Is $x$ an Increasing Function of $t$?

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Discussion Overview

The discussion revolves around whether the function $x$ is an increasing function of $t$ given the differential equation $\frac{dx}{dt} = \frac{1}{x}$ with the initial condition $x = 3$ when $t = 0$. Participants explore the implications of the equation, the behavior of the function, and the graphical representation of the solution.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about the graphical representation of the function, particularly regarding the negative root and its reflection properties.
  • Several participants derive the solution $x(t) = \sqrt{2t + 9}$ and discuss the implications of this solution on the behavior of $x$ as $t$ increases.
  • There is a discussion on why only the positive square root is considered, noting that $\frac{1}{x}$ does not exist for $x = 0$, which affects the solution's domain.
  • One participant suggests that the faint graph drawn by the professor might represent an imaginary part of the solution, although the purpose of its inclusion is questioned.
  • Concerns are raised about the graph not passing through the point (0, 3) and whether this is due to the general solution being presented without initial conditions.
  • Participants question the behavior of the right-hand side of the ODE and its sign, leading to further inquiry about the conditions under which $x$ is increasing.

Areas of Agreement / Disagreement

Participants generally agree that $x$ is an increasing function of $t$ based on the initial conditions and the nature of the differential equation. However, there is no consensus on the graphical representation and the implications of the negative root, leading to ongoing confusion and debate.

Contextual Notes

Participants note limitations in understanding the graphical representation, particularly regarding the treatment of initial conditions and the implications of the negative root. There is also uncertainty about the inclusion of the faint graph and its relevance to the solution.

Who May Find This Useful

This discussion may be useful for students studying differential equations, particularly those interested in the behavior of solutions and their graphical representations in relation to initial conditions.

MermaidWonders
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Question - True or False: If $\frac{dx}{dt}$ = $\frac{1}{x}$ and $x$ = 3 when $t$ = 0, then $x$ is an increasing function of $t$.

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I understand how the graph of $x$ was obtained (the graph on the board), but I really don't understand why she attempted to draw the negative root of $x$ the way she did in those faint dash lines (if you can see them, they actually try to show a the function $-x$). In other words, why is the $-x$ function a reflection in the $y$-axis of function $x$? Why isn't it reflected in the $x$-axis?
 

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MermaidWonders said:
Question - True or False: If $\frac{dx}{dt}$ = $\frac{1}{x}$ and $x$ = 3 when $t$ = 0, then $x$ is an increasing function of $t$.
I understand how the graph of $x$ was obtained (the graph on the board), but I really don't understand why she attempted to draw the negative root of $x$ the way she did in those faint dash lines (if you can see them, they actually try to show a the function $-x$). In other words, why is the $-x$ function a reflection in the $y$-axis of function $x$? Why isn't it reflected in the $x$-axis?
Yes, if dx/dt= 1/x then x dx= dt. Integrating both sides, (1/2)x^2= t+C. Since x= 3 when t= 0, 9/2= C. (1/2)x^2= t+ 9/2 or x^2= 2t+ 9. Solving for x= sqrt(2t+ 9).

Strictly speaking, there is no "y" in the problem so there is no y-axis! You are really asking "why is only the positive square root taken and not x= -sqrt(2t+ 9)?" The difficulty is that 1/x does not exist for x= 0 so dx/dt= 1/x does not exist at x= 0. The initial value, x= 3 when t= 0 is in the upper half plane and the solution cannot be extended past x= 0.
 
I believe the faint graph is the imaginary part of the solution. I don't know why it would be included though.

Personally, I wouldn't even compute the solution to answer the question...given that $x$ is positive at $t=0$, and the fact that the RHS of the ODE is not going to change sign, we must therefore have that $x$ is an increasing function of $t$.
 
Country Boy said:
Yes, if dx/dt= 1/x then x dx= dt. Integrating both sides, (1/2)x^2= t+C. Since x= 3 when t= 0, 9/2= C. (1/2)x^2= t+ 9/2 or x^2= 2t+ 9. Solving for x= sqrt(2t+ 9).

Strictly speaking, there is no "y" in the problem so there is no y-axis! You are really asking "why is only the positive square root taken and not x= -sqrt(2t+ 9)?" The difficulty is that 1/x does not exist for x= 0 so dx/dt= 1/x does not exist at x= 0. The initial value, x= 3 when t= 0 is in the upper half plane and the solution cannot be extended past x= 0.

Wait... so does $x$ depend on $t$? Then why doesn't the graph show the point (0, 3) due to the fact that $x$ = 3 and $t$ = 0? Put another way, why does the graph pass through the origin instead?

- - - Updated - - -

MarkFL said:
I believe the faint graph is the imaginary part of the solution. I don't know why it would be included though.

Personally, I wouldn't even compute the solution to answer the question...given that $x$ is positive at $t=0$, and the fact that the RHS of the ODE is not going to change sign, we must therefore have that $x$ is an increasing function of $t$.

How do you know that "the RHS of the ODE is not going to change sign" everywhere else?
 
$\dfrac{dx}{dt} = \dfrac{1}{x} \implies x = f(t)$

solving the DE with the given initial condition $(0,3)$ yields ...

$x(t) = \sqrt{2t+9}$

$\dfrac{dx}{dt} = \dfrac{1}{\sqrt{2t+9}} = \dfrac{1}{x}$
 
skeeter said:
$\dfrac{dx}{dt} = \dfrac{1}{x} \implies x = f(t)$

solving the DE with the given initial condition $(0,3)$ yields ...

$x(t) = \sqrt{2t+9}$

$\dfrac{dx}{dt} = \dfrac{1}{\sqrt{2t+9}} = \dfrac{1}{x}$

Wait... so why didn't my prof draw the graph like the one you have here (which actually passes through (0, 3))? Is it because her graph is the general solution without taking into account the initial conditions?
 
Now, I'm still confused why my professor drew the very faint dashed graph that is the reflection of the "main" graph there... :(
 
Wait... so why didn't my prof draw the graph like the one you have here (which actually passes through (0, 3))? Is it because her graph is the general solution without taking into account the initial conditions?

I have no idea.

The chalkboard graph is when $C = 0$, $x = \sqrt{2t}$

the reflected graph is $x = \sqrt{2(-t)}$ ... I'd ask your prof for further clarification
 
skeeter said:
I have no idea.

The chalkboard graph is when $C = 0$, $x = \sqrt{2t}$

the reflected graph is $x = \sqrt{2(-t)}$ ... I'd ask your prof for further clarification

OK, I'll do that. Thanks!
 

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