Particle's position as a function of time - Integration

[phyzmatix]

Homework Statement

A particle falling under gravity is subjet to a retarding force proportional to its velocity. Find its position as a function of time, if it starts from rest, and show that it will eventually reach a terminal velocity.

2. The attempt at a solution

To save myself some latex time, I skipped a couple of the intermediary steps here:

F_{tot}=F_{g}-F_{retarding}

m\ddot{x}=-mg-\lambda\dot{x}

\ddot{x}=-g-\frac{\lambda}{m}\dot{x}

let

\gamma=\frac{\lambda}{m}

then

\frac{d}{dt}dx=-gdt-\gamma dx

integrating both sides

\frac{d}{dt}\int_{o}^{x}dx=-g\int_{0}^{t}dt-\gamma\int_{0}^{x}dx

\dot{x}+\gamma x=-gt

This final equation I get is correct (hopefully my procedure is too) and at this point, the question offers a hint:

[HINT: After the first integration, use an integrating factor, i.e. a function f(t) such when the equation is multiplied by f(t), the left-hand side becomes an exact derivative, in fact, the derivative of xf. The final stage requires an integration by parts]

This is where I got stuck. How do I determine the integrating factor f(t)? And what exactly is meant by "the left-hand side becomes an exact derivative, in fact, the derivative of xf"?

Your help is appreciated, thanks!
phyz

 

[tiny-tim]

… And what exactly is meant by "the left-hand side becomes an exact derivative, in fact, the derivative of xf"?

Hi phyz!

(have a lambda: λ )

Wouldn't it be easier to put x' = v, so mv' = -mg - λv?

An exact derivative means something like vcos(t) + v'sin(t) …

it's exactly the derivative of vsin(t)

 

[phyzmatix]

Ho Mr Tim!

I'd love to say that I see where you're going, but that would be a blatant lie!

So please bear with me...

Wouldn't it be easier to put x' = v, so mv' = -mg - λv?

How would this be easier necessarily? I'd still end up with

<br /> \dot{x}+\gamma x=-gt<br />

wouldn't I?

An exact derivative means something like vcos(t) + v'sin(t) …

it's exactly the derivative of vsin(t)

Oh, OK...I get this bit...But how? Grrrr! I hate feeling so daft

 

[phyzmatix]

becomes an exact derivative, in fact, the derivative of xf

So if x is a function of t and we know f is a function of t then the derivative of xf will be

\frac{d}{dt}xf(t)=\dot{x}f(t)+xf&#039;(t)

Right?...but that doesn't bring me any closer to determining f(t)...

*banging head on desk*

 

[phyzmatix]

Hold on...methinks the grey matter is kicking in (albeit slowly)...if what I have above is right, then

f&#039;(t)=\gamma

so

f(t)=\gamma t

?

...hang in there, I need to get to pen and paper!

 

[phyzmatix]

*sigh*

Nope...screwed up...

*back to banging head on desk*

 

[tiny-tim]

How would this be easier necessarily? I'd still end up with

<br /> \dot{x}+\gamma x=-gt<br />

wouldn't I?

It's easier because your equation had x' x and t, but mine only has v' and v, and can be integrated immediately to give v as a function of t.

Oh, OK...I get this bit {exact derivative}...But how? Grrrr! I hate feeling so daft :cry

You need to find a function f(t) to multiply x' + γx,

so that f(t)x' + γf(t)x is an exact derivative, of the form g(t)x' + g'(t)x …

so the equation relating f f' g and g' is … ?

 

[phyzmatix]

I think I may actually have it now...if we let

f(t)=e^{\gamma t}

and we multiply it with

\dot{x}+\gamma x=-gt

throughout, then we get

\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt

and

\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}

is an exact derivative of

x e^{\gamma t}

How's that? Or am I still missing the mark?

Thanks for your help Tiny-Tim! It really is much appreciated!

 

[tiny-tim]

Yes, that's exactly right!

(btw, the general method would be g'/g = γ, so ln(g) = ∫γdt + C, or g = Ae^∫γdt …

in this case, γ is constant, so it's just Ae^γt)

 

[phyzmatix]

Yes, that's exactly right!

(btw, the general method would be g'/g = γ, so ln(g) = ∫γdt + C, or g = Ae^∫γdt …

in this case, γ is constant, so it's just Ae^γt)

Can you believe that I actually understand what you're saying!

OK, so progress has been made...However, it seems as if last night I used up my allowed number of "Eureka" moments for the week and haven't been able to complete the solution

I know that the answer to this problem is

x=(\frac{g}{\gamma^2})(1-e^{-\gamma t})-\frac{gt}{\gamma}

But can't seem to make the connection required to move there from

\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt

At first I thought that, since we now have an exact derivative, we could just integrate both sides with respect to t as follows

\int(\dot{x}e^{\gamma t}+x\gamma e^{\gamma t})dt=-g\int tdt
xe^{\gamma t}=-\frac{\gamma}{2}t^2

Which is clearly going the wrong way Then I thought

\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt
\frac{dx}{dt}e^{\gamma t}+x\gamma e^{\gamma t}=-gt
e^{\gamma t}dx + x\gamma e^{\gamma t}dt=-gt dt
dx=-\gamma x dt -\frac{gt}{e^{\gamma t}}dt

But what then to do with the term

-\gamma x dt

Where am I going wrong?

 

[tiny-tim]

Hi phyzmatix!

(btw, I'm having difficulty reading your x-dots … I wish you'd use dashes instead of dots and maybe not even use latex )

Can you believe that I actually understand what you're saying!

Woohoo! ​

OK, so progress has been made...However, it seems as if last night I used up my allowed number of "Eureka" moments for the week and haven't been able to complete the solution

You need to recharge by having a bath in between.

I know that the answer to this problem is

x=(\frac{g}{\gamma^2})(1-e^{-\gamma t})-\frac{gt}{\gamma}

But can't seem to make the connection required to move there from

\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt
…
Where am I going wrong?

erm …\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-e^{\gamma t} gt

 

[phyzmatix]

Hi phyzmatix!

(btw, I'm having difficulty reading your x-dots … I wish you'd use dashes instead of dots and maybe not even use latex )

Done!

erm …\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-e^{\gamma t} gt

Ye gods! What a muppet!

 

[phyzmatix]

Mr Tim, you're a legend!

Thank you so much for your patience!

 

[tiny-tim]

thanks

and now try it again, starting with v'/(g + γv) = -1, just to see the difference ​