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$x=\cos t+\cos 2t$ and $y = \sin t+\sin 2t$ at $(-1,1)$

I did dy/dt and dx/dt and then dy/dx and got $\displaystyle \frac{\cos t+2\cos 2t}{-\sin t-2\sin 2t}$

Now I'm confused, do I plug in the -1 and 1 to get two answers and then subtract them from each other to the get a final slope? I need to get the slope so that I can make an equation in the y=mx+b form.

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2 Answers 2

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Hint:

Find the value(s) of $t$ which give the point $(-1,1)$.


Note however, that by inspection, the point $(-1,1)$ on the curve corresponds to $t=\frac{\pi}{2}$ in the parametric equations. So you find the slope at $t=\frac{\pi}{2}.$

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so i set my x and y equations equal to -1 and 1 respectively? –  Ryan Jul 18 '11 at 5:00
    
yes Ryan. That's what you'd normally do. All though we can do this by inspection as well. I'll add to my answer shortly. –  Nana Jul 18 '11 at 5:03
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To find the value(s) of t that yield (-1, 1), I computed $x^2+y^2$ and got (modulo my usual errors) $2(1+cos^3 t)$. Setting this equal to 2 gives $\cos(t) = 0$. This gives two possible values of t, only one of which gives the desired values of x and y. As to why I did this, I noticed that the sin/cos terms with args of t and 2t would sum to 2 (in $x^2+y^2$), so I thought I would see what happened to the cross terms. Turns out it helped.

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Good idea. But you do not mean $2(1+\cos^3 t)$. For one thing that would give $\cos t=-1$, and your $\cos t=0$ is right. –  André Nicolas Jul 18 '11 at 5:24
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