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Thread: Maximum Error Problem

  1. #1
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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:
    $\displaystyle A=\frac{1}{2}bh$

    $\displaystyle \partial A = \frac{\partial A}{\partial a} * \partial a + \frac{\partial A}{\partial b} * \partial b
    $

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

    sub b=8.0m
    $\displaystyle =4$

    $\displaystyle \frac{\partial A}{\partial b}$

    $\displaystyle =\frac{1}{2}h$

    sub h=6
    $\displaystyle =3$

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

    $\displaystyle = 0.025$

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

    $\displaystyle = 0.033$

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

    $\displaystyle =7$

    P.S
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  2. #2
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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:
    $\displaystyle A=\frac{1}{2}bh$

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

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

    sub b=8.0m
    $\displaystyle =4$

    $\displaystyle \frac{\partial A}{\partial b}$

    $\displaystyle =\frac{1}{2}h$

    sub h=6
    $\displaystyle =3$

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

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

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

    $\displaystyle = 0.033$
    db also equals .1

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

    $\displaystyle =7$

    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!

    $\displaystyle dA= \frac{1}{2}h db+ \frac{1}{2}b dh= \frac{1}{2}(h db+ b dh)= \frac{1}{2}(6(.1)+ 8(.1))= \frac{1}{2}(1.4)= 0.7$ 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, $\displaystyle \frac{dA}{A}= \frac{dx}{x}+ \frac{dy}{y}$.

    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 $\displaystyle \frac{.1}{.6}= \frac{1}{6}$ and the relative error in base is $\displaystyle \frac{.1}{.8}= \frac{1}{8}$. The relative error in the area is $\displaystyle \frac{1}{6}+ \frac{1}{8}= \frac{4}{24}+ \frac{3}{24}= \frac{7}{24}$. 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
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    ok thanks i think i understand now.
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