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Math Help - derivative of an integral, but with 3 variable?

  1. #1
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    derivative of an integral, but with 3 variable?

    Welp, heres the last of my problems. As you can tell im quite the calc dunce. So we are basing this on Newton’s second law to derive the force balance (E).
    m(dv/dt) = -ks, Newton’s 2nd law, where v = ds/dt.

    So what they are asking to do is replace –ks with –f(s), a continuous, odd function. This looks like

    m(dv/dt) = -f(s)


    from there we multiply both sides by ds/dt.
    Heres what right side looks like:


    -f(s) (ds/dt) = -d/dt [ (INT s(0) to s) f(τ) dτ]


    (note: t and τ are not the same var.)
    I have no idea how to diff/ integrate this. “INT s(0) to s”
    just means the integral from s(0), bottom, to s, the top.
    I guessed maybe the answer is just f(t) but im thinking not. Sorry for the big jumble, its hard type all these complex functions.
    Let me know if this doesn’t make sense. Thank you!
    Last edited by chrsr345; September 26th 2007 at 07:16 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chrsr345 View Post
    Welp, heres the last of my problems. As you can tell im quite the calc dunce. So we are basing this on Newton’s second law to derive the force balance (E).
    m(dv/dt) = -ks, Newton’s 2nd law, where v = ds/dt.

    So what they are asking to do is replace –ks with –f(s), a continuous, odd function. This looks like

    m(dv/dt) = -f(s)


    from there we multiply both sides by ds/dt.
    Heres what right side looks like:


    f(s) (ds/dt) = d/dt [ (INT s(0) to s) f(τ) dτ]


    (note: t and τ are not the same var.)
    I have no idea how to diff/ integrate this. “INT s(0) to s”
    just means the integral from s(0), bottom, to s, the top.
    I guessed maybe the answer is just f(t) but im thinking not. Sorry for the big jumble, its hard type all these complex functions.
    Let me know if this doesn’t make sense. Thank you!
    is this what you mean?

    f(s) \frac {ds}{dt} = \frac {d}{dt} \int_{s(0)}^{s} f( \tau )~d \tau

    for some reason, that formula doesn't make sense to me. the left hand side is weird. and where did \tau come from?
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    thank you jhevon for the tip. i really am just anxious but ill delete it.
    And yes, i also dont know where the different greek t comes from. He was also quite inconsistent too...on the sheet he wrote that the whole thing is negative and another it is positive (maybe forgot the sign). but if it is negative, wouldnt that just equal 0?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chrsr345 View Post
    thank you jhevon for the tip. i really am just anxious but ill delete it.
    And yes, i also dont know where the different greek t comes from. He was also quite inconsistent too...on the sheet he wrote that the whole thing is negative and another it is positive (maybe forgot the sign). but if it is negative, wouldnt that just equal 0?
    since we are not even sure of the formula we have, i think it is best to wait for someone who has some background in this field to answer. topsquark and/or CaptainBlack and/or someone else should be able to answer this by tomorrow or so. hope you don't need it before then
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    is this what you mean?

    f(s) \frac {ds}{dt} = \frac {d}{dt} \int_{s(0)}^{s} f( \tau )~d \tau

    for some reason, that formula doesn't make sense to me. the left hand side is weird. and where did \tau come from?
    This is merely an identity. By the first fundamental theorem of Calculus we have that
    f(x) = \frac{d}{dx} \int_a^xf(\tau) d \tau
    (The variable \tau that is confusing you so much is merely an integration variable. We could call it anything we like because it is a "dummy" variable and has no influence on the final answer.)

    Since we are not taking \frac{d}{ds} of the integral, but \frac{d}{dt} we need to apply the chain rule to the deriviative, leaving us with
    \frac{d}{dt} \int_a^s f(\tau) d \tau = f(s) \cdot \frac{ds}{dt}
    (where I have simply put the constant "a" equal to the constant s_0.)

    -Dan
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