Q15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Newton’s law of cooling states that the rate of change in the temperature T(t) of a body is proportional to the difference between the temperature of the medium M(t) and the temperature of the body. That is, $\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{K}\left[\mathbf{M}\left(\mathbf{t}\right)\mathbf{-}\mathbf{T}\left(\mathbf{t}\right)\right]$ where K is a constant. Let $\mathbf{K}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{04}{\left(\mathbf{min}\right)}^{\mathbf{-}\mathbf{1}}$ and the temperature of the medium be constant, $\mathbf{M}\left(\mathbf{t}\right)\mathbf{=}\mathbf{293} \mathbf{kelvins}$ . If the body is initially at 360 kelvins, use Euler’s method with h = 3.0 min to approximate the temperature of the body after 

(a) 30 minutes. 

(b) 60 minutes.

Step-by-Step Solution

Answer

⦁    The temperature of the body after 30 minutes is approximately 311.7 kelvins.

⦁    The temperature of the body after 60 minutes is approximately 298.2 kelvins.

Euler’s Scheme for an Initial Value problem

Euler’s method is a tool to approximate the solution of an initial value problem at a given point. For a given initial value problem  $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{f}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{,}\mathbf{y}\left({\mathbf{x}}_{\mathbf{0}}\right)\mathbf{=}{\mathbf{y}}_{\mathbf{0}}$, the recursive relation that is referred to as Euler’s scheme is

$$\begin{gathered} {\mathbf{x}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}\mathbf{h} \\ {\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{hf}\left({\mathbf{x}}_{\mathbf{n}}\mathbf{,}{\mathbf{y}}_{\mathbf{n}}\right) \end{gathered}$$

Here, h is the step size.

Approximation of the Temperature of the body after 30 min

The prescribed initial value problem is

$\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{04}\left(\mathbf{293}\mathbf{-}\mathbf{T}\right)\mathbf{,}\mathbf{T}\left(\mathbf{0}\right)\mathbf{=}\mathbf{360}$

The given step size is h = 3.0

Here, 

$$\begin{gathered} \mathbf{f}\left(\mathbf{t}\mathbf{,}\mathbf{T}\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{04}\left(\mathbf{293}\mathbf{-}\mathbf{T}\right) \\ {\mathbf{t}}_{\mathbf{0}}\mathbf{=}\mathbf{0} \\ {\mathbf{T}}_{\mathbf{0}}\mathbf{=}\mathbf{360} \end{gathered}$$

Apply the Euler’s scheme as defined in step 1. 

Putting n = 0, it results, 

$$\begin{gathered} {\mathbf{t}}_{\mathbf{1}}\mathbf{=}{\mathbf{t}}_{\mathbf{0}}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{0}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{3} \\ {\mathbf{T}}_{\mathbf{1}}\mathbf{=}{\mathbf{T}}_{\mathbf{0}}\mathbf{+}\mathbf{3}\mathbf{f}\left(\mathbf{0}\mathbf{,}\mathbf{360}\right) \\ \mathbf{=}\mathbf{360}\mathbf{+}\mathbf{3}\left[\mathbf{0}\mathbf{.}\mathbf{04}\left(\mathbf{293}\mathbf{-}\mathbf{360}\right)\right] \\ \mathbf{=}\mathbf{351}\mathbf{.}\mathbf{96} \end{gathered}$$

Putting n = 1, it gives, 

$$\begin{gathered} {\mathbf{t}}_{\mathbf{2}}\mathbf{=}{\mathbf{t}}_{\mathbf{1}}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{3}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{6} \\ {\mathbf{T}}_{\mathbf{2}}\mathbf{=}{\mathbf{T}}_{\mathbf{1}}\mathbf{+}\mathbf{3}\mathbf{f}\left(\mathbf{3}\mathbf{,}\mathbf{351}\mathbf{.}\mathbf{96}\right) \\ \mathbf{=}\mathbf{351}\mathbf{.}\mathbf{96}\mathbf{+}\mathbf{3}\left[\mathbf{0}\mathbf{.}\mathbf{04}\left(\mathbf{293}\mathbf{-}\mathbf{351}\mathbf{.}\mathbf{96}\right)\right] \\ \mathbf{=}\mathbf{344}\mathbf{.}\mathbf{8848} \end{gathered}$$

Continuing in this process, the result obtained is listed below. 

$\mathit{n}$	${\mathit{t}}_{\mathbf{n}}$	${\mathit{T}}_{\mathbf{n}}$
0	0	360
1	3	351.96
2	6	344.8848
3	9	338.658624
4	12	333.17958912
5	15	328.3580384256
6	18	324.115073814528
7	21	320.3812649567846
8	24	317.0955131619705

9	27	314.204051582534
10	30	311.6595653926299

From the above table, you can see that the approximate value of T (30) is 311.6595653926299. Rounding off to the one-decimal place, the approximate value of T (30) is 311.7. 

Thus, after 30 mins, the temperature of the body is approximately 311.7 kelvins.

Approximation of the temperature of the body after 60 mins

In the previous part, the temperature of the body has already been calculated. In this step, perform the iteration for n = 30 and onwards. 

Putting n = 30 in the iteration scheme, it follows, 

$$\begin{gathered} {\mathbf{t}}_{\mathbf{31}}\mathbf{=}{\mathbf{t}}_{\mathbf{30}}\mathbf{+}\mathbf{h} \\ \mathbf{=}\mathbf{30}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{33} \\ {\mathbf{T}}_{\mathbf{31}}\mathbf{=}{\mathbf{T}}_{\mathbf{30}}\mathbf{+}\mathbf{3}\mathbf{f}\left(\mathbf{30}\mathbf{,}\mathbf{311}\mathbf{.}\mathbf{7}\right) \\ \mathbf{=}\mathbf{311}\mathbf{.}\mathbf{7}\mathbf{+}\mathbf{3}\left[\mathbf{0}\mathbf{.}\mathbf{04}\left(\mathbf{293}\mathbf{-}\mathbf{311}\mathbf{.}\mathbf{7}\right)\right] \\ \mathbf{=}\mathbf{309}\mathbf{.}\mathbf{456} \end{gathered}$$

Putting n = 31, it results

$$\begin{gathered} {\mathbf{t}}_{\mathbf{32}}\mathbf{=}{\mathbf{t}}_{\mathbf{31}}\mathbf{+}\mathbf{h} \\ \mathbf{=}\mathbf{33}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{36} \\ {\mathbf{T}}_{\mathbf{32}}\mathbf{=}{\mathbf{T}}_{\mathbf{31}}\mathbf{+}\mathbf{3}\mathbf{f}\left(\mathbf{33}\mathbf{,}\mathbf{309}\mathbf{.}\mathbf{456}\right) \\ \mathbf{=}\mathbf{309}\mathbf{.}\mathbf{456}\mathbf{+}\mathbf{3}\left[\mathbf{0}\mathbf{.}\mathbf{04}\left(\mathbf{293}\mathbf{-}\mathbf{309}\mathbf{.}\mathbf{456}\right)\right] \\ \mathbf{=}\mathbf{307}\mathbf{.}\mathbf{48128} \end{gathered}$$

Continuing in this process, the result obtained is listed below.

$\mathit{n}$	${\mathit{t}}_{\mathbf{n}}$	${\mathit{T}}_{\mathbf{n}}$
0	30	311.7
1	33	309.456
2	36	307.48128
3	39	305.7435264
4	42	304.214303232
5	45	302.86858684416
6	48	301.6843564228608
7	51	300.642233652117504
8	54	299.725165613863404

9	57	298.918145740199796
10	60	298.20796825137582

From the above table, you can see that the approximate value of T (60) is 298.20796825137582. Rounding off to the one-decimal place, the approximate value of T (60) is 298.2.