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Math Help - Differentials

  1. #1
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    Exclamation Differentials

    Newton's law of gravitation states that the force of F of attraction between tow particles having masses m and M is given by F = (gmM)/(s^2) for a constant g and the distance s between the particles. If s = 20cm, use differentials to approximate the change in s that will increase F by 10%.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by smschaefer View Post
    Newton's law of gravitation states that the force of F of attraction between tow particles having masses m and M is given by F = (gmM)/(s^2) for a constant g and the distance s between the particles. If s = 20cm, use differentials to approximate the change in s that will increase F by 10%.
    You have \frac{dF}{ds}=\frac{-2(gMm)}{s^3}..so dF=\frac{ds\cdot{-2(gMm)}}{s^3} so imput yoru values and calculate ds
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  3. #3
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    What?!?!?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    ..................

    Quote Originally Posted by smschaefer View Post
    What?!?!?
    What do you mean what...that is what you asked for isnt it? Using differentials?
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  5. #5
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    i just dont get it... is there another way you could explain it or expand more on what you did.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    ok

    Quote Originally Posted by smschaefer View Post
    i just dont get it... is there another way you could explain it or expand more on what you did.
    Most of what I am about to say is not technically correct but it will help you understand....since a derivative is measuring slope you have the derivative is \frac{dy}{dx}\Rightarrow{\frac{\Delta{y}}{\Delta{x  }}}...therefore dy\approx{\Delta{y}},dx\approx{\Delta{x}}...so you take the derivative so you can get your changes in x and y...now that you have that seperate...so you are trying to find the change in s( ds) that makes F change 10%...make sense?
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  7. #7
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    that sounds a little bit more familiar, but how did you get -2 and s^3 in the problem?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Because

    Quote Originally Posted by smschaefer View Post
    that sounds a little bit more familiar, but how did you get -2 and s^3 in the problem?
    you have F=\frac{gMm}{s^2}...now to get \frac{dF}{ds} which is what we want we have to differentiate both sides which is how I got that
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  9. #9
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    ok i think i got it...thanks
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Yep

    Quote Originally Posted by smschaefer View Post
    ok i think i got it...thanks
    No problem...thanks
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  11. #11
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    more help

    if i am suppose to plug in s=20 to the equation how do i factor in the 10%.... dF= (ds(-2(gmM))) / 20
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