \alpha $. While it correctly includes $ \csc \alpha $, the relationship with $ r $ is incorrect, as the correct expression is directly proportional to $ r $.

Thus, Option 2 ($r \cosec \alpha$) is the correct answer, as it accurately reflects the derived formula for the radius of curvature for the given curve.