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March 9th, 2019, 10:03 PM  #1 
Senior Member Joined: Apr 2008 Posts: 194 Thanks: 3  percentage error related to differentials
I am a math tutor. One of my students asked me to help him with a few problems. There was a question I could not get the right answer for. (ex) The length x of the side of a square is measured less than 1% error. By approximately what percentage will the calculated area A=x^2 of the square be in error? My attempt We are given dx = 0.01 . The differential is dA = 2x dx = 0.02 x . My answer is wrong. The correct answer is dA = 0.02 x^2 . I don't understand why the differential in the correct answer has x squared instead of just x. Could someone explain how to get the x squared? Thanks a lot. 
March 10th, 2019, 03:27 AM  #2 
Newbie Joined: Feb 2019 From: United Kingdom Posts: 15 Thanks: 3 
Because you’ve substituted one percent of a unit. Substitute 0.01x for the increment dx

March 10th, 2019, 05:59 AM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
 
March 11th, 2019, 07:03 PM  #4 
Senior Member Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications 
Hmm, I get 2% for "By approximately what percentage will the calculated area A=x^2 of the square be in error?" (my bold), and I get $\Delta A \approx 0.02x^2$. I'm not sure which answer is the desired one, nor whether the OP's answer (along with the one from SDK) is correct, for that matter. Let the error of the side be +1%, giving $l$: $\displaystyle \large {l=(1+0.01)x \\ A=l^2 \\ A=(1+0.01)^2x^2 \\ A=(1+0.02+0.0001)x^2 \\ \Delta A \approx 0.02 x^2 \\ \text{% error of A} \approx \frac{0.02x^2}{x^2} \cdot 100 \approx \text{2%} }$ Numerical examples: $\displaystyle \large{ \begin{array} {cccccc} x & x^2 & 1.01x &A(1.01x) & \Delta A & 0.02x^2& \frac{\Delta A}{x^2} \\ 1.0 & 1.0 & 1.01 & 1.0201 & \approx 0.02 & 0.02 & \approx 0.02 \\ 2.0 & 4.0 & 2.02 & 4.0804 & \approx 0.08 & 0.08 & \approx 0.02 \\ 3.0 & 9.0 & 3.03 & 9.1809 & \approx 0.18 & 0.18 & \approx 0.02 \end{array} }$ 
March 12th, 2019, 05:00 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
That's correct. The calculated area (if the measured length of the side of the square has an error of less than 1%) will have an error of less than approximately 2%.


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