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Thread: 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 $\displaystyle t$ or with respect to $\displaystyle x$?

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

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

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


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

    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 $\displaystyle t$ or with respect to $\displaystyle x$?

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

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

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


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

    Then you can simplify a bit.
    I believe that

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

    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

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

    Yeah?
    Correct, as long as $\displaystyle x$ is constant with respect to $\displaystyle t$.
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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 $\displaystyle x$ is constant with respect to $\displaystyle t$.
    .........................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 $\displaystyle x$ is constant, so you should leave it in a form involving $\displaystyle \frac{dx}{dt}$. If $\displaystyle x$ 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 $\displaystyle t$, (D) would be correct.
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