# Relative error problem in vector calculus gradient intro

• carstensentyl
In summary, the conversation is about finding the relative error of the formula for the number pi in terms of the perimeter and area of a circle. Two different approaches are discussed, one involving experimental error and the other involving the differential function. The correct answer is found to be 2dL/L - dA/A. The conversation also reveals that the problem on webassign may be bugged.
carstensentyl
1.
(a) Write a formula for the number in terms of the perimeter L and the area A of a circle.

(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

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.

Last edited:
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.

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

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

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?

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

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!

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

## 1. What is relative error in vector calculus gradient?

Relative error in vector calculus gradient refers to the difference between the true value and the approximate value of a gradient, divided by the true value. It is a measure of the accuracy of the approximation.

## 2. How is relative error in vector calculus gradient calculated?

To calculate relative error in vector calculus gradient, subtract the approximate value from the true value, then divide by the true value. The result is then multiplied by 100 to get a percentage.

## 3. Why is relative error important in vector calculus gradient?

Relative error is important in vector calculus gradient because it allows us to quantify the accuracy of our calculations. It helps us determine how close our approximate value is to the true value, and can be used to identify any potential errors in the calculation.

## 4. How can relative error be minimized in vector calculus gradient?

To minimize relative error in vector calculus gradient, it is important to use precise and accurate data in the calculation. Additionally, double-checking calculations and using more advanced methods, such as numerical methods, can also help reduce relative error.

## 5. What are some common sources of error in vector calculus gradient calculations?

Common sources of error in vector calculus gradient calculations include rounding errors, data input errors, and limitations of the calculation method used. It is important to be aware of these potential sources of error and take steps to minimize them in order to obtain more accurate results.

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