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Question Number 118798 by WasimShaikh last updated on 19/Oct/20

 1)Find (dy/dx)  ;   if   x = at^2  ,  y = 2at    2)

$$\left.\:\mathrm{1}\right){Find}\:\frac{{dy}}{{dx}}\:\:;\:\:\:{if}\:\:\:{x}\:=\:{at}^{\mathrm{2}} \:,\:\:{y}\:=\:\mathrm{2}{at} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\: \\ $$

Answered by Dwaipayan Shikari last updated on 20/Oct/20

x=at^2   (dx/dt)=2at  y=2at  (dy/dt)=2a       (dy/dx)=(dy/dt).(dt/dx)=(1/t)       x=at^2   y=2at  (x/y)=(t/2)⇒(x/(2a))=(t^2 /2)⇒t=±(√(x/a))  So  (dy/dx)=±(√(a/x))

$${x}={at}^{\mathrm{2}} \\ $$$$\frac{{dx}}{{dt}}=\mathrm{2}{at} \\ $$$${y}=\mathrm{2}{at} \\ $$$$\frac{{dy}}{{dt}}=\mathrm{2}{a}\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}}.\frac{{dt}}{{dx}}=\frac{\mathrm{1}}{{t}}\:\:\:\:\:\:\:{x}={at}^{\mathrm{2}} \:\:{y}=\mathrm{2}{at} \\ $$$$\frac{{x}}{{y}}=\frac{{t}}{\mathrm{2}}\Rightarrow\frac{{x}}{\mathrm{2}{a}}=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow{t}=\pm\sqrt{\frac{{x}}{{a}}} \\ $$$${So} \\ $$$$\frac{{dy}}{{dx}}=\pm\sqrt{\frac{{a}}{{x}}} \\ $$

Answered by Olaf last updated on 19/Oct/20

x = at^2  = a((y/(2a)))^2  = (y^2 /(4a))  y = ±(√(4ax))  (dy/dx) = ±((4a)/(2(√(4ax)))) = ±(√(a/x))

$${x}\:=\:{at}^{\mathrm{2}} \:=\:{a}\left(\frac{{y}}{\mathrm{2}{a}}\right)^{\mathrm{2}} \:=\:\frac{{y}^{\mathrm{2}} }{\mathrm{4}{a}} \\ $$$${y}\:=\:\pm\sqrt{\mathrm{4}{ax}} \\ $$$$\frac{{dy}}{{dx}}\:=\:\pm\frac{\mathrm{4}{a}}{\mathrm{2}\sqrt{\mathrm{4}{ax}}}\:=\:\pm\sqrt{\frac{{a}}{{x}}} \\ $$

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