I know how to solve the following problem but I have a confusion concerning the sentence in bold.
A metal bar is taken from an inside room and dropped into a large container of boiling water.The initial temperature of the bar was 20° C. If it is known that the temperature of the the bar increases 2°C in 1 second, How long will it take the bar to reach 98° C?
The solution of the problem is as following
The differential equation is
$\frac{dT}{dt}=k(T_m-T)$
Its solution is
$T=Ce^{-kt}+T_m$
Substitute by $T_m=100$ and by the initial condition $T(0)=20$ in the solution of the DE
$T(0)=100+C$
Hence $C=-80$
Hence $T=100-80e^{-kt}$
To get the constant k, we should substitute by the condition $T(1)=22$ in the solution
$22=100-80e^{-k}$
Hence $k=0.02532$
Then, we can substitute by $T=98$ in the solution to get the required time.
My question is:
Can we deal with the information "the temperature of the the bar increases 2° in 1 second" that this is the rate of change at $T=0$? I mean:
$\frac{dT}{dt}\big|_{t=0}=2$
Hence, we can substitute in the DE
$\frac{dT}{dt}\big|_{t=0}=k(100-T(0))$
$2=k(100-20)$
Hence
$k=0.025$
We got finally approximately the same value for the constant k.
Note:
I think it is not allowed to deal with the given info as a rate because we divided the increase in temperature by the duration which is not exactly the derivative or the rate; we should divide by a duration that is infinitely small.
Also, I think that if the given info is the rate, it should be written in the form 2°c/sec.
I felt confused because when I deal with the given info as a rate, it gives me approximately the same solution.
Best Answer
Your first solution is correct. In the second, you are treating the mixing up instantaneous rate of change and average rate of change. The information given, that the temperature of the the bar increases 2°C in 1 second, is the average rate of change, $$\frac{T(1) - T(0)}{1} = T(1) - T(0) = 2,$$ not the instantaneous rate of change, $\frac{dT}{dt}(0)$. Remember that as the temperature changes, so does the rate at which it changes.
That said, remember that we can use the derivative at a point to give us a linear approximation of our function at a point. For a function $f$, you may have seen this written as $$f(x) \approx f(a) + f'(a)(x - a).$$ For your function $T$, then, we could write $$T(1) \approx T(0) + T'(0)(1 - 0)$$ or after rearranging, $$T(1) - T(0) \approx T'(0)$$ which gives you $$2 \approx \frac{dT}{dt}(0).$$ So, essentially what you did in the second solution, is use an approximation for $\frac{dT}{dt}(0)$, which gave you an approximation of the solution.