What form of calculus needs to be used?

[SirHall]

I have recently been attempting to solve a problem that has been bugging me for quite some time. I've gotten back into calculus and integrals to attempt to solve a little formula I'm trying to build for a simulation test. Over-all, if I have $\int _0^bv^2x\ dx$ I'd expect the outcome to be $\frac{1}{2}b^2v^2$. But my problem is that at x = 0, v = u(u being any constant) while at x = b, v = 0. Therefore when I attempt to solve the integral I get $\frac{1}{2}b^2\ 0^2-\frac{1}{2}0^2u^2$, which of course is equal to zero, but this is impossible. I have the suspicion that I'm using an incorrect 'type of integral'. If what I've written isn't too vague, would anyone happen to know what it I'm doing incorrectly here?

 

[Mark44]

I have recently been attempting to solve a problem that has been bugging me for quite some time. I've gotten back into calculus and integrals to attempt to solve a little formula I'm trying to build for a simulation test. Over-all, if I have $\int _0^bv^2x\ dx$ I'd expect the outcome to be $\frac{1}{2}b^2v^2$. But my problem is that at x = 0, v = u(u being any constant) while at x = b, v = 0. Therefore when I attempt to solve the integral I get $\frac{1}{2}b^2\ 0^2-\frac{1}{2}0^2u^2$, which of course is equal to zero, but this is impossible. I have the suspicion that I'm using an incorrect 'type of integral'. If what I've written isn't too vague, would anyone happen to know what it I'm doing incorrectly here?

Without more information about v, your integral isn't doable. In $\int _0^bv^2x\ dx$, the dx indicates that x is the variable of integration, and presumably v is a function of x. Knowing that v(0) = v(b) = 0 doesn't help, since we don't know what v(x) is.

 

[Buzz Bloom]

Hi SirHall:

I think at least part of your confusion is failing of think about v as a function of x. You have
v(0) = y (a constant)
v(b) = 0.
Since v² is part of the integrand, you need to know what values v(x) has for 0 < x < b.

Hope this helps.

Regards,
Buzz

 

[SirHall]

Without more information about v, your integral isn't doable. In $\int _0^bv^2x\ dx$, the dx indicates that x is the variable of integration, and presumably v is a function of x. Knowing that v(0) = v(b) = 0 doesn't help, since we don't know what v(x) is.

Hi SirHall:

I think at least part of your confusion is failing of think about v as a function of x. You have
v(0) = y (a constant)
v(b) = 0.
Since v² is part of the integrand, you need to know what values v(x) has for 0 < x < b.

Hope this helps.

Regards,
Buzz

I see your point... While I wasn't too happy to do it this way, here we go. I'm attempting to find the total distance a projectile would travel in a fluid given:
m = Mass
u = Initial Velocity
p = Fluid Density
A = Projectile Reference Area
C = Drag Coefficient
b = Total Distance Travelled
x = Distance Travelled
v = Velocity(Some odd form I guess)
So I have the equation: $\frac{1}{2}mu^2\ =\ ρAC\int _0^bv^2\left(x\right)\ dx$ and I need to solve for b. (Btw, this is the drag equation I'm using here.)
But... I don't want to introduce time which is why I have resisted $v\left(x\right)=\frac{x}{t}$ (Where x would usually be written as 'd' or distance).
I also thought of using $v=\frac{v_i+v_f}{2}$ but that simply uses a mean of the velocity, and it doesn't use 'x' as you's mentioned.
So, I guess this is where I'm stuck.

 

[Mark44]

I see your point... While I wasn't too happy to do it this way, here we go. I'm attempting to find the total distance a projectile would travel in a fluid given:
m = Mass
u = Initial Velocity
p = Fluid Density
A = Projectile Reference Area
C = Drag Coefficient
b = Total Distance Travelled
x = Distance Travelled
v = Velocity(Some odd form I guess)
So I have the equation: $\frac{1}{2}mu^2\ =\ ρAC\int _0^bv^2\left(x\right)\ dx$ and I need to solve for b. (Btw, this is the drag equation I'm using here.)
But... I don't want to introduce time which is why I have resisted $v\left(x\right)=\frac{x}{t}$ (Where x would usually be written as 'd' or distance).
I also thought of using $v=\frac{v_i+v_f}{2}$ but that simply uses a mean of the velocity, and it doesn't use 'x' as you's mentioned.
So, I guess this is where I'm stuck.

If you don't know v(x), you can't evaluate the integral. It's as simple as that.
BTW, your notation is ambiguous -- $v^2(x)$. This could be interpreted as either $[v(x)]^2$ or $v^2 \cdot x$. From your earlier post, I assume you mean the latter.

 

[jedishrfu]

You might glean something from here to solve your problem or perhaps figure out what you're missing:

http://hyperphysics.phy-astr.gsu.edu/hbase/airfri2.html

 

[Buzz Bloom]

I don't want to introduce time

Hi @SirHall:

Why do you want to avoid including time in your analysis? I think that is the usual way to approach this kind of problem. Velocity changes because of the forces operating on the object. I get that you are starting some sort of projectile in a fluid (gas or liquid) medium, and the force will be the drag. If F(t) is the time varying drag force, then
v(t) = ∫F(t)/m dt . If the fluid is stationary, then the drag velocity is the same as the projectile velocity, although in the opposite direction.

Hope this helps.

Regards,
Buzz

 

[SirHall]

If you don't know v(x), you can't evaluate the integral. It's as simple as that.
BTW, your notation is ambiguous -- $v^2(x)$. This could be interpreted as either $[v(x)]^2$ or $v^2 \cdot x$. From your earlier post, I assume you mean the latter.

Well, after a little experimentation I think I may have found v(x)! $v\left(x\right)\ =\ \sqrt{2\cdot \frac{-\frac{1}{2}pACx^2}{m}\left|x\right|+u^2}$. So, back to the original reason for this thread, what do I do with v(x)?

Hi @SirHall:

Why do you want to avoid including time in your analysis? I think that is the usual way to approach this kind of problem. Velocity changes because of the forces operating on the object. I get that you are starting some sort of projectile in a fluid (gas or liquid) medium, and the force will be the drag. If F(t) is the time varying drag force, then
v(t) = ∫F(t)dt . If the fluid is stationary, then the drag velocity is the same as the projectile velocity, although in the opposite direction.

Hope this helps.

Regards,
Buzz

Well, if I want to find the distance it took for the object to stop from it's initial velocity to the point it stops, I can't use 't' as a value because I don't know it. Also, I don't see how time would even have an effect on this. If it does, then that would be rather interesting.

Btw, would like to state I seriously do appreciate the help I've received thus far. Never learned that I needed to find v(x). :)

 

[Buzz Bloom]

Hi @SirHall:
I don't know how to advise you any further because I don't fully understand the the physical setup. Below is my guess about it as I mentioned previously:

I get that you are starting some sort of projectile in a fluid (gas or liquid) medium, and the force will be the drag.

If the fluid is stationary, then the drag velocity is the same as the projectile velocity

Is this correct? That is, do you have a projectile that starts at some specified velocity and moves through a stationary fluid, and the projectile slows as it moves due to the drag of the fluid? Or if this is wrong, can you describe a correction?

Regards,
Buzz

 

[SirHall]

Hi @SirHall:
I don't know how to advise you any further because I don't fully understand the the physical setup. Below is my guess about it as I mentioned previously:Is this correct? That is, do you have a projectile that starts at some specified velocity and moves through a stationary fluid, and the projectile slows as it moves due to the drag of the fluid? Or if this is wrong, can you describe a correction?

Regards,
Buzz

You are exactly right. However after looking a little further with our great friend Google Scholar and taking advice given here, I've found my answer. So no further replies are needed. Thanks again for the assistance!