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Math Help - Differential calculus

  1. #1
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    Exclamation Differential calculus

    How do I go from

    <br />
\int \frac{dv}{cv^2 - g} = \int~dt<br />

    to

    <br />
\frac{1}{2\sqrt{cg}}\ln\left(\frac{\sqrt g + v\sqrt c}{\sqrt g - v\sqrt c}\right)=a-t<br />

    It's not just urgent - it's VERY urgent! :/ Big assignment..
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  2. #2
    GAMMA Mathematics
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    Shouldn't it be a + t?
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  3. #3
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    Hmm..

    My calculator gives me a-t..
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  4. #4
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    Hello, Rydbirk!

    How do I go from: . <br />
\int \frac{dv}{cv^2 - g} \:= \:\int \,dt

    . . . to: . \frac{1}{2\sqrt{cg}}\ln\left|\frac{\sqrt g + v\sqrt c}{\sqrt g - v\sqrt c}\right|\;=\;C-t

    The right side is: . \int dt \;=\;t + C


    Let:  cv^2 \:=\:u^2\quad\Rightarrow\quad v \:=\:\frac{1}{\sqrt{c}}u\quad\Rightarrow\quad dv \:=\:\frac{du}{\sqrt{c}}
    Also let: g \:=\:a^2

    The left side becomes: . \int\frac{1}{u^2-a^2}\left(\frac{du}{\sqrt{c}} \right) \;=\;\frac{1}{2a\sqrt{c}}\ln\left|\frac{u-a}{u+a}\right| + C


    Back-substitute: . u \:=\:v\sqrt{c}\;\text{ and }\; a = \sqrt{g}

    . . \frac{1}{2\sqrt{g}\sqrt{c}} \ln\left|\frac{v\sqrt{c} - \sqrt{g}}{v\sqrt{c} + \sqrt{g}}\right|


    And we have: . \frac{1}{2\sqrt{cg}}\ln\left|\frac{v\sqrt{c} - \sqrt{g}}{v\sqrt{c} + \sqrt{g}}\right| \;=\;t + C


    This answer is equivalent to the one your calculator gave.

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  5. #5
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    Thank you!! You saved my day
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