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Math Help - Differential calculus?

  1. #1
    Super Member fardeen_gen's Avatar
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    Differential calculus?

    If ω & A are constant then d[ω√(A^2 - x^2)]/dt =?

    A) ω/{2√(A^2 - x^2)}
    B) -ω/{2√(A^2 - x^2)}
    C) -ωx/{√(A^2 - x^2)}
    D) None of these

    Can u please explain the reasoning too?
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  2. #2
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    If ω & A are constant then d[ω√(A^2 - x^2)]/dt =?

    A) ω/{2√(A^2 - x^2)}
    B) -ω/{2√(A^2 - x^2)}
    C) -ωx/{√(A^2 - x^2)}
    D) None of these

    Can u please explain the reasoning too?
    Are you differentiating with respect to t or with respect to x?

    For \frac{d}{dt}\left[\omega\sqrt{A^2 - x^2}\right] we have

    \frac{d}{dt}\left[\omega\sqrt{A^2 - x^2}\right] = \frac{d}{dt}\left[\omega\left(A^2 - x^2\right)^{1/2}\right]

    = \frac{\omega}2\left(A^2 - x^2\right)^{-1/2}\frac{d}{dt}\left[A^2 - x^2\right]


    = \frac{\omega}2\left(A^2 - x^2\right)^{-1/2}\left(-2x\frac{dx}{dt}\right)

    Then you can simplify a bit.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Reckoner View Post
    Are you differentiating with respect to t or with respect to x?

    For \frac{d}{dt}\left[\omega\sqrt{A^2 - x^2}\right] we have

    \frac{d}{dt}\left[\omega\sqrt{A^2 - x^2}\right] = \frac{d}{dt}\left[\omega\left(A^2 - x^2\right)^{1/2}\right]

    = \frac{\omega}2\left(A^2 - x^2\right)^{-1/2}\frac{d}{dt}\left[A^2 - x^2\right]


    = \frac{\omega}2\left(A^2 - x^2\right)^{-1/2}\left(-2x\frac{dx}{dt}\right)

    Then you can simplify a bit.
    I believe that

    \frac{d}{dt}\bigg[f(\omega,A,x)\bigg]=0

    Yeah?
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  4. #4
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    I believe that

    \frac{d}{dt}\bigg[f(\omega,A,x)\bigg]=0

    Yeah?
    Correct, as long as x is constant with respect to t.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Reckoner View Post
    Correct, as long as x is constant with respect to t.
    .........................I know this............................I was pointing something out........in your post.
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  6. #6
    Super Member fardeen_gen's Avatar
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    Is the answer 0 then? So D is the correct option, right?
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  7. #7
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by fardeen_gen View Post
    Is the answer 0 then? So D is the correct option, right?
    We do not know for sure that x is constant, so you should leave it in a form involving \frac{dx}{dt}. If x happens to be constant the derivative will still work out to be 0.

    So in this case, assuming you typed the problem correctly and we are differentiating with respect to t, (D) would be correct.
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