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Math Help - Calculus

  1. #1
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    Calculus

    A bug is crawling along a parabolla y= x^2 (in quadrant 1) at a rate such that its distance from the origin is increasing at 6 cm/min. At what rate are the x- and y- coordinates of the bug increasing when the bug is at the point (3,9)?
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  2. #2
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    Re: Calculus

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  3. #3
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    Re: Calculus

    We know that y(t) = (x(t))2. Therefore:

    $y'(t) = 2x(t)(x'(t))$

    We also know that the distance $d(t)$ from the origin is:

    $d(t) = \sqrt{(x(t))^2 + (y(t))^2}$.

    Taking $d'(t)$ we get:

    $6 = d'(t) = \dfrac{(2x(t))(x'(t)) + (2y(t))(y'(t))}{2\sqrt{(x(t))^2 + (y(t))^2}}$

    $= \dfrac{(x'(t))(x(t))(1 + 2(x(t))^2)}{x(t)\sqrt{1 + (x(t))^2}}$

    so that:

    $x'(t) = \dfrac{6\sqrt{1 + (x(t))^2}}{(1 + 2(x(t))^2}$

    Substituting in $x(t) = 3$, we obtain:

    $x'(t) = \dfrac{6\sqrt{10}}{19}$ and:

    $y'(t) = 2x(t)(x'(t)) = \dfrac{36\sqrt{10}}{19}$
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  4. #4
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    Re: Calculus

    thanks
    Quote Originally Posted by infkelsier View Post
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