# Raindrops - Euler's Method

Author(s):
David A. Smith and Lawrence C. Moore

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   v replaced by   v1  and   v2: In general, to go from   vk - 1  to   v 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.