# Homework Help: Progation of error in this formula

1. Feb 12, 2005

### thegame

Hi,

u = (1/(2LS))^2

L = .9810 +/- 0.0005
S = 8.35 +/- 0.15

Can anyone give me the formula to calculate the uncertainity on u.

2. Feb 12, 2005

### vincentchan

$$(\delta u)^2 = (\frac{\partial u}{\partial L} \delta L)^2+ (\frac{\partial u}{\partial S} \delta S )^2$$

3. Feb 12, 2005

### thegame

Thanks but a very similar formula is given in my lab manual, but I don't know partial differentiation. So, how would i actually evaluate the uncertainty.

4. Feb 12, 2005

### faust9

Do you know how to do regular differentiation?

5. Feb 12, 2005

### thegame

ya... I am pretty sure uncertainities are evaluated using partial differentation

6. Feb 12, 2005

### vincentchan

just regard it as orinary derivative
for

$$\frac{\partial u}{\partial L}$$

take S as constant and L as independent variable, and do

$$\frac{d u}{d L}$$

7. Feb 12, 2005

### faust9

Given a function $U=x^2+y^2+2xy$.

To find the partials simply pretend the other variables are simply constants:

$$\frac{\partial U}{\partial x}=\frac{\partial}{\partial x}(x^2+y^2+2xy)$$

So, looking at the above $y^2$ will act as a constant. The derivative of a constant is zero so $\frac{\partial}{\partial x}y^2=0$.

Using that logic:

$$\frac{\partial U}{\partial x}=\frac{\partial}{\partial x}(x^2+y^2+2xy)=2x+0+2y$$

Or simply

$$\frac{\partial U}{\partial x}=2x+2y$$

8. Feb 12, 2005

### thegame

Thanks for all your help so far, but I will leave partials to next year calc... Can you guys simplify this for me

9. Feb 12, 2005

### faust9

It's easy to do. Just follow the steps I supplied in my above post. Pretend the other variables are constants and perform differentiation (just like regular except only one one variable with the others held constant) on the equation. Repeat this until all variables have been differentiated.

It's not hard, don't let the fact that you haven't had calc III scare you from this concept.

10. Feb 12, 2005

### faust9

Here: I'll do more work to show you how to do this rather than simply give you the answer...

$\delta x=0.1$ This is the error in x
$\delta y=0.05$ This is the error in y

x=10--value found during an experiment.
y=20--value found during an experiment.

$$U=x^3+3x^2y+3y^2x+y^3$$

the partial of the above (with y magically transformed into a constant) would be:

$$\frac{\partial U}{\partial x}=(3)x^2+(6y)x+(3y^2)+0$$

I put () around all of the constants in the final expression.

$$\frac{\partial U}{\partial y}=0+(3x^2)+(6x)y+(3)y^2$$

So, the error in U would be:

$$U_{error}=\sqrt{(\frac{\partial U}{\partial x}\delta x)^2+(\frac{\partial U}{\partial y}\delta y)^2}$$

which when we substitute the partials into the error eqn yields:

$$U_{error}=\sqrt{([(3)x^2+(6y)x+(3y^2)]\delta x)^2+([(3x^2)+(6x)y+(3)y^2]\delta y)^2}$$

Plug in the numeric values for x/y and the errors:

$$U_{error}=\sqrt{([(3)10^2+(6\cdot 20)10+(3\cdot 20^2)]0.1)^2+([(3\cdot 10^2)+(6\cdot 10)20+(3)20^2]0.05)^2}$$

Plug the above into a calculator and you'll get something greater than the largest single value of error (you will never get a total error smaller than any one of the error components.

Hope this helped.

11. Feb 12, 2005

### thegame

Thanks, your post is really helpful... I don't know how to use latex but I got the final error as 1.36 x 10-4 .. Hopefully I did it right :rofl: