Question

If $x = a \cos \theta$, $y = b \sin \theta$, then $\left[\frac{d^2 y}{d x^2}\right] = $

• A. $\frac{2}{a^2b}$
• B. $\sqrt{2}\left(\frac{b}{a^2}\right)$
• C. $-2\sqrt{2}\left(\frac{b}{a^2}\right)$
• D. $2\sqrt{2}\left(\frac{b}{a^2}\right)$

Hint:

Be extremely careful with the second step in parametric differentiation! A very common mistake is forgetting to multiply by the chain rule component $\frac{d\theta}{dx}$. Always remember: $\frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta}(y')}{\frac{dx}{d\theta}}$. Keep this pattern in mind to avoid missing negative signs or losing factor terms!

Answer 1

The Correct Option is C

Step 1: Differentiate $x$ and $y$ with respect to $\theta$.
$\frac{dx}{d\theta}=-a\sin\theta$ and $\frac{dy}{d\theta}=b\cos\theta$.

Step 2: Get the first derivative.
\[ \frac{dy}{dx}=\frac{b\cos\theta}{-a\sin\theta}=-\frac{b}{a}\cot\theta \]

Step 3: Differentiate again with respect to $x$.
Differentiate the above with respect to $\theta$, then multiply by $\frac{d\theta}{dx}=\frac{1}{-a\sin\theta}$. This gives $\frac{d^2y}{dx^2}=-\frac{b}{a^2}\csc^3\theta$.

Step 4: Put $\theta=\frac{\pi}{4}$.
Here $\csc\frac{\pi}{4}=\sqrt2$, so $\csc^3=2\sqrt2$. \[ \frac{d^2y}{dx^2}=-2\sqrt2\left(\frac{b}{a^2}\right) \] \[ \boxed{-2\sqrt2\left(\frac{b}{a^2}\right)} \]