Solution to Av^2 + B = C dv/dt Differential Equation?

[Cyrus]

Does anyone know the solution to this differential equation. It's not for a class, it came up in another thread about drag forces on a falling body. I don't care how you get to the solution. If you want to post how, that's fine and thanks! If not, I don't care one way or another. I just want to know if there is a solution for this differenetial equation, and if so what is it? I can do the case for Av, but not $Av^2$.


$$Av^2 + B = C \frac {dv}{dt}$$

Thanks guys,

 

[AKG]

$$t/C = \int \frac{dv}{Av^2 + B} = \frac{1}{A}\int\frac{dv}{v^2 + B/A} = \frac{1}{A}\frac{\arctan\left(\frac{v}{\sqrt(B/A)}\right)}{\sqrt{B/A}}$$

Here's a useful website: The Wolfram Integrator

 

[Cyrus]

Ah, that's great thank you very much!

 

[nrqed]

$$t/C = \int \frac{dv}{Av^2 + B} = \frac{1}{A}\int\frac{dv}{v^2 + B/A} = \frac{1}{A}\frac{\arctan\left(\frac{v}{\sqrt(B/A)}\right)}{\sqrt{B/A}}$$

Here's a useful website: The Wolfram Integrator

Of course! (slapping myself in the face!)
I had the same question asked th eother day and I refer to Maple..
Thanks AKG

 

[Cyrus]

Damn, one problem. If I use the case of a falling body, then that means,

$$B = -mg$$

and the square root is complex. It can't take complex time to reach a certain speed. How do I fix this dilemma?

HELP!

 

[nrqed]

Damn, one problem. If I use the case of a falling body, then that means,

$$B = -mg$$

and the square root is complex. It can't take complex time to reach a certain speed. How do I fix this dilemma?

HELP!

Then it is a logarithm instead of an arctan

 

[nrqed]

$$t/C = \int \frac{dv}{Av^2 - B} = \frac{1}{A}\int\frac{dv}{v^2 - B/A} = \frac{1}{2A} {\sqrt{A/B}} ln({x-{\sqrt{B/A}} \over x+ {\sqrt{B/A}}})$$

 

[Cyrus]

You guys are the best!

 

[Cyrus]

Ah, not another problem.

With your new equation, if I set the inital conditions as t=0, v=0, I get:

0= ln (-1)

But that is undefined!


EDIT: My table of integrals says absolute value of ln

Arg, yet another problem,

that fraction is always going to be less than 1,

which means you will always get a negative value from the ln(x).

Would it be ok to just stick another absolute value sign around the ln?

|ln(|x|)| Another problem is when sqrt(B/A) = x, becaue the numerator is now zero and the function is undefined.

?

 

[arildno]

No, remember that your velocity will be non-positive; hence, the absolute value of your fraction inside the log will be greater than 1.
When the velocity goes to $v=-\sqrt{\frac{B}{A}}$, then time goes to (positive) infinity, as it should.

 

[Cyrus]

Why is it not positive?

I had the hunch that when the square root of (B/A) becomes close to V, it takes infinite time, so the sqrt(B/A) would be the terminal velocity.

I still don't see why the velocity is non-positive. If I give the body an initial thrust up, the initial conditions will be a positive upward velocity at t=0.

 

[nrqed]

Why is it not positive?

I had the hunch that when the square root of (B/A) becomes close to V, it takes infinite time, so the sqrt(B/A) would be the terminal velocity.

I still don't see why the velocity is non-positive. If I give the body an initial thrust up, the initial conditions will be a positive upward velocity at t=0.

There are two ways to do the problem: working with the y component of the velocity , $v_y$ or working with the speed .

I am assuming here that the object is moving downward...

Then, using v to represent the *speed*, the equation is
$C { dv \over dt} = - A v^2 + mg$ (assuming A and C positive).

However, if you work with the y component of the velocity (with the positive y-axis pointing upward), you get

$C {dv_y \over dt} = + A v_y^2 - m g$

Since you said B=-mg, it sounds like you are working with the y component and since it is moving downward, v will always be negative

(EDIT: If you throw it *upward* then the equation is

$C {dv_y \over dt} = - A v_y^2 - m g$)
Pat

 

[arildno]

If you have a positive velocity, then your drag will act in the same direction as gravity (giving you an arctan result rather than the artanh result), rather than acting in the opposite direction as in your case. Thus, you would have a different expression to handle a positive velocity.

The general law for a "square-velocity" drag problem is:
$$-A|v|v-mg=m\frac{dv}{dt}, A>0$$
where |v| is the absolute value of the velocity.

 

[Cyrus]

There are two ways to do the problem: working with the y component of the velocity , $v_y$ or working with the speed .

I am assuming here that the object is moving downward...

Then, using v to represent the *speed*, the equation is
$C { dv \over dt} = - A v^2 + mg$ (assuming A and C positive).

However, if you work with the y component of the velocity (with the positive y-axis pointing upward), you get

$C {dv_y \over dt} = + A v_y^2 - m g$

Since you said B=-mg, it sounds like you are working with the y component and since it is moving downward, v will always be negative

(EDIT: If you throw it *upward* then the equation is

$C {dv_y \over dt} = - A v_y^2 - m g$)
Pat

I am not 100% sure what your doing here, so let's step back a moment.

For your first equation using speed, why did you have a +mg term, and not a minus?

For your second equation, why did you switch from + to - signs for the coefficient A?

What is your distinction between speed and velocity? To me, speed is the magnitude, velocity has magnitude and direction.

 

[nrqed]

I am not 100% sure what your doing here, so let's step back a moment.

For your first equation using speed, why did you have a +mg term, and not a minus?

For your second equation, why did you switch from + to - signs for the coefficient A?

What is your distinction between speed and velocity? To me, speed is the magnitude, velocity has magnitude and direction.

Arildno explained it well in post #13 but let me say it in a different way.
First, yes, velocity is a vector, v is the magnitude. And v_y is the y component (they are three different things, right?). the speed v can never be negative, v_y may be positive or negative. In the following, I use a y-axis pointing upward.


In an equation using the y component, gravity will always be -mg. But the drag force is opposite to th emotion so if the object is moving up, the drag force will be downward and vice versa. That explains my signs with the velocity equations.

For the speed, it's a bit different. Let's say you are moving downward. Does gravity tends to increase the speed? Yes, so you put +mg (notice that gravity always tend to *decrease* the y component of the velocity, no matter what the direction of the motion). And the drag force tends to decrease the speed, hence the minus sign. If the object was moving upward and you were writing an equation for the speed, both terms would be negative because both forces will act to decrease the speed.

I hope this helps

Patrick

 

[Cyrus]

I see what you mean now, but god is that awfully sloppy notation

 

[nrqed]

I see what you mean now, but god is that awfully sloppy notation

What? My notation?? I use "v" for speed and v_y for y component of the velocity! That's standard!

Well, this is what I get for a lot of typing

 

[Cyrus]

I am used to calling the velocity, $$\bar{V}$$, and the speed, is $$| \bar{V}|$$, and the y-component of the velocity is $$|V_y|$$

Actually, I am more used to calling the speed, s, not v.

 

[nrqed]

I am used to calling the velocity, $$\bar{V}$$, and the speed, is $$| \bar{V}|$$, and the y-component of the velocity is $$|V_y\hat{y}|$$

Actually, I am more used to calling the speed, s, not v.

Your very first post contains the symbol "v". This is always used to represent the speed so I assumed it was the speed! And I think I made it fairly clear early on that I was talking about the speed.


I have tried hard to make the notation crystal clear by writing v_y and showing all the cases separately. I don't knwo how I could have made things more clear.

 

[Cyrus]

Well, I am used to v for the case where the motion is along one orthagonal direction.

I am just splitting hairs now!

I get what your saying, and you are right!

In any event, I see what you mean now, Thanks for your help!

 

[Cyrus]

could I integrate this solution to get position as a function of time as well?

$$\frac{t}{c} =\frac{1}{2A} {\sqrt{A/B}} ln({v-{\sqrt{B/A}} \over v+ {\sqrt{B/A}}})$$

$$\int \frac{t}{c} dt = \int \frac{1}{2A} {\sqrt{A/B}} ln({v-{\sqrt{B/A}} \over v+ {\sqrt{B/A}}} dv)$$

and say:

$$\frac{t^2}{2C} = \frac{1}{2A} {\sqrt{A/B}}[ - \sqrt{A/B} log(x-\sqrt{A/B})+xlog( \frac{x-\sqrt{A/B}}{x+\sqrt{A/B}}) -\sqrt{A/B}log(x+\sqrt{A/B}) ]$$
?

Actually, I don't think I can just integrate like that. I think I have to somehow isolate the variable v, which is dx/dt, and then move over the dt and integrate. But that looks like a freaking mess. Help!

Oh, I was wrong earlier. What the hell was I saying velocity is s? No no no, displacement is usually s. Sorry, my mind is in pieces right now.

 

[Cyrus]

Damn, the more I think about it, the more I it seems that I can't just integrate the function. Any thoughts?

 

[mathlete]

Hint: Move the constants all over to the side with t, and then take the exponential of both sides

 

[Cyrus]

Yea, but then you have that faction which is not very nice.

 

[nrqed]

could I integrate this solution to get position as a function of time as well?

$$\frac{t}{c} =\frac{1}{2A} {\sqrt{A/B}} ln({v-{\sqrt{B/A}} \over v+ {\sqrt{B/A}}})$$

$$\int \frac{t}{c} dt = \int \frac{1}{2A} {\sqrt{A/B}} ln({v-{\sqrt{B/A}} \over v+ {\sqrt{B/A}}} dv)$$

and say:

$$\frac{t^2}{2C} = \frac{1}{2A} {\sqrt{A/B}}[ - \sqrt{A/B} log(x-\sqrt{A/B})+xlog( \frac{x-\sqrt{A/B}}{x+\sqrt{A/B}}) -\sqrt{A/B}log(x+\sqrt{A/B}) ]$$
?

Actually, I don't think I can just integrate like that. I think I have to somehow isolate the variable v, which is dx/dt, and then move over the dt and integrate. But that looks like a freaking mess. Help!

Oh, I was wrong earlier. What the hell was I saying velocity is s? No no no, displacement is usually s. Sorry, my mind is in pieces right now.

Indeed, you must first isolate v, write this as dx/dt, isolate dx on one side, and integrate both sides. When you isolate v, you will get a ratio of the form (1+exponential)/(1-exponential). This is a simple hyperbolic trig function. Then you can look up an integral table and get x(t).

Patrick