My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Thanks Tree1Thanks
  • 1 Post By SenatorArmstrong
Reply
 
LinkBack Thread Tools Display Modes
November 11th, 2017, 02:00 PM   #1
Senior Member
 
SenatorArmstrong's Avatar
 
Joined: Nov 2015
From: United States of America

Posts: 162
Thanks: 21

Math Focus: Calculus and Physics
thermo

An object is placed in an environment where temperature is kept at a constant
temperature $T_o$. The temperature of the object is initially $T_o$. Now, the object is directly heated by an infra red light beam. Assuming that the temperature of the environment remains $T_o$, the
temperature of the object at time t, T(t), obeys the following differential equation...

$\frac{dT}{dt} = -k(T-T_o) + q$

where k and q are positive constant.

The first term in the right hand side represents cooling by the environment and the second term heating by the beam. Find T(t). What is the final
temperature when time goes to infinity?

I am going to run through how I did this, but it is incorrect. I have a solution for this, but the limits of the integration through the differential equation keep changing to meaningless variables not introduced in the problem or solution which makes for a very confused student.

Anyhow... $T-T_o$ is the change in temperature, so I will call that $\tau$

$T-T_o = \tau \Rightarrow \frac{d\tau}{dt} = -k\tau + q$

This looks like a linear differential equation in the form of

$\frac{dy}{dx} + P(x)y = Q(x)$

I will now find my integrating factor.

$\mu(x) = e^{\int{kdt}} = e^{kt}$

$\frac{d\tau}{dt}*e^{kt} + k\tau*e^{kt} = qe^{kt}$

Then I make not of the product rule going here...

$\frac{d}{dt} \tau e^{kt} = qe^{kt}$ $\Rightarrow$ $\tau e^{kt} = \frac{q}{k}e^{kt}$ $\Rightarrow$ $\tau = T-T_o = \frac{q}{k}$

$T = \frac{q}{k} + T_o$

My question is, why did the solution I provided below use the variable "s" in one of the integrals? Also, why is there a negative in front of the integrating factor? If I had the correct limits of integration, I might have a chance of solving this correctly, but I don't see where these limits of integration are coming from. Can anyone help me out?


Correct solution:

SenatorArmstrong is offline  
 
November 11th, 2017, 02:00 PM   #2
Senior Member
 
SenatorArmstrong's Avatar
 
Joined: Nov 2015
From: United States of America

Posts: 162
Thanks: 21

Math Focus: Calculus and Physics
Sorry about the poor choice in the subject of this post. It was just a place holder, but I hit submit without changing it.
SenatorArmstrong is offline  
November 12th, 2017, 01:21 PM   #3
Senior Member
 
SenatorArmstrong's Avatar
 
Joined: Nov 2015
From: United States of America

Posts: 162
Thanks: 21

Math Focus: Calculus and Physics
Did it a different way and arrived at the same solution. Finally!
Thanks from romsek
SenatorArmstrong is offline  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
thermo



Thread Tools
Display Modes






Copyright © 2017 My Math Forum. All rights reserved.