Results 1 to 2 of 2

Thread: question

  1. #1
    Harezmi&Cabir
    Guest

    question

    forum or my mail thank you. mathmanman@hotmail.com
    Attached Thumbnails Attached Thumbnails question-ads-z.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,119
    Thanks
    718
    Awards
    1
    $\displaystyle x = \frac{cos^3(t)}{\sqrt{cos(2t)}}$

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

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

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

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

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

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

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

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

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum