Finding the Error in z for q = -0.6 \pm 10\% and z = 0.2

[Kyrios]

Homework Statement

For the equation [tex] q = \frac{z(z+2) - 2DH}{z^2} [/tex] [itex] q = -0.6 \pm 10\% [/itex] , and z = 0.2.
D and H are known exactly.
I have to find the error in z that will give an answer of [itex] q = -0.6 \pm 10\% [/itex]

Homework Equations

The Attempt at a Solution

I have considered rewriting the equation in terms of z, which gives
[tex] z = \frac{1 \pm \sqrt{1-2DH(q-1)}}{q-1} [/tex]
but I'm not sure where to go with that, how the plus/minus affects it, and what to do with D and H (if anything).

I've tried to do (error in q * 0.5) / (error in q) which gives an answer for z of [itex] 0.2 \pm 10\% [/itex] but that seems a bit too simple

 

[ehild]

Homework Statement

For the equation [tex] q = \frac{z(z+2) - 2DH}{z^2} [/tex] [itex] q = -0.6 \pm 10\% [/itex] , and z = 0.2.
D and H are known exactly.
I have to find the error in z that will give an answer of [itex] q = -0.6 \pm 10\% [/itex]

Homework Equations

The Attempt at a Solution

I have considered rewriting the equation in terms of z, which gives
[tex] z = \frac{1 \pm \sqrt{1-2DH(q-1)}}{q-1} [/tex]

It is not needed. You know that z=0.2. From that, you can determine DH. You know also q and its error.
The error of z can be obtained by differentiating both sides of the equation
[tex]q = \frac{z(z+2) - 2DH}{z^2}[/tex]
[tex]\Delta q =\frac {dq}{dz}\Delta z[/tex] Substitute z, DH, and ##\Delta q##. Solve for ##\Delta z##.

ehild

 

[Kyrios]

so I get this:

[tex] DH = \frac{z(z+2) - z^2 q}{2} = 0.232 [/tex]
[tex] \frac{dq}{dz} = \frac{4DH - 2z}{z^3} = 66 [/tex]
[tex] \Delta q = \frac{dq}{dz} \Delta z [/tex]
[tex] \Delta z = \frac{0.06}{66} = 9 \times 10^{-4} [/tex]

The error seems really small..?

 

[ehild]

It is almost right. You made some small mistake when calculating DH. Check.

The error is small, but z=0.2, so its relative error is about 0.5 %.

ehild