# Approximating uncertainty

1. Sep 1, 2004

### physicsss

Can someone get me started with finding the approximate uncertainty of the area of a circle with a radius of 2.0 cm?

Thanks,
Yours

2. Sep 1, 2004

### Tide

Ask yourself by what amount will the area of the circle change if you introduced a small error in the radius of the circle?

3. Sep 2, 2004

### physicsss

um...

Can you give me the answer so I know if I'm doing this right? (I need to show work for full credit anyway)

4. Sep 2, 2004

### Tide

Tell us what you did so far and I'm sure someone will provide further guidance! :-)

5. Sep 2, 2004

### physicsss

I found the area of both pi*(2.0+0.1)^2 and pi*(2.0-0.1)^2, which is 13.9 and 11.3, respectively. I also found the area of pi*(2.0)^2, which is 12.6. Then I got stuck.

6. Sep 2, 2004

### Tide

That's a good start! Generally, the error in area will be related to the error in measuring the radius by $\delta A = 2\pi r \delta r[/tex] which is essentially what you have for the specific case that [itex]\delta r = 0.1$ and $r = 2$. ($\delta$ means "error" or "deviation.")

7. Sep 2, 2004

### physicsss

So the final answer is 2.8 X 10^9 +/- 1.3?

Last edited: Sep 2, 2004
8. Sep 2, 2004

### Tide

I would refine it just a little. The radius was given to two significant digits meaning that it is given to the nearest 0.1 centimeters from which I would use $\pm 0.5$ for the error in r.

9. Sep 2, 2004

### physicsss

Thank you for your help, Tide!

10. Sep 2, 2004

I'm being asked to compute the uncertantity of air density that i calculated using pv=nrt. I have a uncertanty of .1 on the barometer, and .5 degrees for temperature, but the equation in my fluid mechanics book is too complicated. I can't draw what they're trying to say. The formula has to do with like 3 terms squared to the ^1/2 power. I just can't get over all the complicated lingo, I know its really easy. My recitation teacher explained it, but I'm having a brain freeze, anybody?

11. Sep 2, 2004

### Chronos

It appears you should be using the formula
$$D =\frac{P}{RT}$$
where:
D = density, kg/m3
P = pressure, Pascals ( multiply mb by 100 to get Pascals)
R = gas constant , J/(kg*degK) = 287.05 for dry air
T = temperature, degK = deg C + 273.15

12. Sep 2, 2004

### Tide

I don't know whether I can make it any easier but essentially you just need to substitute numbers into the equation

$$\frac {\delta n}{n} = \sqrt{ \left( \frac{\delta v}{v} \right)^2 + \left( \frac{\delta p}{p} \right)^2 + \left( \frac{\delta T}{T} \right)^2}$$

The symbol $\delta$ means "deviation" or "error" of a given quantity. Note that generally a quantity such as $\frac {\delta n}{n}$ is called the "relative error."