How Do Derivatives Estimate Measurement Errors?

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SUMMARY

The discussion focuses on estimating measurement errors using derivatives, specifically in the context of a resistor with a nominal value of 5KΩ and a measured value of 4.8KΩ. While a simple percentage calculation yields a -4% error, participants clarify that derivatives can only be applied when a continuous function exists. The conversation emphasizes that derivative-based error estimation requires a theoretical function dependent on continuous physical parameters, distinguishing it from traditional percent error calculations.

PREREQUISITES
  • Understanding of basic calculus concepts, particularly derivatives.
  • Familiarity with measurement error concepts, including percent error.
  • Knowledge of continuous functions in mathematical analysis.
  • Basic principles of electrical resistance and resistor values.
NEXT STEPS
  • Study the application of derivatives in error analysis in physics.
  • Learn about continuous functions and their significance in measurement theory.
  • Explore theoretical models that relate physical parameters to measurement outcomes.
  • Investigate advanced topics in error propagation and estimation techniques.
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Students in physics or engineering, educators teaching measurement theory, and professionals involved in precision measurement and error analysis.

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Homework Statement


I have a resistor which has value written as [itex]5K\Omega[/itex] and I measured the value and it is [itex]4.8K\Omega[/itex]. I want to work out the error.


2. The attempt at a solution
I can use simple percentage calculation to get [itex]\frac{4800-5000}{5000} \times 100 = -4[/itex]% error but I've been told that derivative can also be used to calculate error, how would I go about doing that? :confused: Thanks in advance.
 
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Determining % error using a derivative requires a continuous function that is derivable. Since the data is discrete and only two points are given a derivative in this case cannot be taken.
 
I think you may be confusing percent error with expected error. As chrisk suggests, you need to have a theoretical function that depends continuously on various physical parameters, and then you take the derivative w.r.t. those parameters, and then multiply by the precision of those parameters (basically). This error estimation using derivatives is not a percent error, because it makes no assumption regarding what the result of the measurement should be.
 

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