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anonymous

  • 5 years ago

consider a closed rectangular box with a square base, whose edges have length x. If x is measured with error at most 2% and the height y is measured with error at most 3%, use a differential to estimate the corresponding %error in computing the box's surface area.

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  1. anonymous
    • 5 years ago
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    how much of an estimate can we make? I used x+dx, y+dy and found a term for the error, but it doesnt exactly come out nice.

  2. anonymous
    • 5 years ago
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    the answer should be +/- 5%

  3. anonymous
    • 5 years ago
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    ok I got about 4%

  4. anonymous
    • 5 years ago
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    here is my attempt, given: dx/x=2%, dy/y=3%, and i am looking for ds/s

  5. anonymous
    • 5 years ago
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    using partial derivatives, ds=(4x+4y)dx+(4x)dy

  6. anonymous
    • 5 years ago
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    ds/s=(4x+4y)dx+(4x)dy /(2x^2+4xy), and then i am stuck.. i was able to get dx/x but not dy/y

  7. anonymous
    • 5 years ago
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    k give me a sec to think about this

  8. anonymous
    • 5 years ago
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    Thanks so much!!!!

  9. anonymous
    • 5 years ago
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    k it turns out the same way as the way I did it actually: multiply the first part of the numerator by x/x and the second part by y/y then you have ((4x^2+4xy)dx/x+4xy(dy/y)) we know dx/x=.02, dy/y=.03 pluging in (.08x^2+.08xy+.12xy)/(2x^2+4xy) =(.08x^2+.2xy)/(2x^2+4xy) now here is why I asked how big of an estimate we can make, we round .08 to .1 so that it comes out nicely: .1(x^2+2xy)/(2(x^2+2xy))=.1/2=.05 5%

  10. anonymous
    • 5 years ago
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    AWESOME! i got stuck because i didn't round it off.......

  11. anonymous
    • 5 years ago
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    yeah it's not nice, but they just want an estimate so I geuss it is ok

  12. anonymous
    • 5 years ago
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    Thank you very much :)

  13. anonymous
    • 5 years ago
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    you're welcome :)

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