Question

If a curve is represented parametrically by $x=\sin (t+\frac{7\pi }{12})+\sin (t-\frac{\pi }{12})+\sin (t+\frac{3\pi }{12}),$ $y=\cos (t+\frac{7\pi }{12})+\cos (t-\frac{\pi }{12})+\cos (t+\frac{3\pi }{12})$, then the value of $\frac{d}{dt}(\frac{x}{y}-\frac{y}{x})$ at $t=\frac{\pi }{8}$ is

Answer 1

Given :
$x=\sin (t+\frac{7\pi }{12})+\sin (t-\frac{\pi }{12})+\sin (t+\frac{3\pi }{12})\Rightarrow x=2\sin (t+\frac{\pi }{4})\cos \left(\frac{\pi }{3}\right)+\sin (t+\frac{\pi }{4})\Rightarrow x=2\sin (t+\frac{\pi }{4})$
Similarly
$y=\cos (t+\frac{7\pi }{12})+\cos (t-\frac{\pi }{12})+\cos (t+\frac{3\pi }{12})\Rightarrow y=2\cos (t+\frac{\pi }{4})\cos \left(\frac{\pi }{3}\right)+\cos (t+\frac{\pi }{4})\Rightarrow y=2\cos (t+\frac{\pi }{4})$

Now,
$\frac{x}{y}=\tan (t+\frac{\pi }{4})\frac{y}{x}=\cot (t+\frac{\pi }{4})$
Therefore,
$\frac{d}{dt}(\frac{x}{y}-\frac{y}{x})={\sec }^2(t+\frac{\pi }{4})+{\text{cosec}}^2(t+\frac{\pi }{4})\Rightarrow \frac{d}{dt}(\frac{x}{y}-\frac{y}{x})=\frac{1}{{\cos }^2(t+\frac{\pi }{4}){\sin }^2(t+\frac{\pi }{4})}\Rightarrow \frac{d}{dt}(\frac{x}{y}-\frac{y}{x})=\frac{4}{{\sin }^2(2t+\frac{\pi }{2})}$
Putting $t=\frac{\pi }{8}$, we get
$\Rightarrow \frac{d}{dt}(\frac{x}{y}-\frac{y}{x})=8$