# Thread: Parametric question...confused on trig =/

1. ## Parametric question...confused on trig =/

Question:
A curve has parametric equations
x = 3 cos2 t, y = sin 2t, 0 ≤ t < π.
(i) Show that
= $\displaystyle \frac{dy}{dx} = -\frac{2}{3}cot 2t$
(ii) Find the coordinates of the points where the tangent to the curve is parallel to the x-axis

Ok, the first bit i did no problem, but its the second part because the gradient must be 0 correct?
meaning cot 2t = 0
However you cannot do the inverse of this because when cot t = 0...its an asymptote...

Thanks for any help

2. Originally Posted by AshleyT
Question:
A curve has parametric equations
x = 3 cos2 t, y = sin 2t, 0 ≤ t < π.
(i) Show that
= $\displaystyle \frac{dy}{dx} = -\frac{2}{3}cot 2t$
(ii) Find the coordinates of the points where the tangent to the curve is parallel to the x-axis

Ok, the first bit i did no problem, but its the second part because the gradient must be 0 correct?
meaning cot 2t = 0
However you cannot do the inverse of this because when cot t = 0...its an asymptote...

Thanks for any help
For (i), I find $\displaystyle \frac{dy}{dx} = -\frac{1}{3}\cot 2t$.

As for (ii), you must be confusing with something else: you indeed need to find points where $\displaystyle 0=\frac{dy}{dx}$, which means $\displaystyle 0=\cot 2t=\frac{\cos 2t}{\sin 2t}$. This is 0 when $\displaystyle \cos 2t=0$. We have $\displaystyle 0\leq 2t<2\pi$, hence the solutions are $\displaystyle 2t=\frac{\pi}{2}$ and $\displaystyle 2t=\frac{3\pi}{2}$. It remains to find $\displaystyle (x,y)$ for these $\displaystyle t$.

3. Originally Posted by Laurent
For (i), I find $\displaystyle \frac{dy}{dx} = -\frac{1}{3}\cot 2t$.

As for (ii), you must be confusing with something else: you indeed need to find points where $\displaystyle 0=\frac{dy}{dx}$, which means $\displaystyle 0=\cot 2t=\frac{\cos 2t}{\sin 2t}$. This is 0 when $\displaystyle \cos 2t=0$. We have $\displaystyle 0\leq 2t<2\pi$, hence the solutions are $\displaystyle 2t=\frac{\pi}{2}$ and $\displaystyle 2t=\frac{3\pi}{2}$. It remains to find $\displaystyle (x,y)$ for these $\displaystyle t$.
Thanks! I think i just managed to confuse myself.

For part one though: How?
Cuz you divide
2cos 2t by -3sin 2t
and get -2/3 cot 2t

4. Originally Posted by AshleyT
Thanks! I think i just managed to confuse myself.

For part one though: How?
Cuz you divide
2cos 2t by -3sin 2t
and get -2/3 cot 2t
You divide $\displaystyle 2\cos 2t$ by $\displaystyle -{\color{red}6}\sin 2t$ (there is the same factor 2 when differentiating $\displaystyle \cos 2t$ or $\displaystyle \sin 2t$).