If x = t^2 and y = t^3,then frac{d^2y}{dx^2} | vedclass

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(B) Given $x = t^2$ and $y = t^3$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule for parametric equations:$\frac{dx}{dt} = 2t$ a

Vedclass · Accepted Answer

$\frac{3}{4t}$

Vedclass · Answer

(B) Given $x = t^2$ and $y = t^3$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule for parametric equations:$\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2$.$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3}{2}t$.Now,differentiate $\frac{dy}{dx}$ with respect to $x$ to find $\frac{d^2y}{dx^2}$:$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{3}{2}t\right) = \frac{d}{dt}\left(\frac{3}{2}t\right) \cdot \frac{dt}{dx}$.Since $\frac{dt}{dx} = \frac{1}{dx/dt} = \frac{1}{2t}$,we have:$\frac{d^2y}{dx^2} = \frac{3}{2} \cdot \frac{1}{2t} = \frac{3}{4t}$.

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

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