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, 03:21 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, 04:09 PMVonNemo19
- Jan 6th 2010, 04:25 PMalwaizmini
that's all the info that was given...

- Jan 6th 2010, 04:29 PMVonNemo19
- Jan 7th 2010, 08:42 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, 10:24 AMaidan
Calculus26 provided the most appropriate.

This is another way.

minimum $\displaystyle r = 5 -0.0499999 = 4.950001$

maximum $\displaystyle R = 5 +0.049999 = 5.049999$

minimum $\displaystyle h = 12 - 0.049999 = 11.950001$

maximum $\displaystyle H = 12 + 0.049999 = 12.049999$

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

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

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

difference:$\displaystyle 965.4271 - 919.8741 = 45.553$

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

4.83 %