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Math Help - find cube root 0.9

  1. #1
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    Smile find cube root 0.9

    Hi again guys,

    Two posts from me. But they aren't too hard. For you at least!

    Given that y = x^{-\frac{1}{3}} use the calculus to determine an approximate value for  \frac{1}{\sqrt [3]{0.9}} Normally with these there is a descernable x and \delta x but I can't see a way to giggle 0.9 into anything useful. Can you?
    Last edited by mr fantastic; June 29th 2009 at 05:36 PM. Reason: Fixed exponent in latex code
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  2. #2
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    I would find the first few terms of the Taylor series of  x^{-1/3} centered at a=1.
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  3. #3
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    Quote Originally Posted by s_ingram View Post
    Hi again guys,

    Two posts from me. But they aren't too hard. For you at least!

    Given that y = x^{-\frac{1}{3}} use the calculus to determine an approximate value for  \frac{1}{\sqrt [3]{0.9}} Normally with these there is a descernable x and \delta x but I can't see a way to giggle 0.9 into anything useful. Can you?
    Use the linear approximation: f(x + \delta x) \approx f(x) + \delta f'(x).

    Take f(x) = x^{-1/3}, x = 1 and \delta  = -0.1.
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  4. #4
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    Hello, s_ingram!

    Given that: y \:=\: x^{-\frac{1}{3}} use the calculus to determine an approximate value for  \frac{1}{\sqrt [3]{0.9}}
    We're expected to use: . x = 1\:\text{ and }\:dx = \text{-}0.1

    We have: . dy \:=\:-\dfrac{1}{3}x^{-\frac{4}{3}}dx \:=\:-\frac{dx}{3\sqrt[3]{x^4}}


    If x = 1,\:dx = \text{-}0.1. we have: . y + dy \:=\:1^{-\frac{1}{3}} - \frac{\text{-}0.1}{3\sqrt[3]{1^4}} \:=\:1 + \frac{1}{30} \:=\:1.0333\hdots


    Therefore: . \frac{1}{\sqrt[3]{0.9}} \;\approx\;1.033

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  5. #5
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    AH!!!! So easy! I was sure that being a cube root I had to pull a 2 out, so I was playing with 2 x 9/80 = 2. 0.1125 and I wasn't getting very far! Thanks guys - you are both Fantastic!
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