How do you linearize this equation (terminal velocity in fluid)?

  • Thread starter Ryker
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  • #1
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Homework Statement


Linearize the following equation:

[tex]\dfrac{1}{r} = (\dfrac{2g(\rho - \rho\prime)}{9\eta})(\dfrac{r}{v_{0}}) - \dfrac{K}{R}[/tex]

Here, v0 is the observed velocity dependent upon r, and we are trying to get η and K somewhere in the intercept and slope. So we will try out different r's, calculate the observed velocity and then plot the graph.

I don't see what we are supposed to do to get a linear graph. Take some kind of logarithm?

The Attempt at a Solution


Well, the equation itself doesn't look linear to me, but I don't see what we are supposed to do to get a linear graph. Rearranging it won't do the trick, so do we need to take the logarithm?
 

Answers and Replies

  • #2
gneill
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If you start by assigning new constants a and b, the expression looks something like

1/r = a*(r/vo) - b

which rearranges to

vo = a*r2/(1 + r*b)

Now, believe it or not, this will produce a a plot that's very close to linear except when the denominator misbehaves around r*b = -1. That is, when (K/R)*r is near -1. If your r doesn't go near there, you may be good to go...
 
  • #3
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Hmm, the problem, though, is that this way there is no intercept, but two elements in the slope, which I'd like to measure (one in a, one in b). Is there another way perhaps?
 
  • #4
gneill
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You can always "build" a line with a slope that is equal to the slope of the real curve taken far from the "interesting" bit, passing through a given point on that real curve (so, point slope form for the line, which can be translated to slope intercept form).

The slope of the "real" curve is given by the derivative...
 
  • #5
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I guess this would require implicit differentiation then, but the thing is this is for a lab, and we've never been required to go about this in any such "complicated" (for the lack of a better term) way. I'm pretty sure there is a catch to it, which doesn't include taking derivatives, as we also have to show what corresponds to the slope, what to the intercept. Or perhaps there's no catch at all, and we can just plot this equation by taking the aforementioned variables. But I just can't see how that would produce a linear graph that we could analyze. Any additional thoughts on that perhaps? :smile:
 
  • #6
gneill
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Using the constants a and b that I introduced earlier,

[tex]a = \frac{2g(\rho - \rho')}{9 \eta}[/tex]
[tex]b = \frac{K}{R}[/tex]

then
[tex]v_o = \frac{a r^2}{1 + b r}[/tex]

Taking the derivative w.r.t. r:

[tex]\frac{dv_o}{dr} = \frac{a r (2 + b r)}{(1 + b r)^2}[/tex]

When r becomes relatively large, this has a limit:

[tex]M = \lim_{\substack{x\rightarrow \infty}} \frac{a r (2 + b r)}{(1 + b r)^2} = \frac{a}{b}[/tex]

Now, I don't know what is a reasonable maximum or minimum value for your r variable (or for that matter the other variables involved). But if you can choose a suitable large value for r to find a point (r, vo) that the line passes through (and maybe it can be chosen in terms of the given constants a and b). then it should be possible to fix intercepts that depend upon those constants.
 
  • #7
ehild
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Your equation is of the form 1/r = A*r/v0 -B. Plot 1/r in terms of r/v0. You should get a straight line, the intercept with the vertical axis is at -B and the slope is A.

ehild
 
  • #8
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Your equation is of the form 1/r = A*r/v0 -B. Plot 1/r in terms of r/v0. You should get a straight line, the intercept with the vertical axis is at -B and the slope is A.

ehild
In the end, that was exactly what we did, so I guess I was just overthinking it :smile:
 

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