# Extremes of a variable

1. Sep 12, 2004

### Omid

To get a feeling for how extremes of a variable affect an equation, suppose that the acceleration of a body is given by a = [(at^2 - bt)/(t+c)d] + et^2/(t-c)d where a,b,c,d,e are constants : (a) Determine the value of a when t=0. (b) Find the value of a when t>>c.
(c ) find the value of a when t<<c.

I tried approximation by binomial theorem but I got some meaningless equations (for me ) so I couldnt find such a feeling
Thanks

2. Sep 13, 2004

### Darien

Rather than using an equation, just by looking at your acceleration equation try to see what happens when t >> c and t << c. If you can't visualize it, plug in actual numbers and see what happens.

3. Sep 13, 2004

### HallsofIvy

Staff Emeritus
I don't know why you would use the binomial theorem.

a= [(at^2 - bt)/(t+c)d] + et^2/(t-c)d
(Am I correct in assuming that the "a" on both sides is intended?)

Obviously, if t= 0, a= 0.

If t is "much larger than c" (t>> c) we can ignore the c added or subtracted (If you have $10000000,$1 more or less doesn't matter! On the other hand multiplying by a very small number can be important!) and write
a is very close to [(at^2 - bt)/(t)d] + et^2/(t)d = ((at-b)+et)/d= ((a+e)t-b)/d
so da= at+ et- b. (d-t)a= et-b so a= (et-b)/(d-t).

I'm going to interpret "t<< c" to mean "t is non-zero but much smaller than c but non-zero.

Now, we "ignore" the t in t-c and t+ c and have a= [(at^2 - bt)/cd] + et^2/(-c)d=
a(t^2/cd)- bt/cd+ (et^2/cd) so (1- t^2/cd)a= ((cd- t^2)/cd)a= (et^2- bt)/ce and
a= (et^2- bt)/(cd- t^2).

4. Sep 13, 2004

### Omid

I should have written the "a" in the left hand side as capital, A. Because it is diffrent from the right hand side one.
Now, probably, I've found such a feeling.
Thanks