If x=at2 and y=2at, then find dydx is

x=at2ddt(x)=ddt(at2)=a.ddt(t2)=a.(2t)=2aty=2atddt(y)=ddt(2at)=2addt(t)=2a(1)=2adydx=dydtdxdt=2a2at⇒dydx=1t

If x=at2 and y=2at, then find dydx is

x=at2ddt(x)=ddt(at2)=a.ddt(t2)=a.(2t)=2aty=2atddt(y)=ddt(2at)=2addt(t)=2a(1)=2adydx=dydtdxdt=2a2at⇒dydx=1t

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If $x = a t^{2}$ and $y = 2 a t$ , then find $\frac{d y}{d x}$ is

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$2 + t$

$\frac{1}{t}$

$1 + t$

$1 - t$

answer is A.

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Detailed Solution

$\begin{array}{l} x = a t^{2} \\ \frac{d}{d t} (x) = \frac{d}{d t} (a t^{2}) = a . \frac{d}{d t} (t^{2}) = a . (2 t) = 2 a t \\ y = 2 a t \\ \frac{d}{d t} (y) = \frac{d}{d t} (2 a t) = 2 a \frac{d}{d t} (t) = 2 a (1) = 2 a \\ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 a}{2 a t} \Rightarrow \frac{d y}{d x} = \frac{1}{t} \end{array}$