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March 9th, 2019, 10:03 PM   #1
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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.
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March 10th, 2019, 03:27 AM   #2
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Because you’ve substituted one percent of a unit. Substitute 0.01x for the increment dx
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March 10th, 2019, 05:59 AM   #3
SDK
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Quote:
Originally Posted by davedave View Post
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.
Your answer is correct. The other answer is a typo.
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March 11th, 2019, 07:03 PM   #4
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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:

$\displaystyle \large{
\begin{array} {cccccc}
x & x^2 & 1.01x &A(1.01x) & \Delta A & 0.02x^2& \frac{\Delta A}{x^2} \\
1.0 & 1.0 & 1.01 & 1.0201 & \approx 0.02 & 0.02 & \approx 0.02 \\
2.0 & 4.0 & 2.02 & 4.0804 & \approx 0.08 & 0.08 & \approx 0.02 \\
3.0 & 9.0 & 3.03 & 9.1809 & \approx 0.18 & 0.18 & \approx 0.02
\end{array}
}$
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March 12th, 2019, 05:00 AM   #5
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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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