Calculating Uncertainty on (R^2 + z^2) ^3/2

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In summary, the method for calculating uncertainty in a multi-variable function such as (R^2 + z^2)^(3/2) is to take the absolute value of the partial derivatives with respect to each variable, multiply them by the corresponding uncertainty, and then add them together. This applies to any multi-variable function, including those involving trigonometric functions. It is important to note that the uncertainty should be represented as a positive value, and the derivative of a trigonometric function would be multiplied by the uncertainty in that variable, not the differential.
  • #1
Gogsey
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Can someone tell me how to claculate the uncretinty on

(R^2 + z^2) ^3/2
 
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  • #2
Assuming R and z are independent variables, you do it the same way you calculate the uncertainty in any multi variable function:

[tex]\Delta f(R,z) \approx \left| \frac{\partial f}{\partial R} \right| \Delta R+\left| \frac{\partial f}{\partial z} \right| \Delta z[/tex]
 
  • #3
Isn't that just for (z^2 + R^2)?
 
  • #4
Also if you have a cos/sin function in an equation(multiplying) how would you do it for cos/sin?

is it just for cos x = sinxdx/cosx and cosxdx/sinx for sin x?
 
  • #5
Why would it apply to one multi variable function, but not the others?
 
  • #6
So would I just take the uncertainty on z^2 + R^2, then multiply in by 3/2 and squareroot z^2 + R^2?
 
  • #7
Gogsey said:
Also if you have a cos/sin function in an equation(multiplying) how would you do it for cos/sin?

is it just for cos x = sinxdx/cosx and cosxdx/sinx for sin x?

Do you know how to take derivatives? And what happened to the absolute value brackets?...Errors are never negative.

[tex]\left| \frac{\partial}{\partial x} (\cos x) \right|\Delta x= |-\sin(x)|\Delta x=|\sin(x)|\Delta x \neq \frac{\sin x}{\cos x}dx[/tex]

Also, [itex]\Delta x[/itex] is the uncertainty in x, not the differential 'dx'.
 
  • #8
Gogsey said:
So would I just take the uncertainty on z^2 + R^2, then multiply in by 3/2 and squareroot z^2 + R^2?

No, you would compute the absolute value of the partial derivative of [itex](R^2+z^2)^{3/2}[/itex] with respect to R and multiply it by the uncertainty in R; then do the same with respect to z, and then add them together...
 
  • #9
Ok that's for cos x, but I need (unceratinty in cosx)/cosx. The relative over the value.
 

1. How do you calculate uncertainty on (R^2 + z^2) ^3/2?

The uncertainty on (R^2 + z^2) ^3/2 can be calculated using the following formula: Δ(R^2 + z^2) ^3/2 = (3(R^2 + z^2) ^2/2) * (ΔR/R + Δz/z). This formula takes into account the uncertainties in both R and z.

2. What is the significance of calculating uncertainty on (R^2 + z^2) ^3/2?

Calculating uncertainty on (R^2 + z^2) ^3/2 helps to determine the confidence level of the calculated value. It provides a range of values within which the true value is likely to fall.

3. Can you provide an example of calculating uncertainty on (R^2 + z^2) ^3/2?

Sure! Let's say we have R = 5 cm and z = 3 cm, with uncertainties of ΔR = 0.1 cm and Δz = 0.05 cm. Plugging these values into the formula, we get Δ(R^2 + z^2) ^3/2 = (3(5^2 + 3^2) ^2/2) * (0.1/5 + 0.05/3) = 0.9 cm. This means that the true value is likely to be within 0.9 cm of the calculated value.

4. How does uncertainty on (R^2 + z^2) ^3/2 affect the overall calculation?

The uncertainty on (R^2 + z^2) ^3/2 can affect the overall calculation by increasing the uncertainty in the final result. This is because any uncertainties in the initial values (R and z) will be propagated through to the final result.

5. Are there any assumptions made when calculating uncertainty on (R^2 + z^2) ^3/2?

Yes, there are a few assumptions made when calculating uncertainty on (R^2 + z^2) ^3/2. These include assuming that the uncertainties in R and z are independent and that the uncertainties are normally distributed. It is also assumed that the values of R and z are known with a certain level of accuracy.

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