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The parametric equations are: $x=2\cot\theta$ and $y=2\sin^2\theta$

$$\frac{dy}{dx}=-2\sin^3\theta\cdot \cos\theta$$

And the coordinates are: $(-\frac{2}{\sqrt{3}},3/2)$, $(0,2)$, and $(2\sqrt{3}, 1/2)$

I wasn't quite sure what to do with those coordinates, could someone please make sense of what I am to do, and why it should be done that way?

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Would it help if you wrote $\frac{dy}{dx}$ as $-2\sin^4\theta\cot\theta$? –  Mike Nov 27 '12 at 21:35
    
It's likely that I took my derivative incorrectly; but I don't think that would help. I just can't figure out where to plug the coordinates in, and what it means to plug them in the correct function. –  Mack Nov 27 '12 at 21:37
    
The derivative is correct, @EMACK . You can write it as well as $$\frac{dy}{dx}=-\sin 2t\sin^2t$$ –  DonAntonio Nov 27 '12 at 21:40
    
@EMACK I believe Mike is suggesting that you try to express $\frac{dy}{dx}$ in terms of $x$ and $y$. I would write it out as an answer, but it's his idea. –  Daniel Littlewood Nov 27 '12 at 21:43
    
@DanielLittlewood That would probably be more understandable, because it seems weird to use the $\theta$ value to find the slope. Wouldn't it be difficult to write $\frac{dy}{dx} in terms of y and x? –  Mack Nov 27 '12 at 21:45
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1 Answer

up vote 1 down vote accepted

For example, with the point $\,\displaystyle{\left(-\frac{2}{\sqrt 3}\,,\,\frac{3}{2}\right)}\,$ , and with $\,t=\theta\,$ , for simplicity:

$$2\cot t=x=-\frac{2}{\sqrt 3}\Longrightarrow \tan t=-\sqrt 3\Longrightarrow t=\frac{\pi}{3}+k\pi\,\,,\,k\in\Bbb Z$$

$$\frac{3}{2}=y=2\sin^2t\Longrightarrow \sin t=\pm\frac{\sqrt 3}{2}\Longrightarrow t=\frac{\pi}{3}+k\pi\,\,\,,\,\,k\in\Bbb Z$$

Well, choose one of the infinite ammount of possible $\,t'$s above, say

$$t=\frac{\pi}{3}\,\,(k=0\,)\Longrightarrow\,\left.\frac{dy}{dx}\right|_{t=\pi/3}=-2\sin^3\frac{\pi}{3}\cos\frac{\pi}{3}=-2\left(\frac{\sqrt 3}{2}\right)^3\frac{1}{2}=-\frac{3\sqrt 3}{8}$$

so the tangent line to the curve at this point is

$$y-\frac{3}{2}=-\frac{3\sqrt 3}{8}\left(x+\frac{2}{\sqrt 3}\right)$$

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One of the things I am having trouble with is why you have to use both of the coordinates in the ordered-pair to find the t value, shouldn't the slope be the same? I am just having trouble justifying and making sense of each step. –  Mack Nov 27 '12 at 21:42
    
You have to do that in order to prove the point is actually on the curve: if you do what I did above and you get different sets of values of t's for $\,x,y\,$ then that point is not on the curve. –  DonAntonio Nov 27 '12 at 21:44
    
I think I am beginning to understand. Just one thing, though: how did you decide which root to choose for $\pm\frac{\sqrt{3}}{2}$? –  Mack Nov 27 '12 at 22:18
    
I didn't have to: I just wrote as an example the simplest one. This, of course, may depend heavily on the given curve's domain or whatever. –  DonAntonio Nov 27 '12 at 22:20
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