# A metal rod is 60 cm long and is heated at one end. The temperature at a point on the rod at distance x cm from the heated end is denoted by T. At a point halfway along the rod, T=290 and (dT)/dx = -6?

## (A) In a simple model for the temperature of the rod, it is assumed that $\frac{\mathrm{dT}}{\mathrm{dx}}$ has the same value at all points on rod. For this model, express T in terms of x and hence determine the temperature difference between the ends of the rod. (B) In a more refined model, the rate of change of T with respect to z is taken to be proportional to x. Set up a different equation for T, involving a constant of proportionality k. Solve the differential equation and hence show that, in this refined model, the temperature along the rod is predicted to vary from 380 degrees to 20 degrees

Jan 28, 2018

(a) $T = 470 - 6 x$
(b) $T = - \frac{1}{10} {x}^{2} + 380$

#### Explanation:

(a) If we assume that:

$\frac{\mathrm{dT}}{\mathrm{dx}} = - 6$

then we can directly integrate to get:

$T = - 6 x + C$

However, we know that $T = 290$ half way along the rod, ie when $x = \frac{60}{2} = 30 \setminus c m$, and using this fact we get:

$290 = - 6 \left(30\right) + C$
$\therefore 290 = - 180 + C$
$\therefore C = 470$

$T = 470 - 6 x$

(b) We now assume that:

$\frac{\mathrm{dT}}{\mathrm{dx}} \propto x \implies \frac{\mathrm{dT}}{\mathrm{dx}} = K x$

again, we can directly integrate to get:

$T = \frac{1}{2} K {x}^{2} + C$

We know that $\frac{\mathrm{dT}}{\mathrm{dx}} = - 6$ when $x = 30$. From which we get::

$- 6 = 30 K \implies K = - \frac{1}{5}$

And so:

$T = - \frac{1}{10} {x}^{2} + C$

As before, we know that $T = 290$ when $x = 30$, so: #

$290 = - \frac{1}{10} \left(900\right) + C \implies C = 380$

$T = - \frac{1}{10} {x}^{2} + 380$

Utilising this solution we get:

$x = 0 \implies T = 380$
$x = 60 \implies T = - \frac{1}{10} \left(3600\right) + 380 = 20$

Demonstrating that the temperature changes from ${380}^{o}$ to ${20}^{o}$, QED