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: Understand the parametric setup.
Both $x$ and $y$ depend on a parameter $\theta$. To get $\dfrac{d^2y}{dx^2}$ we differentiate through $\theta$ each time.
Step 2: Differentiate $x$ and $y$ with respect to $\theta$.
$\dfrac{dx}{d\theta} = a(1 + \cos\theta)$ and $\dfrac{dy}{d\theta} = a\sin\theta$.
Step 3: Form the first derivative and simplify with half-angles.
$\dfrac{dy}{dx} = \dfrac{a\sin\theta}{a(1+\cos\theta)} = \dfrac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}} = \tan\dfrac{\theta}{2}$. This neat form makes the next step easy.
Step 4: Differentiate $\dfrac{dy}{dx}$ with respect to $\theta$.
$\dfrac{d}{d\theta}\left(\tan\dfrac{\theta}{2}\right) = \dfrac{1}{2}\sec^2\dfrac{\theta}{2}$.
Step 5: Multiply by $\dfrac{d\theta}{dx}$ to finish the second derivative.
$\dfrac{d^2y}{dx^2} = \dfrac{1}{2}\sec^2\dfrac{\theta}{2} \cdot \dfrac{1}{a(1+\cos\theta)} = \dfrac{1}{2}\sec^2\dfrac{\theta}{2}\cdot\dfrac{1}{a\cdot 2\cos^2\frac{\theta}{2}} = \dfrac{1}{4a}\sec^4\dfrac{\theta}{2}$.
Step 6: Evaluate at $\theta = \dfrac{\pi}{2}$.
Then $\dfrac{\theta}{2} = \dfrac{\pi}{4}$ and $\sec\dfrac{\pi}{4} = \sqrt{2}$, so $\sec^4\dfrac{\pi}{4} = 4$. Hence $\dfrac{d^2y}{dx^2} = \dfrac{1}{4a}\cdot 4 = \dfrac{1}{a}$.
\[ \boxed{\left(\dfrac{d^2y}{dx^2}\right)_{\theta=\pi/2} = \dfrac{1}{a}} \]