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

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(C) Given that $x = at^2$ and $y = 2at$. We need to find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. Differentiating

Vedclass · Accepted Answer

$\frac{1}{t}$

Vedclass · Answer

(C) Given that $x = at^2$ and $y = 2at$. We need to find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. Differentiating $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$. Differentiating $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$. Now,$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$. Since $y = 2at$,we have $t = \frac{y}{2a}$. Substituting $t$ in the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{y/(2a)} = \frac{2a}{y}$. Alternatively,from $x = at^2$,we have $t^2 = \frac{x}{a}$,so $t = \sqrt{\frac{x}{a}}$. Thus,$\frac{dy}{dx} = \frac{1}{\sqrt{x/a}} = \sqrt{\frac{a}{x}}$. Looking at the options,if we express $\frac{1}{t}$ in terms of $x$ and $y$,we note that $\frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$. Given $y = 2at$,then $t = \frac{y}{2a}$. Substituting $t$ into $\frac{dy}{dx} = \frac{1}{t}$ gives $\frac{2a}{y}$. However,checking the ratio $\frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$. If we evaluate $\frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$. Comparing with options,$\frac{1}{t}$ is the derivative. If the question implies expressing in terms of $x$ and $y$,$\frac{dy}{dx} = \frac{2a}{y}$.

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

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