If x= ae^{t}\left ( \sin t+\cos t \right )\: and\: y= ae^{t}\left ( \sin t-\cos t \right ), then prove that
\frac{dy}{dx}= \frac{x+y}{x-y}

 

 

 

 
 
 
 
 

Answers (1)

given: x= ae^{t}\left ( \sin t+\cos t \right )---\left ( i \right )
          y= ae^{t}\left ( \sin t-\cos t \right )---\left ( ii \right )
On differentiating equation (i) with w.r.t  t,we get
\frac{dx}{dt}= a\frac{d}{dt}\left [ e^{t}\left ( \sin t+\cos t \right ) \right ]= a\frac{d}{dt}\left [ e^{t} \sin t+e^{t}\cos t\right ]
\Rightarrow a\left [ e^{t}\cos t+\sin t \, e^{t} +e^{t}\left ( -\sin t \right )+e^{t}\cos t\right ]
\Rightarrow a\left [ e^{t}\cos t+e^{t}\sin t -\sin t+e^{t}+e^{t}\cos t\right ]
= 2ae^{t}\cos t---\left ( iii \right )
On differentiating equation (ii) w.r.t t,
\frac{dy}{dt}= a\frac{d}{dt}\left [ e^{t}\left ( \sin t-\cos t \right ) \right ]\Rightarrow a \frac{d}{dt}\left ( e^{t}\sin t-e^{t} \cos t\right )
      = a\left [ e^{t}\cos t+\sin t\: e^{t} -\left ( -e^{t}\sin t+\cos e^{t} \right )\right ]
    \Rightarrow a\left [ e^{t}\cos t+e^{t}\sin t+\sin t\: e^{t} -e^{t}\cos t\right ]
          = 2ae^{t}\sin t---\left ( iv \right )
On dividing equation(iv)  by (iii) 
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{2ae^{t}\sin t}{2ae^{t}\cos t}
\frac{dy}{dx}= \frac{\sin t}{\cos t}
LHS \frac{dy}{dx}= \tan t---\left ( v \right )
Now, RHS= \frac{x+y}{x-y}
= \frac{ae^{t}\left ( \sin t+\cos t \right )+ae^{t}\left ( \sin t-\cos t \right )}{ae^{t}\left ( \sin t+\cos t \right )-ae^{t}\left ( \sin t-\cos t \right )}
\Rightarrow \frac{2ae^{t}\sin t}{2ae^{t}\cos t}= \tan t \: \: \left ( vi \right )
From equation (v) and (vi) we have
\frac{dy}{dx}= \frac{x+y}{x-y}    Hence proved.

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