Help with the Separation of Variables and Integration

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Homework Help Overview

The discussion revolves around the separation of variables and integration in the context of physics, specifically dealing with forces and motion. Participants are exploring how to set up and solve a differential equation involving changing velocity and forces acting on an object.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various approaches to separating variables and integrating, with some expressing confusion over the complexity of the problem. There are questions about the nature of velocity as a changing variable and the implications of integrating with respect to different variables.

Discussion Status

The discussion is active, with participants providing guidance on using substitutions and the chain rule. There is a recognition of the need to clarify definitions and assumptions, particularly regarding the constants and variables involved in the equations.

Contextual Notes

Some participants mention constraints such as the bounds for velocity and the need to find the change in position (∆x). There is also a reference to the importance of using proper mathematical formatting in the discussion.

  • #61
\begin{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}
 
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  • #62
silento said:
\begin{equation}
\ln\left|\frac{u_f}{u_o}\right| = -\frac{2B}{m}x
\end{equation}
exponentiate both sides, and move ##u_o## over to the right hand side.
 
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  • #63
\left|\frac{u_f}{u_o}\right| = e^{-\frac{2B}{m}x}
 
  • #64
am I typing something wrong?
 
  • #65
silento said:
am I typing something wrong?
Delimiters? And you can lose the absolute value bars. The RHS is always positive.
 
  • #66
\begin{equation}\left| \frac{u_f}{u_o} \right| = e^{-\frac{2B}{m}x}\end{equation}
 
  • #67
silento said:
\begin{equation}\left| \frac{u_f}{u_o} \right| = e^{-\frac{2B}{m}x}\end{equation}
The absolute value bars are no longer necessary. The RHS is always positive. Move ##u_o## over.
 
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  • #68
\begin{equation}u_f = u_o \cdot e^{-\frac{2B}{m}x}\end{equation}
 
  • #69
silento said:
\begin{equation}u_f = u_o \cdot e^{-\frac{2B}{m}x}\end{equation}
Now evaluate ##u_f## and ##u_o## using the corresponding ##v_f## and ##v_o##. Make sure you are evaluating ##u##. I repeat ##u## is not ##v##.
 
  • #70
erobz said:
Now evaluate ##u_f## and ##u_o## using the corresponding ##v_f## and ##v_o##. Make sure you are evaluating ##u##. I repeat ##u## is not ##v##.
Oh I just realized you are trying to solve for ##x##, not ##v##. Anyhow just finish it out, the solve for ##x##. I realized this is undoing a few steps. My bad.
 
  • #71
\begin{equation}
u= a - B \times (54)^2
\end{equation}
 
  • #72
thats uf, a represents the constant
 
  • #73
ui is just a since v=0
 
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  • #74
silento said:
\begin{equation}
uf
= a - B \times (54)^2
\end{equation}
That’s just ##u_f##.don’t forget ##u_o##.
 
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  • #75
I think you can figure it out from here.
 
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  • #76
wait hold on its the other way around
 

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