Isolating a Variable in a Tricky Place, or Raised to a PowerDate: 10/01/2010 at 16:50:25 From: Trevor Subject: Isolating Variables Well, I am trying to solve for specific variables in equations like this: x = y + v(t) + 0.5(a)(t)^2 I can solve this equation for several of the variables -- but there are multiple t's to isolate, and a is in a very complicated place. I've tried to isolate t by moving y and 0.5(a)(t)^2 to the opposite side via subtraction. But that left me with terms in t on both sides of the equality. I tried moving certain terms back to the other side of the problem to isolate t, but even then I had more than one t. It seemed like a Rubik's Cube, the way I would move one part only to mess up another. I would love some help on how to do this. Thanks! Date: 10/03/2010 at 10:57:39 From: Doctor Ian Subject: Re: Isolating Variables Hi Trevor, To isolate a, you would isolate 0.5(a)t^2, and then divide both sides by (0.5t^2). So that should be straightforward. Substitution can be very handy when a variable "is in a very complicated place." That is, you can write x = y + v(t) + 0.5(a)(t)^2 as this = that + other*a this = x that = y + v(t) other = 0.5*t^2 The only restriction here is that a shouldn't appear in any of the expressions for which you're substituting. Once you have that, isolating a is pretty easy: this - that = other*a this - that ----------- = a other Then you can undo the substitutions: [x] - [y + v(t)] ---------------- = a [0.5*t^2] I like to wrap everything in ()'s or []'s to avoid any possible confusion about order of operations. But as far as isolating 'multiple t's,' the difficulty is not so much that that variable appears twice. For example, your equation has just as many t's as this one does ... x = at + bt ... and you could readily combine those terms: x = (a + b)t x/(a + b) = t The difficulty is that this isn't a linear equation, so you really are looking for a different kind of solution -- not one where you isolate the variable, but one where you find a product of binomials, e.g., 0 = t^2 + 5t + 6 = (t + 2)(t + 3) True when t = -2 or -3 Do you recognize x = y + v(t) + 0.5(a)(t)^2 as a quadratic equation in t? In the general case, there are no formulas for isolating a variable in a polynomial -- but there are for quartics, cubics, or quadratics such as yours. Isolating t in this context is the same as using the quadratic formula. You can just apply that directly, or you can look in our archives to see how the formula is derived, and apply those same steps one by one. Does it all make more sense now? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 10/14/2010 at 16:18:12 From: Trevor Subject: Thank you (Isolating Variables) Thank you so much, that is exactly what I needed to know. You explained it perfectly and I actually understand it now. Thanks! |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/