Question

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

Answer 1

$y=a{e}^{\theta }(sin\theta +cos\theta )$
$x=a{e}^{\theta }(sin\theta -cos\theta )$
Applying parametric differentiation,
$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$----(1)
Now, $\frac{dy}{d\theta }=a{e}^{\theta }(cos\theta -sin\theta )+a{e}^{\theta }(sin\theta +cos\theta )$
Applying product rule: we get
$=2a{e}^{\theta }\left(cos\theta \right)$
$\frac{dx}{d\theta }=a{e}^{\theta }(cos\theta +sin\theta )+a{e}^{\theta }(sin\theta -cos\theta )$
$=2a{e}^{\theta }\left(sin\theta \right)$
Substituting the values of $\frac{dy}{d\theta }$ and $\frac{dx}{d\theta }$ in (1),
$\frac{dy}{dx}=\frac{2a{e}^{\theta }cos\theta }{2a{e}^{\theta }sin\theta }=cot\theta$

Now, $\frac{dy}{dx}$ at $\theta =\frac{pi}{4}$

$[cot\theta {]}_{\theta =\frac{pi}{4}}=cot\frac{\pi }{4}=1.$