# Calculating Error in Multiple Independent Variables

1. Jan 9, 2013

### IZlo0110

1. The problem statement, all variables and given/known data

Okay so I was given this question to start calculating error on other problems.

Consider the equation F(x,y,z) = x^4+2y^3+5yz+5. Let the uncertainty in x be represented by the variable dx, the uncertainty in y be represented by the variable dy, and the uncertainty in z be represented by the variable dz.

2. Relevant equations

I was given this equation to work from Δfx(x,y,z) = abs(f(x+Δx,y,z)-f(x,y,z))

3. The attempt at a solution

Okay so I started to use the equation I was given, only substituting dx for Δx. Then I "solved" for Δfx to get Δx*y*z. Only that is not the right answer.

I am not sure how to incorporate the equation and I am unclear on how to proceed with this problem.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 9, 2013

### VantagePoint72

You haven't actually stated the question, you've only given some the parameters for it. Is the question to find the error in F? And what is Δfx? Your notation isn't clear. You might find section 9 of this resource useful, particularly the bit near the end.

3. Jan 9, 2013

### IZlo0110

Okay I apologize, the actual question is to find an algebraic expression for the uncertainty in the function due to the uncertainty in x.

4. Jan 9, 2013

### VantagePoint72

I think you are just making an algebraic error. If you expand out $(x+\Delta x)^4$, you will have many more terms left over after cancellation than just the one you gave. Without seeing more of your work, I can't say where you've gone wrong.

5. Jan 9, 2013

### VantagePoint72

Also, are you supposed to assume the error is small, or arbitrary? It would be very helpful to see the entire question.

6. Jan 9, 2013

### IZlo0110

Okay so here is what I tried to do. Though something is still off and I am not sure what I am doing wrong.

Δfx = abs(f(x+Δx,y,z)-f(x,y,z))
Δfx = abs((x+Δx)yz - x^4-2y^3-5yz-5)
Δfx = abs(xyz+Δxyz-x^4-2y^3-5yz-5)

7. Jan 9, 2013

### VantagePoint72

You've forgotten the power of four in the first term. It's not $(x+\Delta x)$, it's $(x+\Delta x)$.

Edit: and you've forgotten the remaining three terms in $f(x+\Delta x)$, the additional $+2y^3 + 5yz + 5$.

Edit2: And I'm not sure why you're multiplying the first term by $yz$.

8. Jan 9, 2013

### VantagePoint72

From the above, it seems you're confused about what $f(x+\Delta x,y,z)$ means. It means that in your original formula, you need to replace everywhere you see $x$ with $x+\Delta x$.