# Relative Error?

1. Sep 14, 2008

### Ashley11

Relative Error!?

If the relative error for the radius of a circle is 5%, what is the relative error for the circumference?

I know that relative error is the ratio of the absolute error in a measurement to the size of the measurement, but I honestly have no idea how to complete the problem. These are one of the pre-lab problems, where we're supposed to try and figure it out before we're taught it, but I'm stumped on this one.

2. Sep 14, 2008

### LowlyPion

Re: Relative Error!?

What is the formula for the circumference of a circle?

If r is off by 5%, what effect will that have on the relative error of the formula for circumference?

3. Sep 14, 2008

### dlgoff

Re: Relative Error!?

Use the equation for finding the circumference from the radius. Since the radius is the measured variable, only error from it will apply.

4. Sep 14, 2008

### Ashley11

Re: Relative Error!?

Well, I came up with two different answers using the circumference formula: 10% and 31.4%. I'm relatively uncertain if I've done this correctly, but I believe 31.4% to possibly be the correct answer. The problem that I'm stuck on seems to be the simplest!

5. Sep 14, 2008

### fluidistic

Re: Relative Error!?

The circumference of a circle is $$2\pi r$$. As dlgoff said, we assume there is only an error in $$r$$, so what's the error of $$2\pi r$$?
EDIT: I think you got it right, the error would be $$2\pi \cdot 5$$% which seems what you did.

6. Sep 14, 2008

### Ashley11

Re: Relative Error!?

Wait...since circumference is proportional to radius, would that make the relative error of the circumference 5%, as well?

7. Sep 14, 2008

### LowlyPion

Re: Relative Error!?

Try it this way. Think about what the relative error means.

Doesn't it mean that the measured value can be as great as 5%?

That (r+.05r)/r yields an error of 5%

Now substitute that into the equation for the circumference.

C = 2πr

Now on the one hand the absolute error will be 2π greater, but then so would the nominal absolute result.

In calculating your RELATIVE error you compare then the Error Result with the magnitude of the correct result.

In that case it becomes (2π*(r+.05))/2π*r After simplifying though what are we left with? (r+.05)/r? And was that our original error in r?

So what does that say about multiplying Relative errors by a constant?

8. Sep 14, 2008

### LowlyPion

Re: Relative Error!?

There you go you beat me to it.

Good thinking.

9. Sep 14, 2008

### Ashley11

Re: Relative Error!?

Great! Thanks a lot for everyone's help! Great, GREATLY appreciated :)