Find the partial diameter error of the surface area of cylinder

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The discussion focuses on calculating the partial diameter error of a cylinder's surface area, with the formula (ΔSA/ΔD) = 2πHΔD being questioned for accuracy. Participants emphasize the need for careful differentiation and proper notation, particularly regarding the distinction between D and ΔD. One user suggests revising the formula to Partial Diameter Error = (2πH + πD) * ΔD, seeking clarification on its correctness. The conversation highlights the importance of precise mathematical derivation in achieving accurate results. Overall, the thread underscores the necessity of careful formula application and differentiation in mathematical calculations related to geometry.
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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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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