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Math Help - Maximum Error Problem

  1. #1
    Super Member
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    Dec 2008
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    Maximum Error Problem

    Hi
    I am having trouble getting the correct answer to the following question:

    1) The sides of a right angle triangle enclosing the right angle are measured as 6.0m and 8.0m respectively. The maximum errors in each measurement are +- 0.1m
    i) Find approximately the maximum error in the calculated area of the triangle.

    This is what i have done:
    A=\frac{1}{2}bh

    \partial A = \frac{\partial A}{\partial a} * \partial a + \frac{\partial A}{\partial b} * \partial b<br />

    \frac{\partial A}{\partial a}
    =\frac{1}{2}b

    sub b=8.0m
    =4

    \frac{\partial A}{\partial b}

    =\frac{1}{2}h

    sub h=6
    =3

    \partial a = \frac{0.1}{4}

    = 0.025

    \partial b = \frac{0.1}{3}

    = 0.033

    \partial A = \frac{4}{0.025} * 0.025 \frac{3}{0.033} * 0.033

    =7

    P.S
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Paymemoney View Post
    Hi
    I am having trouble getting the correct answer to the following question:

    1) The sides of a right angle triangle enclosing the right angle are measured as 6.0m and 8.0m respectively. The maximum errors in each measurement are +- 0.1m
    i) Find approximately the maximum error in the calculated area of the triangle.

    This is what i have done:
    A=\frac{1}{2}bh

    \partial A = \frac{\partial A}{\partial a} * \partial a + \frac{\partial A}{\partial b} * \partial b<br />
]
    Technical point: standard notation for this would be dA = \frac{\partial A}{\partial a} da + \frac{\partial A}{\partial b} db. Don't use " \partial" with differentials. More importantly, since there is an "h" rather than "a" in the formula, it should be
    dA = \frac{\partial A}{\partial h} dh + \frac{\partial A}{\partial b}db

    \frac{\partial A}{\partial a}
    =\frac{1}{2}b

    sub b=8.0m
    =4

    \frac{\partial A}{\partial b}

    =\frac{1}{2}h

    sub h=6
    =3

    \partial a = \frac{0.1}{4}

    = 0.025
    No, da= 0.1. Where did the "4" come from?

    \partial b = \frac{0.1}{3}

    = 0.033
    db also equals .1

    \partial A = \frac{4}{0.025} * 0.025 \frac{3}{0.033} * 0.033

    =7

    P.S
    Does this make sense to you? Errors of .1 m in the measurement of the sides will cause an error of 7 square meters in the area? The calculated error would be (1/2)(6)(8)= 24 square meters. 7 square meters would be almost 1/3 of that!

    dA= \frac{1}{2}h db+ \frac{1}{2}b dh= \frac{1}{2}(h db+ b dh)= \frac{1}{2}(6(.1)+ 8(.1))= \frac{1}{2}(1.4)= 0.7 square meters.

    Note that is U= x+ y then dU= dx+ dy.

    If A= xy then dA= ydx+ xdy. Dividing both sides by A= xy, \frac{dA}{A}= \frac{dx}{x}+ \frac{dy}{y}.

    Those give the old engineers "rule of thumb":

    "When measurements are added, their errors add. When meaurements are multiplied, their relative errors add."

    Here the relative error in height is \frac{.1}{.6}= \frac{1}{6} and the relative error in base is \frac{.1}{.8}= \frac{1}{8}. The relative error in the area is \frac{1}{6}+ \frac{1}{8}= \frac{4}{24}+ \frac{3}{24}= \frac{7}{24}. Since the calculated area is (1/2)(.6)(.8)= .24, the maximum error is [tex]\frac{7}{24}(.24)= 0.7 square meters.
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  3. #3
    Super Member
    Joined
    Dec 2008
    Posts
    509
    ok thanks i think i understand now.
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