Re: Solving f(t) + C*cos(g(t)) for t
From: Dave Rusin (rusin_at_vesuvius.math.niu.edu)
Date: 03/24/05
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Date: 24 Mar 2005 17:57:01 GMT
In article <1111460198.211154.286390@o13g2000cwo.googlegroups.com>,
Trinition <trin@one.net> wrote:
>I've come up with a formula where I can tell the position of a point on
>the edge of a spinning, moving mass:
>
>x(t) = f(t) + C*cos(g(t))
>
>Where:
> f(t) is the position of the center of mass (quadratic of t)
> g(t) is the orientation of the vector from the center of mass to
>the point (quadratic of t)
>
>Now, what I'd like to do is rearrange this formula to solve for t,
Your ability to do that will depend on
-- the form of the functions f and g
-- what you will accept as an answer.
If for example g(t) or g(t) is a constant, then you can rearrange
terms pretty easily to solve for t. There is some hope that you
can do something if f itself is a function which is a trig function
in some natural way. Otherwise you might as well forget it, if you
want an algebraic answer.
On a more general level, you could invoke the Inverse Function Theorem:
for any functon F of one variable t (such as the entire right-hand
side of your equation) the equation x = F(t) can be interpreted as
a description of a curve in the (t,x) plane. Flip the picture
around the line x = t and it becomes a curve in the (x,t) plane,
which you could now interpret as the graph of some function t = G(x).
Well, or maybe not. (The graphs of) functions are supposed to pass
the "vertical line test", and there's no guarantee that this condition
will be satisfied after flipping the picture over. (Try F(t)=t^2 to
see what I mean.) On the other hand, even when that fails, you can
look at a _portion_ of the graph and recognize that curve as being
the graph of a function, at least at most points. That's what you'd
get anyway if you start using things like "acos" which you proposed
in your algebraic solution. Of course you still could have problems
with the vertical line test even if you tried to restrict to very
small neighborhoods of the origin.
And this is what the Inverse Function Theorem guarantees: if F is
a differentiable function and F(a) = b and F'(a) = c and c is nonzero,
then there is a function G(x) which is defined near x = b and
has the property that F(t) = x iff t = G(x) whenever x is near b
and t is near a. (Moreover, G is differentiable at b and G'(b)=1/c,
but you don't seem interested in that.)
The theorem says nothing at all about whether G can be written in
"formulas".
dave
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