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?
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%