# 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!
###### Who is Participating?
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Commented:
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
0

Experts Exchange Solution brought to you by

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Author 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?
0
Commented:
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
0
Commented:
your differentiation is all right.  It is just that there is a difference between a percent and the value itself
0
Author 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.
0
Commented:
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.
0
Author Commented:
ah, ok, I get it now. Thanks!
0
Commented:
So did you end up with the area having a maximum of 8.15% error?

1.03*1.05 = 1.0815
0
###### It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
Math / Science

From novice to tech pro — start learning today.

Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.