# Homework Help: The approx. uncertainty in r

1. May 4, 2010

### voygehr

1. The problem statement, all variables and given/known data
$$V=\pi r^{2} h$$

The uncertainties in V and h are shown below
V 7%
h 3%

The approximate uncertainty in r is:
A. 10% B. 5% C. 4% D. 2%

2. The attempt at a solution
According to the key the correct answer is B. This answer can be calculated by:
$$r^{2}=\frac{V}{\pi h}$$
which gives 3%+7% = 10% uncertainty
and subsequently
$$r=\sqrt{\frac{V}{\pi h}}$$
giving 10% * 1/2 = 5%

However:
if the value of 5% uncertainty for r is inserted in the original equation it all doesn't make sense:
$$V=\pi r^{2} h$$ should then give an uncertainty for V of 3%+5%+5%=13%

2. May 4, 2010

### rock.freak667

I'd say the answer was 2%, but I can't really explain how your book got to that answer.

$$\frac{dV}{V}=2 \frac{dr}{r}+\frac{dh}{h}$$

3. May 4, 2010

### voygehr

That was my first thought as well. But the key says no.

And as proved in my previous post, the 5% uncertainty does make sense, however if used in the original equation (or by simply reversing the second) it doesn't sum up.

4. Apr 20, 2011

### voygehr

Bump. Any ideas?

5. Apr 20, 2011

### eczeno

you must be clear about what is measured and what is calculated.

if you measure V and h and are wanting to calculate r, then 5% is the correct error. but then your check does not make sense because there you are assuming r and h are measured and V is calculated.

If the question asked what error in r (which is measured) would give rise to a 7% error in V (which is calculated) given a 3% error in h (which is measured), then the answer would be 2%.

hope this helps

6. Apr 20, 2011

### voygehr

Yes, that makes sense. Thanks!