We ended Page 3 with an initial value problem to be solved: Find v = v(t) so that
More generally, our problem is to solve any initial value problem of the form
We will calculate approximate values for the velocity v at n equally spaced points in some fixed time interval. Our procedure is simple: We repeatedly calculate a rise in v as slope x run. Then we add the rise to the current value of v to get the next value of v.
Our goal is to estimate the velocity v(t) at times
Our estimated velocity values at these times will be denoted by
Our method for estimating the velocity values will be recursive, i.e., vk will be calculated from the preceding vk--1 for each k = 1, 2, 3, ... .
How do we obtain v1 from v0, the initial velocity? We will answer this in a geometric fashion. We will look at the graph of velocity versus time on the (t,v)-plane. The following figure shows a graph of the starting situation: the initial velocity v0 is shown as a vertical line segment of length v0 at the starting time t0 = 0.
We now include the graph of v versus t. Our next velocity value, v1, is shown as the length of a vertical line segment at time t1.
However, the value of v1 is not known to us, and hence we will estimate its value. We do this by drawing the tangent line to the graph at t = t0. Follow this tangent line to the point P, the top of a vertical line segment that approximates v1.
We can compute the length of this new line segment: We separate the line segment into two pieces -- the bottom piece having length v0, and the top piece being the rise of a right triangle with Using
we see that rise equals slope times . Hence,
(We use the symbol to mean "almost equal to".)
This is the key to Euler's Method for approximating the solution of an initial value problem. It's valuable because the slope (of the tangent line) equals the derivative dv/dt, which is given by our original differential equation when t = t0 and v = v0:
Substituting this value of the slope into the preceding equation, we find
Great! This gives us a method for going from v0 to v1. But how do we go from v1 to v2? Easy -- we use the same equation, only with v0 and v1 replaced by v1 and v2:
In general, to go from vk - 1 to vk we have
This equation, along with the initial value v0 = 0 and the assignment of a value to the step size , plays the central role in our computations.