Question

If $x = a(\theta + \sin \theta)$ and $y = a(1 - \cos \theta)$, then $\left(\frac{d^2y}{dx^2}\right)_{\text{at } \theta = \pi / 2} =$

• A. $\frac{a}{2}$
• B. $\frac{1}{a}$
• C. $a$
• D. $2a$

Hint:

To simplify the first derivative quickly, use half-angle trigonometric identities: $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ and $1+\cos\theta = 2\cos^2\frac{\theta}{2}$. Thus, $\frac{dy}{dx} = \tan\frac{\theta}{2}$. Differentiating this gives $\frac{1}{2}\sec^2\frac{\theta}{2}$ directly, bypassing the quotient rule completely!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the value of the second derivative $\frac{d^2y}{dx^2}$ evaluated at $\theta = \frac{\pi}{2}$ for a curve defined parametrically in terms of $\theta$.

Step 2: Key Formula or Approach:
For parametric equations $x = f(\theta)$ and $y = g(\theta)$, the first derivative is:
$$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$ The second derivative is obtained by differentiating $\frac{dy}{dx}$ with respect to $\theta$ and applying the chain rule:
$$\frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right) \cdot \frac{d\theta}{dx}$$

Step 3: Detailed Explanation:
First, differentiate $x$ and $y$ with respect to $\theta$:
$$\frac{dx}{d\theta} = a(1 + \cos \theta)$$ $$\frac{dy}{d\theta} = a(0 - (-\sin \theta)) = a\sin \theta$$ Now, calculate the first derivative $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{a\sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta}$$ Next, differentiate $\frac{dy}{dx}$ with respect to $\theta$ using the quotient rule:
$$\frac{d}{d\theta}\left(\frac{\sin \theta}{1 + \cos \theta}\right) = \frac{(1 + \cos \theta)(\cos \theta) - (\sin \theta)(-\sin \theta)}{(1 + \cos \theta)^2}$$ $$= \frac{\cos \theta + \cos^2 \theta + \sin^2 \theta}{(1 + \cos \theta)^2} = \frac{\cos \theta + 1}{(1 + \cos \theta)^2} = \frac{1}{1 + \cos \theta}$$ Now, compute the second derivative $\frac{d^2y}{dx^2}$ by multiplying by $\frac{d\theta}{dx} = \frac{1}{a(1 + \cos \theta)}$:
$$\frac{d^2y}{dx^2} = \frac{1}{1 + \cos \theta} \cdot \frac{1}{a(1 + \cos \theta)} = \frac{1}{a(1 + \cos \theta)^2}$$ Finally, evaluate this expression at $\theta = \frac{\pi}{2}$. We know that $\cos\left(\frac{\pi}{2}\right) = 0$:
$$\left(\frac{d^2y}{dx^2}\right)_{\theta = \pi / 2} = \frac{1}{a(1 + 0)^2} = \frac{1}{a}$$ This matches option (B).

Step 4: Final Answer:
The value of the second derivative at $\theta = \frac{\pi}{2}$ is $\frac{1}{a}$, which corresponds to option (B).