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: Use parametric differentiation.
Given, \[ x=f(\theta),\qquad y=g(\theta) \] Then, \[ \frac{dx}{d\theta}=f'(\theta) \] and \[ \frac{dy}{d\theta}=g'(\theta) \] Therefore, \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{g'(\theta)}{f'(\theta)} \]

Step 2: Differentiate \(\frac{dy}{dx}\) with respect to \(\theta\).
Now, \[ \frac{dy}{dx}=\frac{g'(\theta)}{f'(\theta)} \] Differentiating with respect to \(\theta\), we get \[ \frac{d}{d\theta}\left(\frac{dy}{dx}\right) = \frac{f'(\theta)g''(\theta)-g'(\theta)f''(\theta)}{(f'(\theta))^2} \]

Step 3: Convert derivative with respect to \(\theta\) into derivative with respect to \(x\).
We know that \[ \frac{d}{dx} = \frac{1}{\frac{dx}{d\theta}}\frac{d}{d\theta} \] Since, \[ \frac{dx}{d\theta}=f'(\theta), \] we get \[ \frac{d^2y}{dx^2} = \frac{1}{f'(\theta)} \cdot \frac{f'(\theta)g''(\theta)-g'(\theta)f''(\theta)}{(f'(\theta))^2} \] \[ \frac{d^2y}{dx^2} = \frac{f'(\theta)g''(\theta)-g'(\theta)f''(\theta)}{(f'(\theta))^3} \]

Step 4: Final conclusion.
Therefore, \[ \boxed{ \frac{d^2y}{dx^2} = \frac{f'(\theta)g''(\theta)-g'(\theta)f''(\theta)}{(f'(\theta))^3} } \]