Homework help: Uncertainty with negative power

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Jerry Z
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Question:
A distance R is measured to be 3.400 ± 0.007m. What is the absolute uncertainty in R^−2?

Attempted solution:
Relative uncertainty: 2* (0.007/3.4) = 4.11E-3;
R^-2 = 3.4^-2 = 0.0865 m^-2;
Absolute uncertainty = R^-2 * relative = 0.0865 * 4.11E-3 = 3E-4 m^-2;

Any help would be greatly appreciated!

EDIT: instead of 3E-4, the correct rounding should be 4E-4.
 
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Jerry Z said:
Question:
A distance R is measured to be 3.400 ± 0.007m. What is the absolute uncertainty in R^−2?

Attempted solution:
Relative uncertainty: 2* (0.007/3.4) = 4.11E-3;
R^-2 = 3.4^-2 = 0.0865 m^-2;
Absolute uncertainty = R^-2 * relative = 0.0865 * 4.11E-3 = 3E-4 m^-2;

Any help would be greatly appreciated!
Hello @Jerry Z . Welcome to PF !

What's your question?
 
SammyS said:
Hello @Jerry Z . Welcome to PF !

What's your question?
i cannot get the right answer for the question stated
 
SammyS said:
Perhaps there is a problem with significant figures and/or rounding off in intermediate steps.
So I'm solving it correctly?
 
Jerry Z said:
So I'm solving it correctly?
What you did looks reasonable.

The details of getting uncertainties and applying rules for significant figures vary somewhat from book to book, instructor to instructor, discipline to discipline .

I suggest keeping two extra digits (over what's required for sig. figs.) .

OR

Do the entire calculation at one time with no intermediate steps. (Be especially careful of Order of Operations.)Do final rounding at the end to whatever decimal place is required in your situation.
 
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SammyS said:
What you did looks reasonable.

The details of getting uncertainties and applying rules for significant figures vary somewhat from book to book, instructor to instructor, discipline to discipline .

I suggest keeping two extra digits (over what's required for sig. figs.) .

OR

Do the entire calculation at one time with no intermediate steps. (Be especially careful of Order of Operations.)Do final rounding at the end to whatever decimal place is required in your situation.

Thank you so much! It is indeed the rounding in the end that made the difference.
 
SammyS said:
Great !

So, what was the acceptable answer?

Answer is edited in the original text. Thanks again!
 
This is an example of the merits of keeping everything algebraic as long as possible, only plugging in values at the end. In the present case, when at last plugging in numbers, you would have had 2*0.007/3.43. Putting that into my calculator gives 0.000356.
 
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