Relative error problem in vector calculus gradient intro

[carstensentyl]

1.
(a) Write a formula for the number in terms of the perimeter L and the area A of a circle.

(b) Write the differential for your answer in part (a).

(c) Suppose that L and A are determined experimentally. Write the resulting relative error in using your answer in part (b).

3.
a)pi(A,L)=\frac{L^{2}}{4A}
b)dpi=\frac{L}{2A}dL-\frac{L^{2}}{4A^{2}}dA
c)

I've tried several different values for part c, but can't seem to find a convincing answer. I know that the equation begins with dpi/pi and each part has some value with a coefficient of dL/L + dA/A

[EnumaElish]

I can think of two approaches:

1. Suppose experimenting introduces measurement error into each of L and A. Let L* = L + dL and A* = A + dA be the measured quantities of each of L and A respectively (dL and dA are the absolute errors; dL/L and dA/A are the relative errors). Substitute into dpi(A*,L*)/pi(A*,L*).

2. Write dpi = (L^2/2A)dL/L + (L^2/4A)dA/A; divide through by pi = L^2/4A to obtain dpi/pi.

[carstensentyl]

I made a boo boo and seem to have done something wrong on the webassign program. But thank you for the help. I understand the concept a little better now.

[moose]

Haha, I have the same homework problem on webassign right now. I was a little lost as well, crazy that I found this here.

[carstensentyl]

Turns out that the problem is bugged. Even the teacher couldn't get the right answer.

[moose]

Do you also go to UA or is this a common vector calc problem that happened to be assigned at nearly the same exact time lol?

[-EquinoX-]

So what is the correct answer to this question ? I am also doing the same problem and didn't get the answer

[shuasmith]

I go to UA and this problem gave me so much grief. However, I found the right answer!
so d(pi)/pi is the same as taking the differential function and dividing by the original:
d(pi)/pi=(L/(2A)*dL-L^2/(4A^2)*dA)/(L^2/(4A^2)
Simplify, and this gives you:
2dL/L-1dA/A

Hope no one else has to struggle like I did!

[nkrenek]

After coming down to my last submission you're post was extremely helpful! Also a UA student here.