Find the partial diameter error of the surface area of cylinder

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SUMMARY

The discussion focuses on calculating the partial diameter error of the surface area of a cylinder using the formula (ΔSA/ΔD) = 2πHΔD. Participants emphasize the importance of correctly applying derivatives and clarifying that D should not be confused with ΔD. The correct formula for partial diameter error is confirmed as Partial Diameter Error = (2πH + πD) * ΔD. Accurate differentiation and careful attention to constants are crucial for obtaining the correct values.

PREREQUISITES
  • Understanding of calculus, specifically differentiation
  • Familiarity with the formula for the surface area of a cylinder
  • Knowledge of the symbols used in calculus (e.g., Δ for change)
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Review the differentiation rules for functions involving constants
  • Study the surface area formula for cylinders and its derivatives
  • Practice problems involving partial derivatives in real-world applications
  • Explore error analysis techniques in calculus
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Students and professionals in mathematics, engineering, and physics who are working with calculus and need to understand error analysis in geometric contexts.

abobik
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Homework Statement
Hello everyone. I need help.
How Can I find partial diameter error of surface area of cylinder if I have diameter absolute error, average diameter and average height. Just write a formula. Please!
Relevant Equations
Using derivative
(ΔSA/ΔD) = 2πHΔD
Something is wrong I guess as I get wrong value.
 
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You didn’t do the derivative correctly. Write the formula more carefully and then take the derivative more carefully. Hint: D is not ##\Delta D## (but also careful with the constants)
 
Cutter Ketch said:
You didn’t do the derivative correctly. Write the formula more carefully and then take the derivative more carefully. Hint: D is not ##\Delta D## (but also careful with the constants)
Thanks! But, actually, I don't really get it.. Now, the formula is Partial Diameter Error = (2πH + πD) * ΔD?
 
abobik said:
Thanks! But, actually, I don't really get it.. Now, the formula is Partial Diameter Error = (2πH + πD) * ΔD?
Start by posting the formula you are using for the area.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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