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Maximum Error Using Differentials

  1. Mar 9, 2016 #1

    njo

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    1. The problem statement, all variables and given/known data
    2 adjacent sides of a parallelogram measure 15ft and 10ft w/ max errors of ±0.1ft
    angle is 45° w/ max error of ±0.5°

    What is the maximum error in the calculated value of the area or the parallelogram?

    2. Relevant equations
    A = area = xysinθ


    3. The attempt at a solution
    x = 15ft dx = dy = ±0.1ft
    y = 10ft dθ = ±0.5°
    θ = 45°

    dA = (∂A/∂x)dx + (∂A/∂y)dy + (∂A/∂θ)dθ
    dA = 10sin45°(±0.1) + 15sin45°(±0.1) + 150cos45°(±0.5°)
    dA = ±56.3207 ft^2

    Not sure if this is correct. Tried to follow examples on my notes. Any help I would greatly appreciate. Thanks
     
    Last edited: Mar 9, 2016
  2. jcsd
  3. Mar 9, 2016 #2

    SteamKing

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    The error in the measurements of the length of the sides is ±0.01 feet, not ±0.1 feet.

    When accounting for the tolerances of angular measure, you should always use the equivalent radian measure rather than degrees.

    The area of the parallelogram is 106.07 ft2. An error of 56.32 ft2 would render this calculation essentially meaningless.
    You've got some changes to make.
     
  4. Mar 9, 2016 #3

    njo

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    What changes? the differential equation is wrong right?
     
  5. Mar 9, 2016 #4

    SteamKing

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    No, the differential equation appears to be correct. You just plugged the wrong numbers into it, as I explained.
     
  6. Mar 9, 2016 #5

    njo

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    Sorry I didn't see your whole post on my phone at first. That's a typo. All should be +- 0.1 ft not .01 ft
     
  7. Mar 9, 2016 #6

    Ray Vickson

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    Not correct, even when the ±0.1 is the correct figure.

    If the two sides have lengths A and B and the angle is θ, the smallest/largest values of A are A1=14.9 and A2 =15.1, the smallest/largest values of B are B1=9.9 and B2 = 10.1, while the smallest/largest values of θ are θ1= 44.5° and θ2 = 45.5°. The smallest/largest values of the area are A1*B1*sin(θ1) and A2*B2*sin(θ2).

    Compute these directly, and see how the results compare with your differential approximation. For example, you can compute Δsin(θ) = sin(45.5°)-sin(45°) exactly as written, and compare that with your calculus-based expression.
     
    Last edited: Mar 9, 2016
  8. Mar 9, 2016 #7

    njo

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    A = area = xysinθ
    dA = (∂A/∂x)dx + (∂A/∂y)dy + (∂A/∂θ)dθ

    So the margin for the area is about 5ft^2. How is my differential equation incorrect? I took each partial and multiplied by the rate of change.
     
  9. Mar 9, 2016 #8

    Mark44

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    Your total differential is correct, but the value you used for ##\Delta \theta## is not correct. It should be a real number, not be in degrees.
     
  10. Mar 9, 2016 #9

    njo

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    That makes sense. Although I get an answer less than A2*B2*sin(θ2) - A1*B1*sin(θ1)

    Δθ = .5pi/180 ≅ .0087 ∴ 10sin45°(±0.1) + 15sin45°(±0.1) + 150cos45°(±.pi/360) ≅ ±2.6934
     
    Last edited: Mar 9, 2016
  11. Mar 9, 2016 #10

    Ray Vickson

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    You should be able to figure out this for yourself, given all the hints you received. Have you actually done what I suggested, that is, compute Δsin(θ) = sin(45.5°)-sin(45°) and then compared that with you calculus-based result?
     
    Last edited: Mar 9, 2016
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