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!

 

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!

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?

 

Thanks for the reply.

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

I think that would make sense!

 

Thanks for the reply.

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

I think that would make sense!

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?

 

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

 

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

Hmmm, I got something different. We should both double check.

 

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

 

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

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

 

Ahh, great, thanks.

It all adds up now!

Thanks very much for your help !

 

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.

 

Thanks for the reply MrChips.

I agree the name of the thread is probably misleading if that's the point you're making. Although I thought I was safe to assume that 1 / (∫1/v dt) is, in words, the reciprocal of ∫1/v dt.

I'll think of a better thread name next time!

 

Ahh, great, thanks.

It all adds up now!

Thanks very much for your help !

There is even a more general solution. It should be clear now that any scale factor in front will cancel out automatically. What about A(t-B)^n? The above shows that n=1/2 and B=0 works. Are there other possible solutions?