using differentials to calculate area percentage error
The length and the width of a rectangle are measured with error at most 3% and 5% respectively. Use differentials to approximate percentage error in calculating area.
I got stuck after I got to:
A = L*W
dA = L*dw + w*dL, where dL = 0.03, and dw = 0.05, I'm not sure how much further this problem can go without some more numbers to plug into the variables. Unless I'm missing something, wouldn't this be the "final answer"?
I asked my professor and she was saying to divide through by A, but that does not change anything as you will still end up with dA/(L*w) = 0.05/w + 0.03/L... Any guidance is much appreciated it. Kind of need this answered within the next couple of hours, so sooner the btter. Thanks!
but dL does not equal .03 the .03 is the percentage dL = .03 * L
same for dw dw = .05 * w
thus
dA/A = 0.03 + 0.05
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ZenotureAuthor Commented:
if that is the case, then maybe my understanding of the problem is flawed. After reading the problem, I came up with the equation by differentiating sides and coming up with what I posted above. Why are we assuming dL = 0.03*L and dw = 0.05*w?
The problem did not say that dL = 0.03. The problem said that the error percentage was 3 percent
Therefore the value of the error itself is 0.03 * L ie dL = 0.03 * L as I said
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your differentiation is all right. It is just that there is a difference between a percent and the value itself
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ZenotureAuthor Commented:
Yes, I understand my differential is right, but I'm trying to see the physical meaning behind it if that makes any sense. I know how to do it mathematically/mechanically, but I wanted to know what the thinking process behind it was to get to the answer you did.
the dL is a small change in the L. If the real value of L was a bit off from the measured value, it would change the area by a little bit (a dL) Your possible L might be off by 3%. That means that your dA might be a bit off too. You calculated the possible bit off.
So did you end up with the area having a maximum of 8.15% error?
1.03*1.05 = 1.0815
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same for dw dw = .05 * w
thus
dA/A = 0.03 + 0.05