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: Understanding the Question:
The question gives parametric equations for $x$ and $y$ in terms of the parameter $\theta$. We need to calculate the second derivative $\frac{d^2 y}{d x^2}$ evaluated specifically at $\theta = \frac{\pi}{4}$.

Step 2: Key Formula or Approach:
First, we find the first derivative using the parametric differentiation rule: $$ \frac{d y}{d x} = \frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} $$ To compute the second derivative, we differentiate $\frac{d y}{d x}$ with respect to $\theta$ and apply the chain rule factor: $$ \frac{d^2 y}{d x^2} = \frac{d}{d \theta}\left(\frac{d y}{d x}\right) \cdot \frac{d \theta}{d x} $$

Step 3: Detailed Explanation:
Let's find the derivatives of $x$ and $y$ with respect to $\theta$: $$ \frac{d x}{d \theta} = \frac{d}{d\theta}(a \cos\theta) = -a \sin\theta $$ $$ \frac{d y}{d \theta} = \frac{d}{d\theta}(b \sin\theta) = b \cos\theta $$ Now, compute the first derivative $\frac{d y}{d x}$: $$ \frac{d y}{d x} = \frac{b \cos\theta}{-a \sin\theta} = -\frac{b}{a} \cot\theta $$ Next, differentiate this result with respect to $x$ to find the second derivative: $$ \frac{d^2 y}{d x^2} = \frac{d}{d\theta}\left(-\frac{b}{a} \cot\theta\right) \cdot \frac{d \theta}{d x} $$ Knowing that the derivative of $\cot\theta$ is $-\csc^2\theta$: $$ \frac{d^2 y}{d x^2} = \left(-\frac{b}{a} (-\csc^2\theta)\right) \cdot \frac{1}{-a \sin\theta} $$ $$ \frac{d^2 y}{d x^2} = \frac{b \csc^2\theta}{a} \cdot \frac{1}{-a \sin\theta} = -\frac{b}{a^2} \csc^3\theta $$ Now, let's evaluate this expression at $\theta = \frac{\pi}{4}$. Since $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, its reciprocal is $\csc\left(\frac{\pi}{4}\right) = \sqrt{2}$: $$ \left[\frac{d^2 y}{d x^2}\right]_{\theta=\frac{\pi}{4}} = -\frac{b}{a^2} \cdot (\sqrt{2})^3 $$ Evaluating the radical cube $(\sqrt{2})^3 = 2\sqrt{2}$: $$ = -2\sqrt{2}\left(\frac{b}{a^2}\right) $$

Step 4: Final Answer:
The value of the second derivative at the given point is $-2\sqrt{2}\left(\frac{b}{a^2}\right)$, which corresponds to option (C).