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Math Help - Simple real world calculus problem

  1. #1
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    Angry Simple real world calculus problem

    Simple real world calculus problem-untitled.png

    How do you do part a) ?
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  2. #2
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    Re: Simple real world calculus problem

    Quote Originally Posted by princessmath View Post
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    How do you do part a) ?
    Hi there,

    You have x^2 + y^2 = 200

    differentiating them implicitly gives you

    (x^2)' + (y^2)' = (200)' which gives you

    2x(dx/dt) + 2y = 0 ...

    now, plug in the values and solve for dx/dt,

    hope this help,

    dokrbb
    Last edited by dokrbb; April 15th 2013 at 05:32 AM.
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  3. #3
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    Re: Simple real world calculus problem

    Quote Originally Posted by dokrbb View Post
    Hi there,

    You have x^2 + y^2 = 200

    differentiating them implicitly gives you

    (x^2)' + (y^2)' = (200)' which gives you

    2x(dx/dt) + 2y = 0 ...

    now, plug in the values and solve for dx/dt,

    hope this help,

    dokrbb
    This is incorrect. x and y are both functions of t, so when you differentiate both sides, you should get

    \displaystyle \begin{align*} x^2 + y^2 &= 100 \\ \frac{d}{dt} \left( x^2 + y^2 \right) &= \frac{d}{dt} \left( 100 \right) \\ \frac{d}{dx}\left( x^2 \right) \frac{dx}{dt} + \frac{d}{dy}\left( y^2 \right) \frac{dy}{dt} &= 0 \\ 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} &= 0 \\ 2x\,\frac{dx}{dt} + 2 \, \sqrt{ 100 - x^2 } \,\frac{dy}{dt} &= 0 \end{align*}

    NOW you can substitute in the known values and solve for \displaystyle \frac{dx}{dt}.
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