Question

Find the value of $\frac{dy}{dx}$ a $\theta =\frac{n}{4}$ , if $x=a{e}^{\theta }(\sin \theta -\cos \theta )$ and $y=a{e}^{\theta }(\sin \theta -\cos \theta )$

Answer 1

We have,

$x=a{e}^{\theta }(\sin \theta -cos\theta )$ …….. (1)

$y=a{e}^{\theta }(\sin \theta +cos\theta )$ ……… (2)

On differentiating to equation (1) w.r.t $\theta$, we get

$\frac{dx}{d\theta }=a[e^{\theta }(\sin \theta -\cos \theta )+e^{\theta }(\cos \theta -(-\sin \theta ))]$

$\frac{dx}{d\theta }=a{e}^{\theta }[(\sin \theta -\cos \theta )+(\cos \theta +\sin \theta )]$

$\frac{dx}{d\theta }=2a{e}^{\theta }\left(\sin \theta \right)$

Similarly,

On differentiating to equation (2) w.r.t $\theta$, we get

$\frac{dy}{d\theta }=a[e^{\theta }(\sin \theta +\cos \theta )+e^{\theta }(\cos \theta +(-\sin \theta ))]$

$\frac{dy}{d\theta }=a{e}^{\theta }[(\sin \theta +\cos \theta )+(\cos \theta -\sin \theta )]$

$\frac{dy}{d\theta }=2a{e}^{\theta }\left(\cos \theta \right)$

$\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{2a{e}^{\theta }\left(\cos \theta \right)}{2a{e}^{\theta }\left(\sin \theta \right)}$

$\frac{dy}{dx}=\cot \theta$

Put $\theta =\frac{\pi }{4}$

So,

${\frac{dy}{dx}|}_{\theta =\frac{\pi }{4}}=\cot \frac{\pi }{4}=1$

Hence, the value is $1$.