00:01
In this problem, we want to find a function f of t that describes the amount of sulfate present in the dog after t minutes.
00:10
We are told that at any time the rate of change of f of t is proportional to the value of f, which means that we have that f prime of t is going to be equal to some constant k times f of t.
00:29
So we want to solve for f and also determine our constant.
00:32
And to determine this constant, we are given two initial conditions.
00:37
We are told that initially 8 units of sulfate were injected into a dog.
00:44
So we have that at a time of t is equal to 0, f is equal to 8.
00:53
And after 50 minutes, only 4 units remained in the dog.
00:58
So we have that f of 50 is equal to 4.
01:08
So with these initial conditions and this differential equation, we should be able to find the formula f.
01:13
Let's first solve our differential equation.
01:16
Let's bring our f term to the left -hand side.
01:20
And we write this prime as the f over d t...