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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- Jan 6th 2010, 04:21 PMalwaizminimax. 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! - Jan 6th 2010, 05:09 PMVonNemo19
- Jan 6th 2010, 05:25 PMalwaizmini
that's all the info that was given...

- Jan 6th 2010, 05:29 PMVonNemo19
- Jan 7th 2010, 09:42 AMCalculus26
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% - Jan 7th 2010, 11:24 AMaidan