Question

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

• A. \[ \frac{ \frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{d^2x}{dt^2}\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^3} \]
• B. \[ \frac{ \frac{dx}{dt}y-x\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^2} \]
• C. \[ \frac{ x\frac{dy}{dt}-y\frac{dx}{dt} } {x^2} \]
• D. \[ \frac{ \frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{d^2x}{dt^2}\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^2} \]

Hint:

For parametric curves: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}} {\frac{dx}{dt}} \] and \[ \frac{d^2y}{dx^2} = \frac{ \frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{d^2x}{dt^2}\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^3}. \] This formula is frequently used in JEE and entrance examinations.

Answer 1

The Correct Option is A

Concept: For parametric equations \[ x=f(t), \qquad y=g(t), \] the first derivative is \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}} {\frac{dx}{dt}}. \] The second derivative is obtained by differentiating with respect to \(t\) and then dividing by \(\dfrac{dx}{dt}\).

Step 1: Find the first derivative. \[ \frac{dy}{dx} = \frac{y'}{x'} \] where \[ x'=\frac{dx}{dt}, \qquad y'=\frac{dy}{dt}. \]

Step 2: Differentiate with respect to \(t\). Using the quotient rule, \[\begin{aligned} \frac{d}{dt}\left(\frac{dy}{dx}\right) &= \frac{x'y''-x''y'}{(x')^2} \end{aligned}\] where \[ x''=\frac{d^2x}{dt^2}, \qquad y''=\frac{d^2y}{dt^2}. \]

Step 3: Find \(\dfrac{d^2y}{dx^2}\). \[ \frac{d^2y}{dx^2} = \frac{ \dfrac{d}{dt} \left(\dfrac{dy}{dx}\right) } {\dfrac{dx}{dt}} \] Therefore, \[\begin{aligned} \frac{d^2y}{dx^2} &= \frac{ x'y''-x''y' } {(x')^3} \end{aligned}\] Substituting back, \[\begin{aligned} \frac{d^2y}{dx^2} = \frac{ \frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{d^2x}{dt^2}\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^3} \end{aligned}\] \[\begin{aligned} \boxed{ \frac{ \frac{dx}{dt}\frac{d^2y}{dt^2} - \frac{d^2x}{dt^2}\frac{dy}{dt} } {\left(\frac{dx}{dt}\right)^3} } \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.