using differentials to calculate area percentage error
Posted on 2011-10-11
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!