Frustrated by simple algebra problem

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The discussion revolves around expressing the length of a metal rod, L, as a linear function of temperature, t, using the formula L - L0 = aL0(t - t0). Participants explore different approaches to identify the slope and y-intercept while grappling with the roles of constants L0 and t0. Clarification is provided that L0 and t0 are constants representing initial conditions and should not be ignored in the equation. The correct interpretation of the y-intercept and slope is debated, with emphasis on maintaining the integrity of the original formula. Ultimately, the challenge lies in understanding how to manipulate the equation to fit the linear form y = mx + b while recognizing the nature of the constants involved.
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Homework Statement


For small changes in temperature, the formula for the expansion of a metal rod under a change of temperature is:
L-Lo = aLo(t - to)
L= length at temp. t
Lo= initial length at temp. to
a= constant that depends on metal
A) express L as a linear function of t. Find the slope and y intercept(hint: treat the other quantities as constants.)


Homework Equations


Y=mx+b
Y-Yo=m(x-xo)



The Attempt at a Solution


Solution 1?: the slope is aLo as stated by the equation, and the y intercept is Lo...
Distribute
L= aLot-aLoto+Lo
L=Lo(at-ato+1)
Doesnt work, still has variable to, lo, etc.

Solution 2?: assume a=1
L=Lo(t-to+1)
Slope= Lo? Y intercept also lo?

Solution 3: Lo, a and to=0
L=0... doesn't work

Or, pretend all constants=1
L-1=1(t-1)
L=t
Seems too specific.

I don't really know from which other angle to tackle this problem, I would appreciate some ideas, although I can imagine the answer should be obvious to me. I don't know how to turn this fully into y=mx+b. Thanks in advance.
 
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Are to, lo variables? Or are they constants which look like variables?
 
SteamKing said:
Are to, lo variables? Or are they constants which look like variables?

They are constants i suppose, because they are the initial condition which doesn't change. But i don't understand what "treat them as constants" means... does it mean i could just ignore them?

Tried another way..
From L=aLot-aLoto+lo
Y intercept= aLoto+lo
Slope=aLo
 
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aerospacedout said:
They are constants i suppose, because they are the initial condition which doesn't change. But i don't understand what "treat them as constants" means... does it mean i could just ignore them?

Tried another way..
From L=aLot-aLoto+lo
Y intercept= aLoto+lo
Slope=aLo
That y intercept is -aL0t0 + L0 .

Either way might be correct.

The initial result is correct if the y-axis is at t = t0 .

The result is the post quoted here is correct if the y-axis is at t = 0 .

The slope is the same in both cases.

The result you have in your Original Post makes more sense to me.
 
aerospacedout said:
They are constants i suppose, because they are the initial condition which doesn't change. But i don't understand what "treat them as constants" means... does it mean i could just ignore them?

Tried another way..
From L=aLot-aLoto+lo
Y intercept= aLoto+lo
Slope=aLo

Your final answer here is not quite correct. If we use 'intercept' to have its normal meaning, then you need Intercept = L_0 - a L_0 t_0 (not L_0 + a L_0 t_0 as you wrote).

"Treating them as constants" means that they are not the variables in this problem, but certainly you cannot ignore them. The only "variables" here are L and t. In a specific problem the parameters L_0 and t_0 would be given some numerical values, but that would still allow L and t to vary.
 
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