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

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(C) Given $x = at^2$ and $y = 2at$. First,differentiate $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$. Next,differentiate $y$ wi

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$\frac{1}{t}$

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(C) Given $x = at^2$ and $y = 2at$. First,differentiate $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$. Next,differentiate $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$. Now,use the chain rule for parametric differentiation: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$. Since $t = \frac{y}{2a}$,we can also express this as $\frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$. Alternatively,from $y = 2at$,we have $t = \frac{y}{2a}$. Substituting $t$ into $x = at^2$: $x = a(\frac{y}{2a})^2 = a(\frac{y^2}{4a^2}) = \frac{y^2}{4a}$. $y^2 = 4ax$. Differentiating both sides with respect to $x$: $2y \frac{dy}{dx} = 4a$. $\frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}$. Substituting $y = 2at$: $\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.

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

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