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If $x=t^2$,$y=t^3,then\;\large \frac{d^2y}{dx^2}$ is

\[\begin{array}{1 1}(A)\;\frac{3}{2} & (B)\;\frac{3}{4t}\\(C)\;\frac{3}{2t} & (D)\;\frac{3}{t}\end{array}\]

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Toolbox:
  • To find $\large\frac{dy}{dx}$ in case of parametric functions,we obtain a relationship by eliminating the parameter $t$ and then differentiate w.r.t $x$.
  • Hence $\large\frac{dy}{dx}=\Large{\large\frac{dy}{dt}}{\large\frac{dx}{dt}}$
Step 1:
$x=t^2,y=t^3$
$\large\frac{dx}{dt}$$=2t$
$\large\frac{dy}{dt}$$=3t^2$
Therefore $\large\frac{dy}{dx}=\frac{3t^2}{2t}$
$\qquad\qquad\quad=\large\frac{3}{2}$$t$
Step 2:
$\large\frac{d^2y}{dx^2}=\large\frac{d}{dt}\big(\large\frac{dy}{dx}\big)\frac{dt}{dx}$
$\Rightarrow \large\frac{d}{dt}\big(\large\frac{3}{2}$$t\big)\large\frac{1}{2t}$
$\Rightarrow \large\frac{3}{2}\times \frac{1}{2t}$
Therefore $\large\frac{d^2y}{dx^2}=\frac{3}{4t}$
Hence $B$ is the correct option.
answered Jul 4, 2013 by sreemathi.v
 

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