# Desperate Help Needed

1. Mar 9, 2009

### Gogsey

Can someone tell me how to claculate the uncretinty on

(R^2 + z^2) ^3/2

2. Mar 9, 2009

### gabbagabbahey

Assuming R and z are independent variables, you do it the same way you calculate the uncertainty in any multi variable function:

$$\Delta f(R,z) \approx \left| \frac{\partial f}{\partial R} \right| \Delta R+\left| \frac{\partial f}{\partial z} \right| \Delta z$$

3. Mar 9, 2009

### Gogsey

Isn't that just for (z^2 + R^2)?

4. Mar 9, 2009

### Gogsey

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. Mar 9, 2009

### gabbagabbahey

Why would it apply to one multi variable function, but not the others?

6. Mar 9, 2009

### Gogsey

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. Mar 9, 2009

### gabbagabbahey

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

$$\left| \frac{\partial}{\partial x} (\cos x) \right|\Delta x= |-\sin(x)|\Delta x=|\sin(x)|\Delta x \neq \frac{\sin x}{\cos x}dx$$

Also, $\Delta x$ is the uncertainty in x, not the differential 'dx'.

8. Mar 9, 2009

### gabbagabbahey

No, you would compute the absolute value of the partial derivative of $(R^2+z^2)^{3/2}$ 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. Mar 9, 2009

### Gogsey

Ok thats for cos x, but I need (unceratinty in cosx)/cosx. The relative over the value.