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Question Number 102342 by bemath last updated on 08/Jul/20
If { ((x=2t+sin 2t)),((y=e^(sin 2t) )) :} prove that (1/y).(dy/dx) = tan ((π/4)−t)
$${If}\:\begin{cases}{{x}=\mathrm{2}{t}+\mathrm{sin}\:\mathrm{2}{t}}\\{{y}={e}^{\mathrm{sin}\:\mathrm{2}{t}} }\end{cases} \\ $$$${prove}\:{that}\:\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dx}}\:=\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−{t}\right) \\ $$
Commented by Dwaipayan Shikari last updated on 08/Jul/20
There is some error in question if x=2t−cos2t then the prove is true
$${There}\:{is}\:{some}\:{error}\:{in}\:{question}\:\:{if}\:\:\:{x}=\mathrm{2}{t}−{cos}\mathrm{2}{t}\:\:{then}\:{the}\:{prove}\:{is}\:{true} \\ $$
Commented by bobhans last updated on 08/Jul/20
yes sir. i think something error in question
$${yes}\:{sir}.\:{i}\:{think}\:{something}\:{error}\:{in}\:{question} \\ $$
Answered by bobhans last updated on 08/Jul/20
⇒sin 2t = ln(y) ; 2 cos 2t = (1/y).(dy/dt) (dx/dt) = 2+2cos 2t ⇒(dx/dt) = 2+(1/y).(dy/dt) ...(1) (dy/dx) = (dy/dt)×(dt/dx) = (dy/dt)×(1/(dx/dt)) (dy/dx)= 2y cos 2t × (1/(2+2cos 2t)) (1/y) (dy/dx) = ((2cos 2t)/(2(1+cos 2t))) = ((1−2sin^2 t)/(2cos^2 t)) =(1/2)sec^2 t−tan^2 t = (1/2)(tan^2 t−1)−tan^2 t
$$\Rightarrow\mathrm{sin}\:\mathrm{2}{t}\:=\:\mathrm{ln}\left(\mathrm{y}\right)\:;\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{t}\:=\:\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dt}} \\ $$$$\frac{{dx}}{{dt}}\:=\:\mathrm{2}+\mathrm{2cos}\:\mathrm{2}{t}\:\Rightarrow\frac{{dx}}{{dt}}\:=\:\mathrm{2}+\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dt}}\:…\left(\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}\:=\:\frac{{dy}}{{dt}}×\frac{\mathrm{1}}{\frac{{dx}}{{dt}}}\: \\ $$$$\frac{{dy}}{{dx}}=\:\mathrm{2}{y}\:\mathrm{cos}\:\mathrm{2}{t}\:×\:\frac{\mathrm{1}}{\mathrm{2}+\mathrm{2cos}\:\mathrm{2}{t}} \\ $$$$\frac{\mathrm{1}}{{y}}\:\frac{{dy}}{{dx}}\:=\:\frac{\mathrm{2cos}\:\mathrm{2}{t}}{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}{t}\right)}\:=\:\frac{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {t}}{\mathrm{2cos}\:^{\mathrm{2}} {t}}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} {t}−\mathrm{tan}\:^{\mathrm{2}} {t}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}\:^{\mathrm{2}} {t}−\mathrm{1}\right)−\mathrm{tan}\:^{\mathrm{2}} {t} \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 08/Jul/20
if x=2t−cos2t (dx/dt)=2+2sin2t logy=sin2t (1/y).(dy/dt)=2cos2t (1/y).(dy/dx)=((2cos2t)/(2+2sin2t))=(((1−tan^2 t)/(1+tan^2 t))/(1+((2tant)/(1+tan^2 t))))=((1−tan^2 t)/((1+tant)^2 ))=((1−tant)/(1+tant))=tan((π/4)−t)
$${if}\:\:{x}=\mathrm{2}{t}−{cos}\mathrm{2}{t} \\ $$$$\frac{{dx}}{{dt}}=\mathrm{2}+\mathrm{2}{sin}\mathrm{2}{t} \\ $$$${logy}={sin}\mathrm{2}{t} \\ $$$$\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dt}}=\mathrm{2}{cos}\mathrm{2}{t} \\ $$$$\frac{\mathrm{1}}{{y}}.\frac{{dy}}{{dx}}=\frac{\mathrm{2}{cos}\mathrm{2}{t}}{\mathrm{2}+\mathrm{2}{sin}\mathrm{2}{t}}=\frac{\frac{\mathrm{1}−{tan}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}}{\mathrm{1}+\frac{\mathrm{2}{tant}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}}=\frac{\mathrm{1}−{tan}^{\mathrm{2}} {t}}{\left(\mathrm{1}+{tant}\right)^{\mathrm{2}} }=\frac{\mathrm{1}−{tant}}{\mathrm{1}+{tant}}={tan}\left(\frac{\pi}{\mathrm{4}}−{t}\right) \\ $$

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