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Math Help - Find the value of h given the following relation?

  1. #1
    Super Member fardeen_gen's Avatar
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    Find the value of h given the following relation?

    If:
    ∫(vo to 0) (v/(-g - (kv^2/m) dv = ∫(0 to h) dh
    , where k and g are constants,

    Find h.

    (What is the integral of type ∫ (x/((ab - cx^2)/a)) dx, where a, b and c are constants?)

    Last edited by fardeen_gen; January 9th 2009 at 05:06 AM. Reason: Typo
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    If:
    (What is the integral of type ∫ (x/((ab - cx^2)/a)) dx, where a, b and c are constants?)
    \int {\frac{x}{{\left( {{{\left( {ab - c{x^2}} \right)} \mathord{\left/{\vphantom {{\left( {ab - c{x^2}} \right)} a}} \right.\kern-\nulldelimiterspace} a}} \right)}}dx}  = \frac{a}{{2c}}\int {\frac{{2cx}}{{ab - c{x^2}}}dx}  = - \frac{a}{{2c}}\int {\frac{{d\left( {ab - c{x^2}} \right)}}{{ab - c{x^2}}}}  =  - \frac{a}{{2c}}\ln \left| {ab - c{x^2}} \right| + C.
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  3. #3
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by DeMath View Post
    \int {\frac{x}{{\left( {{{\left( {ab - c{x^2}} \right)} \mathord{\left/{\vphantom {{\left( {ab - c{x^2}} \right)} a}} \right.\kern-\nulldelimiterspace} a}} \right)}}dx} = \frac{a}{{2c}}\int {\frac{{2cx}}{{ab - c{x^2}}}dx} = - \frac{a}{{2c}}\int {\frac{{d\left( {ab - c{x^2}} \right)}}{{ab - c{x^2}}}} = - \frac{a}{{2c}}\ln \left| {ab - c{x^2}} \right| + C.
    Is -2cx dx same as d(ab -cx^2)?How?
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  4. #4
    o_O
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    Since: \frac{d}{dx} \left(ab - cx^2\right) = -2cx

    In the standard u-sub notation, it is just recognizing that: {\color{blue}u = ab - cx^2} \ \Rightarrow \ du = -2cx \ dx \ \Leftrightarrow \ {\color{red}-\frac{du}{2c} = x dx}

    So: a \int \frac{{\color{red}x}}{{\color{blue}ab-x^2}} \ {\color{red}dx} \ = \ a \int \frac{{\color{red}-\frac{du}{2c}}}{{\color{blue}u}} \ = \ -\frac{a}{2c} \int \frac{du}{u} = \ \cdots
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