Will_Sawin comments on Inverse Speed - Less Wrong

14 Post author: komponisto 27 March 2011 05:57AM

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Comment author: Will_Sawin 18 April 2011 02:11:13AM *  2 points [-]

My understanding has been that if you know how to set up a problem as a formula then you know how to make it out of toothpicks and rubber bands so you understand it.

This requires a certain ability to manipulate formulas:

2xy/(x+y)=2/(1/x+1/y)=1/(((1/x)+(1/y))/2)

as I'm sure you know.

(but I am 1-in-a-million atypical on this)

So like, the formula for concentrations goes like this:

If we add 1volume of xconcentration to 2volume of yconcentration then:

we are adding 1volume of stuff to 2volume of stuff and getting 1volume+2volume of stuff.

We are adding xconcentration x 1volume to yconcentration x 2volume of special stuff and getting xconcentration x 1volume+yconcentration x 2volume of stuff

so our final concentration is the amount of special stuff divided by the amount of stuff, or

(xconcetration x 1volume+ yconcentration x 2volume)/(1volume+2volume)=zconcentration

So, I mean, that's all there is. That's the formula, that's how the problem works, that's all that's at issue.

It's equivalent to the problem:

You spend 2volume time at yconcentration speed. How much time must you spend at xconcentration speed to attain an average speed of zconcentration?