What is the formula for error analysis in multiplication and division?

  • Thread starter Thread starter zak8000
  • Start date Start date
  • Tags Tags
    Error
Click For Summary
Error analysis in multiplication and division involves using relative errors instead of absolute errors. For multiplication, the mean is calculated by multiplying the values, and the error is derived from the square root of the sum of the squares of the relative errors. In the provided example, for x*y, the result is 15 ± 13.45, while for x/y, the result is 0.6 ± 0.538. Understanding the transition between absolute and relative errors is crucial for accurate calculations. This approach ensures proper error representation in scientific reporting.
zak8000
Messages
69
Reaction score
0

Homework Statement


hi there i am currently working on a lab report which involves intense error anaylsis and i stink at it so i have made up the following two questions to understand how to perform error analysis with multiplication and division

if x=3+-2 m y=5+-3 s

then what is x*y answer in the form a+-b where a is the mean and b is the error

and also what is x/y in the form a+-b where a is the mean and b is the error

please help!


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Hi, the basic rules for error analysis are that for addition and subtraction working with absoulte errors ie plus or minus 1cm would be an example of an absolute error, you just add or subtract the values, and the new error in both cases is given by the square root of the sum of the squares of the absoute errors. ie sqrt[(error in a)^2 + (error in b)^2]



For multiplication or division, you just multiply or divide the values and the new error is given by the same formula but instead of using absolute errors, use relative errors, ie percentage errors, so depending on what sort of errors you have to start with you may just have a bit of switching between absolute and relative errors where

relative error = (absolute error)/value

so considering the values you gave

x*y = 3*5 +- sqrt[(2/3)^2+(3/5)^2)*3*5 = 15+-sqrt(181/225)*15 = 15 +-13.45

and x/y= 3/5 +- sqrt[(2/3)^2+(3/5)^2)*3/5 = 0.6 +- 0.538

Hope this helps
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
9K
  • · Replies 5 ·
Replies
5
Views
2K