Turning points of parametric curve Find the slope of the curve at $t=\frac{1}{4}\pi$.


$$\begin{cases}x=\sin t+\cos t \\ y=\frac{1}{2}\sin(2t)\end{cases}$$

$$\frac{dy}{dx}=\frac{\cos(2t)}{-\sin t+\cos t}$$

$$\left.\frac{dy}{dx}\right|_{\frac{1}{4}\pi}=\lim_{t\to \frac{1}{4}\pi}\frac{\cos(2t)}{-\sin t+\cos t}$$

Multiplying by the conjugate of the denominator won't help, so I applied L'Hôpitals Rule

$$\lim_{t\to \frac{1}{4}\pi}\frac{\cos(2t)}{-\sin t+\cos t}=\lim_{t\to \frac{1}{4}\pi}\frac{-2\sin(2t)}{-\cos t-\sin t}=\sqrt{2}$$

Is there another way to evaluate this limit?
You may use the following

$$\cos (2t) = \cos (t + t) = \cos t\cos t - \sin t\sin t = {\cos ^2}t - {\sin ^2}t = \left( {\cos t - \sin t} \right)\left( {\cos t + \sin t} \right)$$

and hence you can write

$$\eqalign{
  &amp; \mathop {\lim }\limits_{t \to \frac{\pi }{4}} \frac{{\cos (2t)}}{{\cos t - \sin t}} = \mathop {\lim }\limits_{t \to \frac{\pi }{4}} \frac{{\left( {\cos t - \sin t} \right)\left( {\cos t + \sin t} \right)}}{{\cos t - \sin t}}  \cr 
  &amp; \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ = \mathop {\lim }\limits_{t \to \frac{\pi }{4}} \left( {\cos t + \sin t} \right)  \cr 
  &amp; \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ = \frac{{\sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2} = \sqrt 2  \cr} $$

That's all. :)
Let  X=1+3sint   and  Y=2-5cost
determine the turning points of this curve in the interval (0;2)
2.classify the turning points in 4.1 as either maximum or minimum