Solving Simple Differential Equations: Particle Velocity and Position

[resurgance2001]

Homework Statement

Given that a particle has an initial velocity v0 and then undergoes an acceleration a = - bv. , where b is a constant, obtain an expression for v = v(t) and x = x (t) [/B]

Homework Equations

Not sure

The Attempt at a Solution

If I integrate a = - b v

I think I get v + vo = - b x + xo

But then I get stuck

The book actually gives the answer. x(t) = xo + (vo/b) (1 - e^-bt)

But I can't figure out how they got to that. The book kind of assumes that one can solve this and I should be able to but I am stuck. Thanks

 

[Ray Vickson]

Homework Statement

Given that a particle has an initial velocity v0 and then undergoes an acceleration a = - bv. , where b is a constant, obtain an expression for v = v(t) and x = x (t) [/B]

Homework Equations

Not sure

The Attempt at a Solution

If I integrate a = - b v

I think I get v + vo = - b x + xo

But then I get stuck

The book actually gives the answer. x(t) = xo + (vo/b) (1 - e^-bt)

But I can't figure out how they got to that. The book kind of assumes that one can solve this and I should be able to but I am stuck. Thanks

If you knew the velocity function $v(t)$, how would you get the acceleration $a = a(t)$? So, if you are told that $a = -b v$ with constant $b$, what sort of equation do you get?

 

[resurgance2001]

If you know the velocity, you can get the acceleration by differenting. dv/dt = - b v ??

 

[resurgance2001]

So v (t) = - 1/b dv/dt ?

 

[resurgance2001]

Thanks Ray -

 

[resurgance2001]

If my function for v(t) is correct, I try integrating this to get x.

I starting with v(t) = - 1/b dv/dt

Integrating both sides, I get

x(t) + xo = -1/b v (t) + v0.

But I know this is incorrect. I can't see where the e^-bt. In their answer comes from. Thanks

 

[PeroK]

If my function for v(t) is correct, I try integrating this to get x.

I starting with v(t) = - 1/b dv/dt

Integrating both sides, I get

x(t) + xo = -1/b v (t) + v0.

But I know this is incorrect. I can't see where the e^-bt. In their answer comes from. Thanks

It's not incorrect, but you've expressed v in terms of x, not in terms of t. Although, you've been careless with x0 and v0.

If you try $v(t) = e^{-bt}$ you'll find it satifies your equation.

To get v(t) you need to separate v and t before you integrate.

 

[resurgance2001]

Thanks Perok.

I think I've got it. At least I can show that the answer they gave is a solution.Starting with the answer they gave:

x(t) = x0 + (vo/b) (1 - e^-bt)

I differentiate this once to get:

v(t) = vo e^-bt

I differentiate this a second time:

a(t) = -b vo e^-bt Substuting for [vo e^-bt ] gives:

a(t) = -bv (t) as required.

So now I can see that it would (theoretically) be possible for me (next time) to solve the problem.

Thanks

 

[Ray Vickson]

Thanks Perok.

I think I've got it. At least I can show that the answer they gave is a solution.Starting with the answer they gave:

x(t) = x0 + (vo/b) (1 - e^-bt)

I differentiate this once to get:

v(t) = vo e^-bt

I differentiate this a second time:

a(t) = -b vo e^-bt Substuting for [vo e^-bt ] gives:

a(t) = -bv (t) as required.

So now I can see that it would (theoretically) be possible for me (next time) to solve the problem.

Thanks

The DE dv/dt = -bv is among the simplest, and is one of the first you meet in a differential equations course. The solution is easy, because somebody already solved it for us in the past, and it is well documented in just about every textbook: v = const* exp(-b t). The reason is simple: (d/dt) exp(a*t) = a* exp(a*t), from Calculus 101; basically, this is just the exponential derivative (d/dx) exp(x) = exp(x), with a change of variable to x = a*t.

 

[resurgance2001]

Yes - thanks. I should have got it sooner, but it is about 7 years since I did my course in applied maths and am pretty rusty. Cheers

 

[Chestermiller]

From your differential equation,
$$\frac{dv}{v}=-bdt$$
Both sides of this equation are exact differentials.

Chet

 

[resurgance2001]

Hi Chet,

Can you explain a bit more about what you mean when you say that both sides are exact differentials?

Thanks

Peter

 

[Chestermiller]

Hi Chet,

Can you explain a bit more about what you mean when you say that both sides are exact differentials?

Thanks

Peter

What is $\int{\frac{dv}{v}}$?

What is $\int{dt}$?

Chet

 

[resurgance2001]

I am not sure. The integral of dV/V, is it
ln V ?

And the integral of dt is t?

So t = lnV

Which seem to make sense given that

V = Vo = e^-bt (I've got to figure out latex!) But I still don't get the term 'exact differentials. Cheers

 

[PeroK]

I am not sure. The integral of dV/V, is it
ln V ?

And the integral of dt is t?

So t = lnV

Which seem to make sense given that

V = Vo = e^-bt (I've got to figure out latex!)But I still don't get the term 'exact differentials. Cheers

Given that you simply wrote down the integral of each term, is that not a strong hint of what an exact differential is?

A good example is $mv\frac{dv}{dt}$

Can you see what that is an (exact) derivative of?

You definitely need to revise derivatives, integration and differentials if you're trying to study DE's.

 

[Chestermiller]

I am not sure. The integral of dV/V, is it
ln V ?

And the integral of dt is t?

So t = lnV

No. Including the constant of integration C, the result should read:

lnv=-bt + C

Inverting this gives:
$$v=e^{(C-bt)}$$

Hope this makes some sense.

Chet

 

[resurgance2001]

Hi Perok

You wrote:

mvdv/dx and then ask, what is this a derivative of.

This is my attempt:

I start by separating the variables:

mvdv = dx

Then integrating both sides, I get:

ma + vo = t + to

I am not sure if this is what you meant.

 

[resurgance2001]

Hi Chet

Thanks I get that. I am still stuck though on this new term (at least it is a term that I am unfamiliar with) "exact differential". But don't worry - it's probably not so important right now. The main thing is that I am getting back into doing this maths. Cheers and thanks again.

 

[PeroK]

Hi Perok

You wrote:

mvdv/dx and then ask, what is this a derivative of.

This is my attempt:

I start by separating the variables:

mvdv = dx

Then integrating both sides, I get:

ma + vo = t + to

I am not sure if this is what you meant.

First, I gave you an expression, not an equation. You've effectively set that expression = 1.

In any case:

$mv\frac{dv}{dt} = \frac{d}{dt}(\frac{1}{2}mv^2)$

So, that expression is the rate of change of Kinetic Energy. I.e. it's an exact derivative of the expression for KE. It's a good one to remember.

As I said before, you need to revise your calculus!

 

[resurgance2001]

Thanks - I guessed that it would be something really obvious like energy but missed it.

You are right, I do need to revise my calculus. Like I said it is 7 years since I did this stuff and now I am trying to study quantum field theory.

 

[resurgance2001]

Can you give an example of a derivative that is not exact? Sometimes to get a better understanding of what something is one needs to see an example of something which is not the thing in question. Or is it simy because there is no constant? I mean if you just integrate mv over dt , would you not get 1/2 m[v][/2] + C ? Is that why you wrote it mv dv/dt ? Thanks

 

[resurgance2001]

That should have said 1/2 mv^2 + C (iphone doesn't seem to like latex

 

[Mark44]

Can you give an example of a derivative that is not exact? Sometimes to get a better understanding of what something is one needs to see an example of something which is not the thing in question. Or is it simy because there is no constant? I mean if you just integrate mv over dt , would you not get 1/2 m[v][/2] + C ? Is that why you wrote it mv dv/dt ? Thanks

PeroK wrote this several posts back:

mv\frac{dv}{dt} = \frac{d}{dt}(\frac{1}{2}mv^2)

If you integrate the left side, with respect to t, you have this
$$\int mv\frac{dv}{dt} ~dt = \int mv ~ dv = \frac 1 2 mv^2 + C$$
In the expression on the left, the dt's effectively cancel to produce the expression in the middle.