We'll solve equations like \( T'(t) = k \big[ T(t) - T_{\text{air}}\big] \) later with calculus, but the direction field gives us a foothold for solving our problem right now without integrating. The process we show here will also help us solve systems of first-order equations and higher-order equations.

Technically, the purple arrows in our direction field are vectors; the components at the point \( (t,T)\) in the \(tT\)-plane are \( h \big\langle 1, k [ T-T_{\text{air}} ] \big\rangle,\) where \( h > 0. \)

We can follow the arrow from \(\big(0,T(0)\big)\) to its very tip, which approximates \( \big(0+h, T(0+h)\big)\) when \( h\) is tiny. At the tip of this vector, we find another direction arrow, which we follow to its tip, and so on. These hops along the arrows give us \[ T(nh) \approx T_{\text{air}} + [1+ k h ]^{n} \big(T(0)-T_{\text{air}}\big),\] which is an approximation for the temperature at the times \(t_{n} = n h, n = 0, 1, 2, 3, \dots\).

The Python codex below computes the first ten approximations and displays the relative error of the approximation for \( T(nh)\) compared to the exact result we'll find soon. You can set the hop size \( h > 0,\) the initial temperature “To”, and the air temperature “Ta”. We set \(k=-1.\)

The larger the relative error, the worse the approximation; the smaller the relative error, the better! Play around with different values for the constants and then select the correct option.

import sys
import numpy

h=10;
To=0.5;
Ta=1;   
n = 1;

while n <= 10: 
        c = numpy.absolute(Ta+(To-Ta)*numpy.power(1-numpy.absolute(h),n)-Ta-(To-Ta)*numpy.exp(-n*numpy.absolute(h)))/numpy.absolute(Ta-(To-Ta)*numpy.exp(-n*numpy.absolute(h)))
        n = n+1;
        print(c)