Question

If \[ x=f(\theta) \] and \[ y=g(\theta), \] then \[ \frac{d^2y}{dx^2}= \]

• A. \(\frac{g''(\theta)}{f'(\theta)}\)
• B. \(\frac{f''(\theta)}{x'(\theta)}\)
• C. \(\frac{f'(\theta)g''(\theta)-g'(\theta)f''(\theta)}{(f'(\theta))^3}\)
• D. \(\frac{g'(\theta)f''(\theta)-g''(\theta)f'(\theta)}{(g'(\theta))^3}\)

Hint:

For parametric equations \(x=f(\theta)\), \(y=g(\theta)\), first find \[ \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta} \] and then use \[ \frac{d^2y}{dx^2} = \frac{d}{d\theta}\left(\frac{dy}{dx}\right)\div \frac{dx}{d\theta}. \]

Answer 1

The Correct Option is C

Step 1: Set up the parametric derivatives.
We have $x = f(\theta)$ and $y = g(\theta)$. We want to find $\frac{d^2 y}{dx^2}$ in terms of derivatives of $f$ and $g$ with respect to $\theta$.
Step 2: Find the first derivative $dy/dx$.
Using the chain rule for parametric forms: $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{g'(\theta)}{f'(\theta)}$, provided $f'(\theta) \ne 0$.
Step 3: Differentiate $dy/dx$ with respect to $x$ to get $d^2y/dx^2$.
$\frac{d^2 y}{dx^2} = \frac{d}{dx}\left(\frac{g'(\theta)}{f'(\theta)}\right) = \frac{1}{f'(\theta)} \cdot \frac{d}{d\theta}\left(\frac{g'(\theta)}{f'(\theta)}\right)$, since $\frac{d}{dx} = \frac{1}{f'(\theta)}\frac{d}{d\theta}$.
Step 4: Apply the quotient rule to differentiate $g'(\theta)/f'(\theta)$ with respect to $\theta$.
$\frac{d}{d\theta}\left(\frac{g'(\theta)}{f'(\theta)}\right) = \frac{g''(\theta)f'(\theta) - g'(\theta)f''(\theta)}{[f'(\theta)]^2}$.
Step 5: Combine the results.
$\frac{d^2 y}{dx^2} = \frac{1}{f'(\theta)} \cdot \frac{g''(\theta)f'(\theta) - g'(\theta)f''(\theta)}{[f'(\theta)]^2} = \frac{f'(\theta)g''(\theta) - g'(\theta)f''(\theta)}{[f'(\theta)]^3}$.
Step 6: State the final answer.
This matches option 3 exactly. \[\boxed{\dfrac{f'(\theta)g''(\theta) - g'(\theta)f''(\theta)}{[f'(\theta)]^3}}\]