If \(x = \displaystyle\frac{a}{1+t}\) and \(y =\displaystyle\frac{a}{(1+t)^2}\) where \(a \gt 0\), then \(\displaystyle\frac{d^2y}{dx^2}\) at \(t = 1\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{2}{a}\)
The parameter t can be eliminated. From the first equation,
\( \displaystyle \frac{1}{1+t} = \frac{x}{a} \)
so
\( \displaystyle y = a\left(\frac{1}{1+t}\right)^2 = a \cdot \frac{x^2}{a^2} \)
\( \displaystyle y = \frac{x^2}{a} \)
Differentiating twice with respect to x:
\( \displaystyle \frac{dy}{dx} = \frac{2x}{a} \)
\( \displaystyle \frac{d^2y}{dx^2} = \frac{2}{a} \)
This does not depend on t, so at \(t = 1\) it is \(\displaystyle\frac{2}{a}\).
The same result follows by the parametric method:
\( \displaystyle \frac{dx}{dt} = -\frac{a}{(1+t)^2} \)
\( \displaystyle \frac{dy}{dt} = -\frac{2a}{(1+t)^3} \)
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{1+t} \)
\( \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{2}{1+t}\right) \cdot \frac{dt}{dx} \)
\( \displaystyle = -\frac{2}{(1+t)^2} \cdot \left(-\frac{(1+t)^2}{a}\right) \)
\( \displaystyle = \frac{2}{a} \)