(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

At what point does the curve [tex]x = (1-2cos^2(t)[/tex], [tex]y=(tan(t))(1-2cos^2(t))[/tex] cross itself? Find equations of both tangents at that point.

2. Relevant equations

3. The attempt at a solution

To begin, I figured that x=y when it crosses itself, so I set x=y and got [tex]1 = tan(t)[/tex], so [tex]t=\pi/4[/tex]

Next, I decided to solve for dy/dx:

[tex]dx/dt=4cos(t)sin(t)[/tex]

[tex]dy/dt=4sin^2(t)+sec^2(t)-2[/tex]

[tex]dy/dx=\frac{(dy/dt)}{(dx/dt)}[/tex]

At that point, I plugged [tex]\pi/4[/tex] into my eqn for dy/dx and got that the slope at that point is [tex]1[/tex]

Additionally, I solved for x and y at [tex]\pi/4[/tex] and got that both x and y are 0 when [tex]t=\pi/4[/tex]. Using these points and the slope, I got that the equation of the tangent line is:

[tex]y=x[/tex]

BUT

That's justoneof them. If the curve is crossing itself once, it clearly has two tangent lines at that point, but I'm not sure how to find it. Is the slope of that line just [tex]-1[/tex]? I guess it might be, but it doesn't seem right to just assume that it is.

Should I have said that [tex]t=n\pi + \pi/4[/tex]? If that is the case, would I just say that [tex]t=5\pi/4[/tex] AND [tex]t=\pi/4[/tex]? Of course, that assumes that I did the rest of the problem properly, and I'm not even sure if I did. I was very ill all last week, so I have no lecture/recitation notes to go by, and my book doesn't have very many examples for me to practice from.

I'm sorry if this is a bad question. :( Thanks for looking!

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# Homework Help: Parametric equations and tangent(s)

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