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

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(A) Given $x = at^2$ and $y = 2at$.First, find $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.Then, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2

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$-\frac{a}{xy}$

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(A) Given $x = at^2$ and $y = 2at$.First, find $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.Then, $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$.Now, differentiate with respect to $x$:$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{t}) = \frac{d}{dt}(\frac{1}{t}) \cdot \frac{dt}{dx} = (-\frac{1}{t^2}) \cdot \frac{1}{2at} = -\frac{1}{2at^3}$.We know $x = at^2$, so $t^2 = \frac{x}{a}$. Also $y = 2at$, so $t = \frac{y}{2a}$.Substitute $t^2 = \frac{x}{a}$ and $t = \frac{y}{2a}$ into the expression:$\frac{d^2y}{dx^2} = -\frac{1}{2a(t^2)(t)} = -\frac{1}{2a(\frac{x}{a})(\frac{y}{2a})} = -\frac{1}{\frac{xy}{a}} = -\frac{a}{xy}$.

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

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