Homework Help: 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."