# I need a formula

1. Mar 23, 2006

### 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, thats 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,

2. Mar 23, 2006

### 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

3. Mar 23, 2006

### Cyrus

Ah, thats great thank you very much!

4. Mar 23, 2006

### nrqed

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

5. Mar 23, 2006

### 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!

6. Mar 23, 2006

### nrqed

Then it is a logarithm instead of an arctan

7. Mar 23, 2006

### nrqed

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

8. Mar 23, 2006

### Cyrus

You guys are the best! :tongue:

9. Mar 24, 2006

### 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.

?

Last edited: Mar 24, 2006
10. Mar 24, 2006

### 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.

11. Mar 24, 2006

### 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.

12. Mar 24, 2006

### nrqed

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

13. Mar 24, 2006

### 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.

Last edited: Mar 24, 2006
14. Mar 24, 2006

### Cyrus

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.

15. Mar 24, 2006

### nrqed

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

16. Mar 24, 2006

### Cyrus

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

17. Mar 24, 2006

### nrqed

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 :grumpy:

18. Mar 24, 2006

### 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.

19. Mar 24, 2006

### nrqed

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. :grumpy: :yuck:

20. Mar 24, 2006

### 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!!

Last edited: Mar 24, 2006