Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Moving Variables in Equations

  1. Mar 2, 2012 #1

    Drakkith

    User Avatar

    Staff: Mentor

    Say we have an equation that we need to find a variable for. How about s=vt+1/2at^2, and we want to solve for v. How do you know what operation to do when you move variables around? My book says to subtract both sides by (at^2)/2 first. Why do you subtract the at^2 instead of divide?
     
  2. jcsd
  3. Mar 2, 2012 #2

    danago

    User Avatar
    Gold Member

    I like to think about it in terms of taking everything away from the right hand side so all that is left is the v. "vt+1/2at^2" is essentially saying:

    1. First multiply v by t
    2. Then add the quantity 0.5at^2

    So the first step in "isolating" the v is to reverse the last thing that happened i.e. subtract 0.5at^2 from both sides to get rid of the 0.5at^2 on the right hand side.

    It may not be of much help to you, but that is how I thought about it when I started algebra.

    I guess another way to think about it is to try division, and see why it won't work. If you instead divide both sides by 0.5at^2 you get:

    [tex]\frac{s}{0.5at^2} = \frac{vt+0.5at^2}{0.5at^2}[/tex]

    Or, after simplifying:

    [tex]\frac{s}{0.5at^2} = \frac{v}{0.5at}+1[/tex]

    It hasn't really helped!
     
  4. Mar 2, 2012 #3

    chiro

    User Avatar
    Science Advisor

    Hey Drakkith.

    I'm assuming you have values for the other variables.

    The basic goal of solving for an unknown variable is get the variable in an explicit form. In other words if our variable we want to find is v, we want to get our variable in the form v = blah: this is called the explicit form.

    If we can't get our value in v = blah where blah is everything else but v which is known, then you will get what is called an implicit equation where you might end up with something like av^2 + v + 2 = 0 or SQRT(v) + v^2 - v = 0 where you then have to solve for v which means you will have to use what is called a root finding algorithm to get possible solutions for v.

    Now since you have an explicit kind of equation, all you have to do is get v in terms of everything else. This requires you to do basic algebra.

    With basic algebra the idea is simple: what you do to one side you must do to the other: that's all there is to it.

    So if you wanted to get rid of the 1/2at^2 term on the RHS you need to subtract the term on the left side. This means that by doing this you get:

    s - 1/2at^2 = vt.

    The only thing you have to be careful about when doing this kind of thing is to make sure that when you divide both sides by something (including a variable and not just a constant), that the variable is not 0: if you do this you will get into some big problems.

    Other than that you can apply basic algebra on variables in the same way that you do with known constant numbers.
     
  5. Mar 2, 2012 #4

    Drakkith

    User Avatar

    Staff: Mentor

    Thanks guys, but I think I managed to answer my own question. Per here: http://www.algebra-class.com/equations-with-fractions.html, it says:
    So, with s=vt+1/2at^2, the terms on the right side are vt and +1/2at^2, meaning I need to subtract the WHOLE term, not just part of it, and definitely not divide part of it. I honestly don't ever remember learning what a Term was, but it has been a while since I learned math.
     
  6. Mar 2, 2012 #5

    chiro

    User Avatar
    Science Advisor

    You can subtract/add/multiply/divide whatever term you want (as long as you don't divide by zero if its variable (like t) or constant = 0) but the goal is to get your unknown into unknown = blah where blah doesn't have the unknown variable in it.

    Once you realize this, you will get a 'gut feeling' of how to get the unknown = blah equation from the original equation you have been given.

    But yeah basically the rule is whatever you do to one side, you have to do the exact same thing to the other side: no if's or but's.
     
  7. Mar 2, 2012 #6

    Drakkith

    User Avatar

    Staff: Mentor

    What? Why would I be able to do any operation I want? Did I misunderstand you?

    Of course. I know that part, I was confused about why I had to subtract the at^2 instead of divide. Now that I know it's all one term it makes much more sense.
     
  8. Mar 2, 2012 #7

    chiro

    User Avatar
    Science Advisor

    That was just a general comment for algebra problems and not specifically for your problem that was posted: just a tip to help you for you future problems.

    Again the important thing is that for algebraic equations, you can't divide by zero so you have to check that your variable that you are dividing by if you ever divide by that variable (or term: you can divide by a term that has many variables in it) is never zero: if it is zero then you can't do it: there is a whole other discussion on why but what it boils down to is that you get answers that 'don't make sense'.


    Of course. I know that part, I was confused about why I had to subtract the at^2 instead of divide. Now that I know it's all one term it makes much more sense.[/QUOTE]

    Well the goal is to get an expression in terms of v.

    Basically you have an equation of s = vt + blah and in order to get vt in terms of something else you have to - blah from both sides giving you s - blah = vt.

    If you divided blah on both sides you would instead get:

    s/blah = vt/blah + 1 (assuming blah is not zero).

    This has not separated blah from v which is what you are setting out to do.

    So basically if you are trying to solve for something (its v in this case but it could be whatever), you want to try and get v by itself on one side for these kinds of problems. (Later if you pursue further math you'll find that you will get an equation in terms of v but don't worry about that right now).

    Once you do a few of these kinds of problems you'll see why you have to do what you have to do and see why another operation is not going to help you solve the problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Moving Variables in Equations
  1. 4 variable equation (Replies: 4)

Loading...