# Frustrated by simple algebra problem

1. Feb 22, 2014

### aerospacedout

1. The problem statement, all variables and given/known data
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.)

2. Relevant equations
Y=mx+b
Y-Yo=m(x-xo)

3. 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... doesnt 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.

2. Feb 22, 2014

### SteamKing

Staff Emeritus
Are to, lo variables? Or are they constants which look like variables?

3. Feb 22, 2014

### aerospacedout

They are constants i suppose, because they are the initial condition which doesnt change. But i dont 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

Last edited: Feb 22, 2014
4. Feb 22, 2014

### SammyS

Staff Emeritus
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.

5. Feb 22, 2014

### Ray Vickson

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.

Last edited: Feb 22, 2014