# Reciprocals of integrals

Discussion in 'Homework Help' started by acrofts84, Mar 13, 2012.

1. ### acrofts84 Thread Starter New Member

Aug 12, 2010
9
1
Hi,

I'm almost certain this is a silly question, but is the following correct :

dv/dt = 1 / (∫1/v dt)

(the differential of v is the same as the reciprocal of the integral of the reciprocal of v ...if that helps )

Thanks very much!

2. ### steveb Senior Member

Jul 3, 2008
2,431
469
No, that's not true in general. Try v=1, and that is clear. It does make an interesting equation though. Can you solve it?

Last edited: Mar 13, 2012
acrofts84 likes this.
3. ### acrofts84 Thread Starter New Member

Aug 12, 2010
9
1

Is it that dV = V, so it is correct only when V is infinitely small ?

I think that would make sense!

4. ### steveb Senior Member

Jul 3, 2008
2,431
469
That's an interesting thought. I'll have to think about that a little to see if it holds up rigorously. I can see how you are thinking that.

However, I was thinking along more traditional lines. This is basically a type of differential equation (or, technically an integro-differential equation).

Try the general function v=At^n as a test function, where A and n are arbitrary constants. If you plug this in, you will find constraints on A and n. What values or range of values work?

5. ### acrofts84 Thread Starter New Member

Aug 12, 2010
9
1
I find that An = -An, so surely this would only work when A or n are zero? So v is independent of t ?

6. ### steveb Senior Member

Jul 3, 2008
2,431
469
Hmmm, I got something different. We should both double check.

7. ### acrofts84 Thread Starter New Member

Aug 12, 2010
9
1
My algebraic integration clearly isn't as good as it should be! ^

An = -A(n+1)

so this would suggest that the constraint is that n = -1/2.

Although I've tried to do it again with v = At^(-1/2) but it didn't add up unless A was 0

8. ### steveb Senior Member

Jul 3, 2008
2,431
469
You are getting closer. Try v=At^(1/2). Basically, A can be anything but n needs to be 1/2.

acrofts84 likes this.
9. ### acrofts84 Thread Starter New Member

Aug 12, 2010
9
1
Ahh, great, thanks.

Thanks very much for your help !

steveb likes this.
10. ### MrChips Moderator

Oct 2, 2009
19,137
6,150
You are taking the usage and meaning of the words "reciprocal" and "inverse" out of context.

In basic arithmetic, the words mean the same thing and means "to raise to the power -1", i.e. the reciprocal of f is 1/f.

This does not apply to derivatives and integrals and all other functions.

Aug 12, 2010
9
1