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Math Help - Newton's Method

  1. #1
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    Newton's Method

    I'm working on two problems that I need help on.

    (1) The radius of a circle is to be measured and its area computed. If the radius can be measured to an accuracy of 0.001 in and the area must be accurate to 0.1in^2, estimate the maximum radius for which this process can be used.

    (2) If pV=20 and p is measured as 5\pm0.02, estimate V.

    Here are the formulas I have been using for these problems (out of Schaum's Calculus, 4th ed.):

    f(x+\Delta x) \sim f(x)+\dot f(x)*\Delta x)

    \displaystyle x_{n+1}=x_n+\frac{f(x_n)}{f\prime(x_n)}

    I'm not requesting that someone work out the problems for me, just help me with where to start. Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by dbakeg00 View Post
    I'm working on two problems that I need help on.

    (1) The radius of a circle is to be measured and its area computed. If the radius can be measured to an accuracy of 0.001 in and the area must be accurate to 0.1in^2, estimate the maximum radius for which this process can be used.
    Let the radius be $$ r and error in measuring it be $$ e so the measured radius is \hat{r}=r+e, then the computed area from the measumenet is:

    \hat{A}=\pi \hat{r}^2=\pi (r+e)^2=\pi r^2+ \pi (2r\; e+e^2)=A+ \pi (2r\; e+e^2)

    So the error in the area is \pi (2r\; e+e^2) and if e=0.001 units and the error is the area is 0.1 sq-units then:

    \pi (2\times 0.001 r+0.001^2)=0.1

    so:

    r=\frac{0.1}{2 \times 0.001}-\frac{0.001}{2}\approx 50

    Alternatively you use A=f(r)=\pi r^2, then:

    A =f(r) \approx f(\hat{r}) - ef'(\hat{r})=\hat{A} -e f'(\hat{r})

    Then if $$ u is the maximum permissible error in $$ A you solve u=|ef'(\hat{r})|

    Which is equivalent to the earlier method when the e^2 terms are ignored. This later method has the advantage that it will still work for more complicated functions (and in particular the function in the next question).

    CB
    Last edited by CaptainBlack; November 10th 2010 at 11:58 PM.
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