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

same for dw dw = .05 * w

thus

dA/A = 0.03 + 0.05

Solved

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!

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!

8 Comments

same for dw dw = .05 * w

thus

dA/A = 0.03 + 0.05

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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