question

Harezmi&Cabir · January 16th 2007, 12:55 PM

forum or my mail thank you. mathmanman@hotmail.com

**topsquark** · January 16th 2007, 02:47 PM

$x = \frac{cos^3(t)}{\sqrt{cos(2t)}}$

So
$dx = \left ( \frac{3cos^2(t) \cdot -sin(t) \cdot \sqrt{cos(2t)} - cos^3(t) \cdot \frac{1}{2} \frac{1}{\sqrt{cos(2t)}} \cdot -sin(2t) \cdot 2}{cos(2t)} \right ) dt$

$dx = \left ( \frac{\frac{sin(2t)cos^3(t)}{\sqrt{cos(2t)}} - 3sin(t)cos^2(t) \sqrt{cos(2t)}}{cos(2t)} \right ) dt$

$y = \frac{sin^3(t)}{\sqrt{cos(2t)}}$

So
$dy = \left ( \frac{3 sin^2(t) \cdot cos(t) \cdot \sqrt{cos(2t)} - sin^3(t) \cdot \frac{1}{2} \frac{1}{\sqrt{cos(2t)}} \cdot -sin(2t) \cdot 2}{cos(2t)} \right ) dt$

$dy = \left ( \frac{\frac{sin(2t)sin^3(t)}{\sqrt{cos(2t)}} + 3sin^2(t)cos(t) \sqrt{cos(2t)}}{cos(2t)} \right ) dt$

So
$\frac{dx}{dy} = \frac{\left ( \frac{\frac{sin(2t)cos^3(t)}{\sqrt{cos(2t)}} - 3sin(t)cos^2(t) \sqrt{cos(2t)}}{cos(2t)} \right ) dt}{\left ( \frac{\frac{sin(2t)sin^3(t)}{\sqrt{cos(2t)}} + 3sin^2(t)cos(t) \sqrt{cos(2t)}}{cos(2t)} \right ) dt}$

$\frac{dx}{dy} = \frac{\frac{sin(2t)cos^3(t)}{\sqrt{cos(2t)}} - 3sin(t)cos^2(t) \sqrt{cos(2t)}}{\frac{sin(2t)sin^3(t)}{\sqrt{cos(2 t)}} + 3sin^2(t)cos(t) \sqrt{cos(2t)}}$

$\frac{dx}{dy} = \frac{sin(2t)cos^3(t) - 3sin(t)cos^2(t)cos(2t)}{sin(2t)sin^3(t) + 3sin^2(t)cos(t)cos(2t)}$

It can possibly be simplified more, but I'll leave it here.

-Dan

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