00:01
So in this question we're given a differential equation.
00:03
Dt by dt is minus k, t minus t -a, where k is a positive constant, k greater than zero.
00:16
And, yeah, so that's, t, t, a is the ambient temperature.
00:24
So let's solve this equation.
00:27
The integral of dt over t minus t -a is minus k -t, and we're integrating from t -0 to t of t.
00:39
So we're going to have log of t of t minus t a over t -nought minus t a is minus k t, where t -nought is the value of the temperature when the time is zero.
00:55
So now we get the temperature at a time t is going to be t a plus t -nought minus t a e to the minus k t.
01:06
So that's our differential equation.
01:14
And we know that at the time of death, so t of time of death is t .d, which is 37 degrees c.
01:36
And we are told that the ambient temperature is 20 degrees c.
01:45
And we're told that when the corpse is discovered, its temperature at t, its temperature of t0 is t0, which is 29 .5 degrees c, and two hours later, it's 23 .5 degrees c.
02:03
So t of 2 is 23 .5 degrees c.
02:08
So let's write this down.
02:10
So t of t is going to be, the ambient temperature is 20 degrees c.
02:21
Plus the difference between the ambient temperature and the temperature that we find in, which is 9 .5 degrees c, e to the minus kt.
02:33
And we're told that t of 2 is 23 .5 degrees c.
02:37
So this is going to give us the fact that 9 .5 e to the minus 2k is 3 .5...