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Math Help - max. percentage error

  1. #1
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    max. percentage error

    if r=5cm and h=12cm to the nearest mm, what should we expect the max. percentage error in calculating V=(pi)(r^2)h to be?

    Thanks!
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by alwaizmini View Post
    if r=5cm and h=12cm to the nearest mm, what should we expect the max. percentage error in calculating V=(pi)(r^2)h to be?

    Thanks!
    Not enough info.
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  3. #3
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    that's all the info that was given...
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by alwaizmini View Post
    that's all the info that was given...
    Well, then my advice is to move on to the next problem and quit sweating this one. The numbers that you were given will help to calculate the volume, but the problem gives no type of information regarding error.
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  5. #5
    MHF Contributor Calculus26's Avatar
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    Actually there is enough info Using the differential

    Note dv/V is the relative error and dV/V*100 is max %error

    dr = .1cm dh = .1cm

    V=(pi)(r^2)h to be?


    dV = 2pi rh dr +pir^2dh

    dV/V = [2pi rh dr +pir^2dh]/(pi)(r^2)h

    dV/V = 2dr/r + dh/h

    dV/V = 2*.1/5 + .1/12= .0483

    so max % eror is 4.83%
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  6. #6
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    Quote Originally Posted by alwaizmini View Post
    if r=5cm and h=12cm to the nearest mm, what should we expect the max. percentage error in calculating V=(pi)(r^2)h to be?

    Thanks!
    Calculus26 provided the most appropriate.
    This is another way.

    minimum r = 5 -0.0499999 = 4.950001
    maximum R = 5 +0.049999 = 5.049999

    minimum h = 12 - 0.049999 = 11.950001
    maximum H = 12 + 0.049999 = 12.049999

    minimum  V = \pi 4.950001^2 \cdot 11.950001 = 919.8741

    maximum  V = \pi 5.049999^2 \cdot 12.049999 = 965.4271

    average Volume =  \dfrac{919.8741 + 965.4271}{2} = 942.6506

    difference:  965.4271 - 919.8741 = 45.553

    ratio:  \dfrac{45.553}{942.6506} = 0.048324

    4.83 %
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