Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm solving a differential equation problem set and I bumped into the following DE problem where I got few question marks:

The temperature of a body at time $t$ is $T(t)$ and the temperature of its surrounding environment is $T_{env}$. In a small change in time $t$ the temperature change of the body $T(t)$ is proportional to the change in the amount of time $t$ and to the to difference between the temperature of the body $T(t)$ and the temperature of the environment $T_{env}$. Using this information (A) solve the differential equation for temperature $T(t)$, (B) Solve $T(t)$ when the initial conditions are: $T(0) = 10\,^{\circ}\mathrm{C}$ and $T_{env} = 50\,^{\circ}\mathrm{C}$

Now I actually already have an answer for this, it is (by the Newton's law in temperature change):

$$\frac{dT}{dt} = k(T(t)-T_{env})$$

My question is about the form of the question itself. If you were to present this question to a person with no prior knowledge of Newton's law in temperature change, would the phrase:

In a small change in time $t$ the temperature change of the body $T(t)$ is proportional to the change in the amount of time $t$ and to the to difference between the temperature of the body $T(t)$ and the temperature of the environment $T_{env}$

give all the necessary information for the problem solver to be able to deduce (without no prior knowledge) that the connection is:

$$dT = k*dt*(T(t)-T_{env})$$

The problem states that:

The temperature change of the body $T(t)$ is proportional to the change in the amount of time $t$ and to the to difference between the temperature of the body $T(t)$ and the temperature of the environment $T_{env}$

But it doesn't explicitly state that the connection is:

$$dT = k*dt*(T(t)-T_{env})$$

It doesn't say that:

The temperature change of the body $T(t)$ is proportional to the multiplication of the change in the amount of time $t$ and to the to difference between the temperature of the body $T(t)$ and the temperature of the environment $T_{env}$

Without saying this explicitly can someone explain why couldn't it as well be e.g.:

$$dT = k\left[dt + (T(t)-T_{env})\right]$$

Of course the correct answer is more intuitive, but if you were a newbie in the field, how would you recognize this connection? Or would the problem need to be phrased more carefully?

For example:

If I would follow the same style as the problem I gave above, I could give another problem by:

A variable $x$ is proportional to variables $a$, $b$ and $c$. Solve $x$

Don't you think the connection should be told more explicitly to the reader if it is assumed that there is no prior knowledge of the context?

Thnx for any help :)

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Your doubt is lecit: let us check what does it happen if we consider the case $dT=kdt+kT-k\tilde{T}$. A Physicist would probably start with the difference equation (let us put $k=1$ for simplicity)

$$\Delta T(t)=\Delta t+T(t)-\tilde{T},~~(*)$$

for all $t\geq 0$ and check if its solutions are somehow compatible with the physical reality, where $\Delta T(t)=T(t)-T(t_1)$ and $\Delta t=t-t_1$. Choosing for example $t_1=0$, i.e. the starting moment of our analysis of the physical system, we obtain

$$T(t)-T(0)=t+T(t)-\tilde{T},$$

which leads to the "nonsense" $t=T(0)-\tilde{T}$. We can then deduce that the difference equation $(*)$ leads to no sound theory of the evolution of $T(t)$.

We are left to the interpretation $dT=kdt(T-\tilde{T})$ of the text in the OP. Even in this case we have to compare the analytical results with the physical reality: after all, we did not specify the sign of the constant $k\neq 0$...

share|improve this answer
    
+1 Thank you for your help :) –  jjepsuomi Dec 4 '13 at 11:33
1  
you are welcome! Try to solve the DE and analyse it w.r.t. the sign of $k$ and the difference $T(0)-\tilde{T}$ ;-) –  Avitus Dec 4 '13 at 11:34
1  
I will :) thnx ;D –  jjepsuomi Dec 4 '13 at 11:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.