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Particle's position as a function of time - Integration

  • Thread starter phyzmatix
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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:

[tex]F_{tot}=F_{g}-F_{retarding}[/tex]

[tex]m\ddot{x}=-mg-\lambda\dot{x}[/tex]

[tex]\ddot{x}=-g-\frac{\lambda}{m}\dot{x}[/tex]

let

[tex]\gamma=\frac{\lambda}{m}[/tex]

then

[tex]\frac{d}{dt}dx=-gdt-\gamma dx[/tex]

integrating both sides

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

[tex]\dot{x}+\gamma x=-gt[/tex]

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
 

Answers and Replies

  • #2
tiny-tim
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… And what exactly is meant by "the left-hand side becomes an exact derivative, in fact, the derivative of xf"?
Hi phyz! :smile:

(have a lambda: λ :wink:)

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

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

it's exactly the derivative of vsin(t) :smile:
 
  • #3
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Ho Mr Tim! :smile:

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

So please bear with me...

Wouldn't it be easier to put x' = v, so mv' = -mg - λv? :wink:
How would this be easier necessarily? I'd still end up with

[tex]
\dot{x}+\gamma x=-gt
[/tex]

wouldn't I? :confused:

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

it's exactly the derivative of vsin(t) :smile:
Oh, OK...I get this bit...But how? Grrrr!!! I hate feeling so daft :cry:
 
  • #4
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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

[tex]\frac{d}{dt}xf(t)=\dot{x}f(t)+xf'(t)[/tex]

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

*banging head on desk*
 
  • #5
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Hold on...methinks the grey matter is kicking in (albeit slowly)...if what I have above is right, then

[tex]f'(t)=\gamma[/tex]

so

[tex]f(t)=\gamma t[/tex]

?????

...hang in there, I need to get to pen and paper! :biggrin:
 
  • #6
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*sigh*

Nope...screwed up...

*back to banging head on desk*
 
  • #7
tiny-tim
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How would this be easier necessarily? I'd still end up with

[tex]
\dot{x}+\gamma x=-gt
[/tex]

wouldn't I? :confused:
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. :wink:
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 … ? :smile:
 
  • #8
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I think I may actually have it now...if we let

[tex]f(t)=e^{\gamma t}[/tex]

and we multiply it with

[tex]\dot{x}+\gamma x=-gt[/tex]

throughout, then we get

[tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt[/tex]

and

[tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}[/tex]

is an exact derivative of

[tex]x e^{\gamma t}[/tex]

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

Thanks for your help Tiny-Tim! It really is much appreciated! :smile:
 
Last edited:
  • #9
tiny-tim
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Yes, that's exactly right! :biggrin:

(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 Aeγt)
 
  • #10
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Yes, that's exactly right! :biggrin:

(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 Aeγt)
Can you believe that I actually understand what you're saying! :biggrin:

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 :frown:

I know that the answer to this problem is

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

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

[tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt[/tex]

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

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

Which is clearly going the wrong way :grumpy: Then I thought

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

But what then to do with the term

[tex]-\gamma x dt[/tex] :confused:

Where am I going wrong?
 
  • #11
tiny-tim
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Hi phyzmatix! :smile:

(btw, I'm having difficulty reading your x-dots … I wish you'd use dashes instead of dots and maybe not even use latex :redface:)
Can you believe that I actually understand what you're saying! :biggrin:
:biggrin: Woohoo! :biggrin:
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 :frown:
You need to recharge by having a bath in between. :wink:
I know that the answer to this problem is

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

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

[tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-gt[/tex]

Where am I going wrong?
erm :redface: …[tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-e^{\gamma t} gt[/tex] :wink:
 
  • #12
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Hi phyzmatix! :smile:

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

erm :redface: …[tex]\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-e^{\gamma t} gt[/tex] :wink:
Ye gods! What a muppet! :redface:
 
  • #13
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Mr Tim, you're a legend! :biggrin:

Thank you so much for your patience! :smile:
 
  • #14
tiny-tim
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thanks :smile:

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

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