If-x-2t-sin-2t-y-e-sin-2t-prove-that-1-y-dy-dx-tan-pi-4-t-

Question Number 102342 by bemath last updated on 08/Jul/20

I f {\begin{cases} x = 2 t + \sin 2 t \\ y = e^{\sin 2 t} \end{cases}

p r o v e t h a t \frac{1}{y} . \frac{d y}{d x} = \tan (\frac{π}{4} - t)

Commented by Dwaipayan Shikari last updated on 08/Jul/20

T h e r e i s s o m e e r r o r i n q u e s t i o n i f x = 2 t - c o s 2 t t h e n t h e p r o v e i s t r u e

Commented by bobhans last updated on 08/Jul/20

y e s s i r . i t h i n k s o m e t h i n g e r r o r i n q u e s t i o n

Answered by bobhans last updated on 08/Jul/20

\Rightarrow \sin 2 t = \ln (y); 2 \cos 2 t = \frac{1}{y} . \frac{d y}{d t}

\frac{d x}{d t} = 2 + 2 \cos 2 t \Rightarrow \frac{d x}{d t} = 2 + \frac{1}{y} . \frac{d y}{d t} \dots (1)

\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = \frac{d y}{d t} \times \frac{1}{\frac{d x}{d t}}

\frac{d y}{d x} = 2 y \cos 2 t \times \frac{1}{2 + 2 \cos 2 t}

\frac{1}{y} \frac{d y}{d x} = \frac{2 \cos 2 t}{2 (1 + \cos 2 t)} = \frac{1 - 2 \sin^{2} t}{2 \cos^{2} t}

= \frac{1}{2} \sec^{2} t - \tan^{2} t

= \frac{1}{2} (\tan^{2} t - 1) - \tan^{2} t

Answered by Dwaipayan Shikari last updated on 08/Jul/20

i f x = 2 t - c o s 2 t

\frac{d x}{d t} = 2 + 2 s i n 2 t

l o g y = s i n 2 t

\frac{1}{y} . \frac{d y}{d t} = 2 c o s 2 t

\frac{1}{y} . \frac{d y}{d x} = \frac{2 c o s 2 t}{2 + 2 s i n 2 t} = \frac{\frac{1 - {t a n}^{2} t}{1 + {t a n}^{2} t}}{1 + \frac{2 t a n t}{1 + {t a n}^{2} t}} = \frac{1 - {t a n}^{2} t}{{(1 + t a n t)}^{2}} = \frac{1 - t a n t}{1 + t a n t} = t a n (\frac{π}{4} - t)