Maximum eror in calculated surface area

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SUMMARY

The discussion focuses on calculating the maximum error in the surface area of a sphere based on a measured circumference of 73.000 cm with a possible error of 0.50000 cm. The correct approach involves using the formula for surface area (SA=4πr²) and the relationship between circumference and radius (c=2πr). The user initially miscalculated by attempting to find the error in volume instead of surface area, highlighting the importance of applying the correct equations in error analysis.

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  • Understanding of linear approximation in calculus
  • Familiarity with the formulas for circumference and surface area of a sphere
  • Basic knowledge of differentiation and error analysis
  • Ability to manipulate algebraic equations
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  • Learn about error propagation techniques in measurements
  • Explore the relationship between volume and surface area in geometric shapes
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Students in mathematics or physics courses, educators teaching calculus or geometry, and anyone interested in understanding error analysis in measurements.

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Homework Statement


The circumference of a sphere was measured to be 73.000 cm with a possible error of 0.50000 cm. Use linear approximation to estimate the maximum error in the calculated surface area

Homework Equations


SA=4\pi\(r^2 Eq1.
dV=8\pi\(rdr Eq2.
c=2\pi\(r Eq3.

The Attempt at a Solution


I solved for the radius by r=\frac{73}{2\pi}
I then plugged r into Eq2, I set dr= \frac{.5}{2\pi}
I get something like 134.98 and it is wrong.
 
Last edited:
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Never mind, I figured it out.
 
You are calculating the error in the measured volume - not in the measured surface area.
 

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