# Uncertainty Questions

1. Sep 22, 2007

### kayleech

These are my measurements:
D1 = 5.381 +/- 0.001 cm
D2 = 2.070 +/- 0.001 cm
t = 0.304 +/- 0.005 cm
m = 40.04 +/- 0.005 g

I then calculated the uncertainties of D1 and D2 squared:
D1^2 = 28.95 +/- 0.01 cm^2
D2^2 = 4.285 +/- 0.004 cm^2
D1^2 - D2^2 = 24.67 +/- 0.01 cm^2

Up until there, I'm pretty sure everything is correct.

Than I had to calculate the volume of the washer by using the equation:
(pi x (D1^2 - D2^2) x t)/4

And I basically got stuck there. I'm not sure how to calculate my uncertainties when multiplying by t and so on.

I tried it out by first adding the uncertainties of Dtotal and t (when multiplied) by using this equation:
Dtotal^2 x t +/- Dtotal^2 x t x (change in Dtotal/Dtotal + change in t/t)

That gave me a volume of 5.9 +/- 0.1 cm^3... which I then used to calculate a density of 6.8 +/- 0.1 g/cm^3.

Does this seem correct or what am I doing wrong? I also don't know how to express my answers while doing the calculations; as in, I have random numbers from subtracting my givens (while subtracting my uncertainties) that I don't end up using.

I'm really confused.

2. Sep 22, 2007

### Mindscrape

How do you plus or minus something? Dtotal^2 x t +/- Dtotal^2?

Do you know how partial derivatives work? You want to use

$$\left( \sum_{i=1}^n \left( \frac{\partial f}{\partial x_i} \Delta x_i \right)^2 \right)^{1/2}$$

where f is the function you are evaluating, in this case the volume, the x_i's are each variable (d1, d2, ...), and the delta xi's are the uncertainties in each variable.

3. Sep 23, 2007

### kayleech

I'm confused how to even use that equation though!

4. Sep 23, 2007

### kayleech

5. Sep 23, 2007

### Mindscrape

Okay, as long as you got it. They have the total differential a little bit below, I was posting the magnitude of it (because an error vector is worthless). I like taking the magnitude of the total differential because it comes directly from vector calculus, and always works.