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anonymous
 3 years ago
Topic: \(Calculus \ 2\), polar conversions
Q: Identify name & Cartesian equation for: r = 2\(\tan \theta \sec \theta\)
Any graphing calculator can quickly show this is a parabola the one I've sketched below, but I'm trying to understand the process here. These are the facts I know of:
\(r=\sqrt{x^2+y^2}\)
\(\theta = \tan^{1}(\frac{y}{x})\)
x = r cos θ
y = r sin θ
What's the trick here? Can somebody show me? (I'd be more than happy to give out a medal if you can do so) There are a few other problems related to this one I'm working on so a technique to be learned is the goa
anonymous
 3 years ago
Topic: \(Calculus \ 2\), polar conversions Q: Identify name & Cartesian equation for: r = 2\(\tan \theta \sec \theta\) Any graphing calculator can quickly show this is a parabola the one I've sketched below, but I'm trying to understand the process here. These are the facts I know of: \(r=\sqrt{x^2+y^2}\) \(\theta = \tan^{1}(\frac{y}{x})\) x = r cos θ y = r sin θ What's the trick here? Can somebody show me? (I'd be more than happy to give out a medal if you can do so) There are a few other problems related to this one I'm working on so a technique to be learned is the goa

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large \sqrt{x^2+y^2}=\tan(\tan^{1}\frac{y}{x})\sec(\tan^{1}\frac{y}{x})\] But then um... lol what? dw:1342214986994:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1342215821623:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Omg that was it?! D: Good work @myko
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