# max. percentage error

• Jan 6th 2010, 03:21 PM
alwaizmini
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!
• Jan 6th 2010, 04:09 PM
VonNemo19
Quote:

Originally Posted by alwaizmini
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.
• Jan 6th 2010, 04:25 PM
alwaizmini
that's all the info that was given...
• Jan 6th 2010, 04:29 PM
VonNemo19
Quote:

Originally Posted by alwaizmini
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.
• Jan 7th 2010, 08:42 AM
Calculus26
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 AM
aidan
Quote:

Originally Posted by alwaizmini
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 $\displaystyle r = 5 -0.0499999 = 4.950001$
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 %