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Old Jan 26th 2009, 12:31 AM   #1
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Unhappy error analysis

the question is: when a body moves with velocity V1 for first half of the journey and V2 for the remaining journey, its average velocity is given by 2/average V = 1/V1 +1/V2. if V1 and V2 are (15 +- 0.5)m/s and (30 +- 0.1)m/s, find the average velocity within error limits.
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Old Jan 26th 2009, 12:39 AM   #2
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Originally Posted by the curious View Post
the question is: when a body moves with velocity V1 for first half of the journey and V2 for the remaining journey, its average velocity is given by 2/average V = 1/V1 +1/V2. if V1 and V2 are (15 +- 0.5)m/s and (30 +- 0.1)m/s, find the average velocity within error limits.
Which method of error propagation were you instructed to use?
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Old Jan 26th 2009, 11:41 AM   #3
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Perhaps your study materials already worked a problem with another quantity that was defined like the average velocity and you can imitate it (resistances in ohms or something like that).

I suppose you could find the extremes of v by trying to solve for the min and max of
$\displaystyle \frac {2 v_1 v_2} {(v_1 + v_2) }$ subject to the conditions that
$\displaystyle 15 - 0.5 \le v_1 \le 15 + 0.5 $
and
$\displaystyle 30 - 05 \le v_2 \le 30 + 0.5 $

Which would be a two variable calculus problem.

That way assumes the 0.5 is an absolute bound. If 0.5 is the standard deviation of a normal distribution then we have a different problem.
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Old Jan 26th 2009, 09:07 PM   #4
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combination of errors using the chain rule.
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Old Jan 26th 2009, 09:40 PM   #5
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That's a good idea, t.c.

$\displaystyle V(v_1,v2) = \frac {2 v_1 v_2}{ v_1 + v_2} $
Compute $\displaystyle dV = \frac { \partial V} {\partial v1 } dv_1 + \frac {\partial V}{ \partial v_2} dv_2 $

Let $\displaystyle dv_1 = 0.5 $ $\displaystyle dv_2 = 0.5 $

Evaluate the partial derivatives at $\displaystyle v_1 = 15 $ $\displaystyle v_2 = 30 $


Call the resulting value of $\displaystyle dV $ the "plus or minus" error in $\displaystyle V $.
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Old Jan 27th 2009, 09:33 PM   #6
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Please explain the step marked as X in the following solution done by my teacher.
Sol:
V = average velocity;
it is given that V = 2V1V2/ V1+V2
therefore V = 2 * 15 * 30 / 15 + 30
= 20 m/s.
Now the error in V
2/V =1/V1 +1/V2
2dV/V^2 = dV1/V1^2 + dV2/V2^2 -------> X
2dV/400 = 0.5/225 + 0.1/900
dV = 0.46
so V = (20 +- 0.4)m/s.
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Old Jan 27th 2009, 10:27 PM   #7
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Newton

Sorry, im busy wd students but let u know, latter if u dont find a sol. Bye
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Old Jan 27th 2009, 11:50 PM   #8
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I think the step is "implicit differentiation". We pretend V, V1 and V2 are all functions of some variable, which I will call x.

In the equation $\displaystyle \frac 2 V = \frac 1 V_1 + \frac 1 V_2 $
Take the derivative of both sides with repsect to x.

The derivative of left side with respect to x is $\displaystyle 2(-\frac 1 V_2) dV $
where $\displaystyle dV $ is $\displaystyle \frac {dV}{dx} $.

The derivative of the right hand side with respect to x is
$\displaystyle (- \frac 1 V_1^2)dV1 + (- \frac 1 V_2^2)dV2 $

where $\displaystyle dV1 = \frac {dV_1}{dx} $ and $\displaystyle dV2 = \frac {dV_2}{dx} $

All the derivatives involve he pattern
$\displaystyle f(x) = \frac 1 {x^n} = x^{-n} $
By the chain rule, the derivative of $\displaystyle f(V(x)) = f'(V(x)) V'(x) $
$\displaystyle =( -n V(x)^{-n-1} )V'(x) = -n \frac 1 {V(x)^{n-1}} V'(x) $
$\displaystyle = -n \frac 1 {V(x)^{n-1}} dV $

So
$\displaystyle 2(-\frac 1 V_2) dV = (- \frac 1 V_1^2)dV1 + (- \frac 1 V_2^2)dV2 $

Multiply both sides by (-1) and you get the teacher's equation.

The we solve that equation for $\displaystyle dV $
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