If \( x=2t^{2} \) and \( y=3t^{3} \), then the value of \( \displaystyle\frac{d^{2}y}{dx^{2}} \) at \( t=3 \) is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{3}{16} \)
Since x and y are given in terms of the parameter t,
\( \displaystyle \frac{dx}{dt}=4t,\quad \frac{dy}{dt}=9t^{2} \)
\( \displaystyle \frac{dy}{dx}=\frac{9t^{2}}{4t} \)
\( \displaystyle =\frac{9t}{4} \)
For the second derivative, differentiate \( \frac{dy}{dx} \) with respect to t and divide by \( \frac{dx}{dt} \) again:
\( \displaystyle \frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}\left(\frac{9t}{4}\right)}{dx/dt} \)
\( \displaystyle =\frac{9/4}{4t} \)
\( \displaystyle =\frac{9}{16t} \)
At t = 3:
\( \displaystyle \frac{d^{2}y}{dx^{2}}=\frac{9}{48} \)
\( \displaystyle =\frac{3}{16} \)