Help with the Separation of Variables and Integration

[silento]

Homework Statement

Hello! This is a combination of physics and calculus. I have a non constant drag force I’m trying to integrate in relation to velocity and position. I’ve started it, however I’m getting stuck on the separating and integrating. I’ve tired moving equations over to different sides however it’s just not working. I’m trying to get v to the right side and dx and m to the left. Everything besides the position (x) are known constants

Relevant Equations

See pic below.

 

[silento]

Tried a few approaches but all became very complex and couldn’t seperate the dx from dv

 

[mathhabibi]

solve for dv/dx and integrate with respect to x.

 

[silento]

solve for dv/dx and integrate with respect to x.

I may have forgot to add that v isn't a single constant. it is changing from 120 to 0 which will be the bounds for the right side But ∆x is what I'm solving for in the end

 

[mathhabibi]

I may have forgot to add that v isn't a single constant

which is why you should integrate.

 

[mathhabibi]

But ∆x is what I'm solving for in the end

In the post you said you are solving for v. Be more clear next time. What even is $\Delta x$?

 

[silento]

change in position is ∆x

 

[silento]

Its similar to what I am doing here. But, there is now two additional forces in additon to the drag force but only the drag force is non constant both the usFN and urFN are just constants

 

[mathhabibi]

I'm not sure

View attachment 340969
Its similar to what I am doing here. But, there is now two additional forces in additon to the drag force but only the drag force is non constant

It will take me some time but from what I am seeing, after integrating you want to solve for $x_f-x_i$, is that correct?

 

[silento]

yes correct!

 

[silento]

thank you!

 

[mathhabibi]

No problem. I am a bit sketchy with physics but is it true that $F_n=m\frac{dv}{dt}$?

 

[silento]

m(dv/dt) is the net force. sigma F

 

[mathhabibi]

okay I started my tutoring session will be back later.

 

[silento]

F_n is known as the normal force. see you later!

 

[erobz]

Your equation is of the form:

$$ \varphi - \beta v^2 = m v\frac{dv}{dx}$$

Where $\varphi$ is a constant.

Make the substitution $u = \varphi - \beta v^2$, and differentiate $u$ w.r.t. $x$.

Show us what you get.

 

[silento]

Your equation is of the form:

$$ \varphi - \beta v^2 = m v\frac{dv}{dx}$$

Where $\varphi$ is a constant.

Make the substitution $u = \varphi - \beta v^2$, and differentiate $u$ w.r.t. $x$.

Show us what you get.

should there be one more of the constant symbols out front or does it not matter? Or can the two constants be combined into one number?

 

[erobz]

should there be one more of the constant symbols out front or does it not matter? Or can the two constants be combined into one number?

If they are both constant just combine them $\varphi$ to work with less symbols. You can add them back in at the end.

 

[Mark44]

I may have forgot to add that v isn't a single constant.

That is obvious, as your equation includes the derivative $\frac{dv}{dx}$.

Also, most of the images you've posted are of poor quality due to bad lighting. This is one of the main reasons we ask members to post using LaTeX rather than images of hand-written work.

It's not that difficult to use the version of LaTeX that is supported on this site. The equation you wrote in post #1 of this thread is
$$\mu_sF_n - \mu_rF_n = mv\frac{dv}{dx} + \frac 1 2 Cd Apv^2$$
This is how what you wrote appears to me.

The code I wrote is $$\mu_sF_n - \mu_rF_n = mv\frac{dv}{dx} + \frac 1 2 Cd Apv^2$$.

At the lower left corner of the text entry pane there is a link to our LaTeX tutorial.

 

[silento]

Your equation is of the form:

$$ \varphi - \beta v^2 = m v\frac{dv}{dx}$$

Where $\varphi$ is a constant.

Make the substitution $u = \varphi - \beta v^2$, and differentiate $u$ w.r.t. $x$.

Show us what you get.

I'm trying to understand this. So when differentiating u, it would be 2Bv. how do I do it with respect to x?

 

[erobz]

I'm trying to understand this. So when differentiating u, it would be 2Bv. how do I do it with respect to x?

Chain rule.

 

[silento]

would I add a dx behind 2Bv? (I apologize, my calc skills are a bit rusty so please bear with me)

 

[erobz]

would I add a dx behind 2Bv? (I apologize, my calc skills are a bit rusty so please bear with me)

Look up the Chain Rule. $u$ is a function of $v$ and $v$ is a function of $x$.

 

[silento]

Look up the Chain Rule. $u$ is a function of $v$ and $v$ is a function of $x$.

(du/dx)=(du/dv)*(dv/dx)

 

[Mark44]

So when differentiating u, it would be 2Bv.

You've omitted the minus sign, plus what @erobz says below.

Look up the Chain Rule. u is a function of v and v is a function of x.

 

[erobz]

Please, try to learn the math formatting language here as @Mark44 already asked. See LaTeX Guide. It's not hard.

 

[silento]

Please, try to learn the math formatting language here as @Mark44 already asked. See LaTeX Guide. It's not hard.

Will do

 

[silento]

would it just be

?

 

[erobz]

would it just be View attachment 340971?

Nope. You have the factor $\frac{du}{dv} = -2 \beta v$ correct. What is the derivative of $v$ w.r.t $x$? Don't over think this.

 

[silento]

Nope. You have the factor $\frac{du}{dv} = -2 \beta v$ correct. What is the derivative of $v$ w.r.t $x$? Don't over think this.

apologies. had to go to work. Ima keep tackling this now

 

[silento]

hold on I think I got something

 

[erobz]

du= -2Bvdv

I was looking for this:

$$ \frac{du}{dx} = - 2 \beta v \frac{dv}{dx} $$

Which is what basically what you wrote (Don't drop the independent variable from the notation).

 

[silento]

Alright...that makes sense. Interested to see where we go from here

 

[erobz]

Alright...that makes sense. Interested to see where we go from here

look at the RHS of that result, and compare it to the RHS of your original equation. Notice anything?

 

[silento]

looks like we moved that drag force over to the right hand side

 

[erobz]

looks like we moved that drag force over to the right hand side

I meant examine similarities between post 16 and post 32.

 

[silento]

. however -2B is in place of the mass

 

[erobz]

View attachment 340974 . however -2B is in place of the mass

luckily, that's just a constant... Quite literally solve for $v \frac{dv}{dx}$ in terms of $\frac{du}{dx}$ in post 32, and plug it into post 16 equation. Don't forget about the $u$ substitution made in 16 either. If you do it correctly, watch the equation transform into a butterfly.

I 've got to try to get to bed though. Good luck.

 

[silento]

 

[erobz]

View attachment 340975

Try again. everything should be in terms of $u,x$..no more $v$ in the equation.

 

[silento]

From post 32

then I plugged that into the answer from post 16 in place of vdv/dx. I don't see where I'm going wrong

 

[erobz]

From post 32 View attachment 340976 then I plugged that into the answer from post 16 in place of vdv/dx. I don't see where I'm going wrong

in post 16 eqn, substitute on the LHS to get rid of $v$ and teh constant... remember I defined $u = \varphi - \beta v^2$.

So you end with a simple ODE in terms of variable $u$ and $x$.

 

[silento]

Oh right!

 

[silento]

wait a minute. I think I see where you are going with this

 

[silento]

correct me in i'm wrong, but we are going to move the du to the LHS. Then we can sub v back in for u. Then velocity would be the only variable on the LHS along with the constants. On the right hand side it would just be 1/-2Bdx which is just constants???

 

[erobz]

correct me in i'm wrong, but we are going to move the du to the LHS. Then we can sub v back in for u. Then velocity would be the only variable on the LHS along with the constants. On the right hand side it would just be 1/-2Bdx which is just constants???

Solve for $u(x)$ from the resulting ODE. After you’ve done that you can change back to $v$ and it’s values etc…

 

[silento]

Im sorry but what's ODE?

 

[erobz]

Im sorry but what's ODE?

Ordinary differential equation. The equation we just developed. Linear. separable, ODE.

 

[silento]

solve for u(x)? I guess I was wrong. I was thinking I could just do

. Then I can plug u back in and change u to v. So, all the v would be on the left while the x is on the right. I can then take the integral of both sides and solve for xf-xi

 

[silento]

(1/2) is to the power. After I get u(x) then I change it back to v then do I integrate? 120 to 0 is bounds for velocity and I need to find ∆x which is just xf-xi which are the bounds for dx(position)