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!
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%
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 %