Approximate uncertainty in area of circle

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SUMMARY

The discussion focuses on calculating the approximate uncertainty in the area of a circle with a radius of 5.3 * 104 cm using the formula A = πr2. The calculated area is 8.8 * 109 cm2, with an assumed uncertainty of 0.1 * 104 cm for the radius. The user initially miscalculated the uncertainty in area due to a rounding error, ultimately correcting it to 3 * 108 cm2 for the uncertainty in area.

PREREQUISITES
  • Understanding of the formula for the area of a circle (A = πr2)
  • Knowledge of significant figures and their application in calculations
  • Basic calculus concepts, specifically differentiation (dA/dr)
  • Familiarity with scientific notation
NEXT STEPS
  • Review the principles of significant figures in calculations
  • Study the relationship between differential calculus and area calculations
  • Explore error propagation techniques in measurements
  • Learn about the implications of rounding errors in scientific computations
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Students in mathematics or physics, educators teaching geometry and calculus, and anyone interested in precision in scientific calculations.

chops369
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Homework Statement


What is the approximate uncertainty in the area of a circle of radius 5.3 * 104 cm? Express your answer using one significant figure.


Homework Equations


A = pi*r2


The Attempt at a Solution


Using the given radius, I found the area to be 8.8 * 109 cm2.

And since the uncertainty is not given, I'm assuming that it's 0.1 * 104 cm.

Using this, the upper limit for the radius is 5.4 * 104 cm, which makes the upper limit for the area 9.2 * 109 cm2.

Subtracting 9.2 - 8.8 = 0.4 * 109 cm2. But this is apparently not the correct answer, and I can't figure out why. Unless I'm forgetting some fundamental aspect of significant figures, the only other way I can express this in one sig fig is to write out the actual number, i.e. 400000000; but this is also incorrect.

What am I doing wrong?
 
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If the area is A = pi*r2, then the difference in area is delta-A (sorry, I forgot all my latex). Can you relate delta-A to delta-r (the radius)? If you've taken calculus, then instead of delta-A it would be dA/dr.
 
I just figured it out.

Turns out it was a rounding error on my part, so what I rounded up to 4 * 108 should have actually been rounded down to 3 * 108.
 
chops369 said:
Using this, the upper limit for the radius is 5.4 * 104 cm, which makes the upper limit for the area 9.2 * 109 cm2.

That's the upper bound, now what's the lower bound?
 

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