Question

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

Answer 1

Answer: 8

We have
$x=\sin \left(t+\frac{7\pi }{12}\right)+\sin \left(t-\frac{\pi }{12}\right)+\sin \left(t+\frac{3\pi }{12}\right)=2\sin \left(t+\frac{\pi }{4}\right)\cos \left(\frac{\pi }{3}\right)+\sin \left(t+\frac{\pi }{4}\right)=2\sin \left(t+\frac{\pi }{4}\right)$
|||ly
$y=\cos \left(t+\frac{7\pi }{12}\right)+\cos \left(t-\frac{\pi }{12}\right)+\cos \left(t+\frac{3\pi }{12}\right)=2\cos \left(t+\frac{\pi }{4}\right)\cos \left(\frac{\pi }{3}\right)+\cos \left(t+\frac{\pi }{4}\right)=2\cos \left(t+\frac{\pi }{4}\right)$
Now, $\frac{x}{y}=\tan \left(t+\frac{\pi }{4}\right)=\frac{1+\tan t}{1-\tan t}$ and $\frac{y}{x}=\frac{1}{\tan \left(t+\frac{\pi }{4}\right)}=\frac{1-\tan t}{1+\tan t}$
$\therefore \left(\frac{x}{y}-\frac{y}{x}\right)=\left(\frac{1+\tan t}{1-\tan t}\right)-\left(\frac{1-\tan t}{1+\tan t}\right)=\frac{(1+\tan t{)}^2-(1-\tan t{)}^2}{1-{\tan }^{2}t}=\frac{4\tan t}{1-{\tan }^{2}t}=2\tan 2t$
Hence ${\frac{d}{dt}\left(\frac{x}{y}-\frac{y}{x}\right)=\frac{d}{dt}(2\tan 2t)=4{\sec }^{2}2t]}_{t=\frac{\pi }{8}}=4{\sec }^2\frac{\pi }{4}=8$