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Linear Approximations

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the linear approximation of [itex]\sqrt[3]{27.02}[/itex]

    2. Relevant equations


    3. The attempt at a solution

    So what I did was work with[itex]\sqrt[3]{27}[/itex] since that's an easily known value. So my f(x)=x[itex]^{1/3}[/itex] and my f'(x)=1/3x[itex]^{-2/3}[/itex]. From there, I worked f(27) = 3, and f'(27)=1/27.

    Then I used the above equation. 3+1/27(x-27). For my value of x, I used 27.02, and got .0004.

    Something tells me I'm doing something incorrectly, though...

    Thanks for the help!
  2. jcsd
  3. Dec 6, 2011 #2


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    Homework Helper

    Are you saying 3+1/27(x-27) = .0004? It has to larger than 3.
  4. Dec 6, 2011 #3
    My mistake -- calculated that incorrectly. I actually get 3.00074. Was my method correct in that case?
  5. Dec 6, 2011 #4


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    Homework Helper

    Your method looks ok to me. A good sanity check is to calculate the cube root of 27.02 and compare to the approximation.
    Last edited: Dec 6, 2011
  6. Dec 6, 2011 #5
    Thank goodness for sanity checks. Thanks a lot!
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