# Particle's position as a function of time - Integration

## 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"?

phyz

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tiny-tim
Homework Helper
… 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)

Ho Mr Tim!

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

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

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

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

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

$$\frac{d}{dt}xf(t)=\dot{x}f(t)+xf'(t)$$

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

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

$$f'(t)=\gamma$$

so

$$f(t)=\gamma t$$

?????

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

*sigh*

Nope...screwed up...

*back to banging head on desk*

tiny-tim
Homework Helper
How would this be easier necessarily? I'd still end up with

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

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 … ?

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!

Last edited:
tiny-tim
Homework Helper
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 Aeγt)

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 Aeγt)
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 :grumpy: 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
Homework Helper
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 …$$\dot{x}e^{\gamma t}+x\gamma e^{\gamma t}=-e^{\gamma t} gt$$

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!

Mr Tim, you're a legend!

Thank you so much for your patience!

tiny-tim
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