Turning points of parametric curve

Question

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?

Answer 1

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

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

That's all. :)

Answer 2

You could try applying an identity to express $\sin(2t)$ in terms of $\sin(t)$ and $\cos(t)$, then eliminating $t$ to express $y$ as a function of $x$ (using the identity $\sin^2(t) + \cos^2(t) = 1$). Then you could differentiate $y$ with respect to $x$.

Answer 3

You can also use the very first terms of Taylor expansions around $t=\frac \pi 4$. $$\cos(2t)=-2 \left(t-\frac{\pi }{4}\right)+\cdots$$ $$\cos(t)=\frac{1}{\sqrt{2}}-\frac{t-\frac{\pi }{4}}{\sqrt{2}}+\cdots$$ $$\sin(t)=\frac{1}{\sqrt{2}}+\frac{t-\frac{\pi }{4}}{\sqrt{2}}+\cdots$$ $$\cos(t)-\sin(t)=-\sqrt{2} \left(t-\frac{\pi }{4}\right)+\cdots$$ and then the result.

Answer 4

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

Answer 5

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