If x = at^2 and y = 2at,then find y_2 (where | vedclass

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

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$-\frac{1}{2at^3}$

Vedclass · Answer

(A) Given $x = at^2$ and $y = 2at$. First,find the first derivative $\frac{dy}{dx}$: $\frac{dx}{dt} = 2at$ $\frac{dy}{dt} = 2a$ $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t} = t^{-1}$. Now,find the second derivative $y_2 = \frac{d^2y}{dx^2}$: $y_2 = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}$ $y_2 = \frac{d}{dt}(t^{-1}) \cdot \frac{1}{2at}$ $y_2 = (-t^{-2}) \cdot \frac{1}{2at} = -\frac{1}{t^2} \cdot \frac{1}{2at} = -\frac{1}{2at^3}$. Thus,the correct option is $A$.

If $x = at^2$ and $y = 2at$,then find $y_2$ (where $t \neq 0$).

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