{"id":10234,"date":"2017-06-13t15:35:33","date_gmt":"2017-06-13t22:35:33","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=10234"},"modified":"2022-06-14t15:35:33","modified_gmt":"2022-06-14t22:35:33","slug":"ap-calculus-bc-review-eulers-method","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/ap-calculus-bc-review-eulers-method\/","title":{"rendered":"ap calculus bc review: euler’s method"},"content":{"rendered":"
what is euler’s method? in this post we review this technique for approximating solutions to certain kinds of differential equations and work out a few examples based on what you may see on the ap calculus bc exam.<\/p>\n
euler’s method is a step-based method for approximating the solution to an initial value problem<\/em> of the following type.<\/p>\n <\/p>\n the given time t<\/em>0<\/sub> is the initial time<\/strong>, and the corresponding y<\/em>0<\/sub> is the initial value<\/strong>.<\/p>\n first, you must choose a small step size h<\/em> (which is almost always given in the problem statement on the ap exam).<\/p>\n then the solution at later times t<\/em>1<\/sub>, t<\/em>2<\/sub>, t<\/em>3<\/sub>, … can be found using the following algorithm.<\/p>\n <\/p>\n it’s helpful to understand a little of the theory behind euler’s method. in fact, a good grasp of the theory helps you in other areas of calculus as well.<\/p>\n the key is that the y<\/em>\u00a0‘ always measures the rate of change<\/strong> of the function y<\/em>. so even if you don’t know what y<\/em> is, if you can tell how fast it’s changing, then that info can help you to build up the values of y<\/em>.<\/p>\n now if y<\/em> changes from y<\/em>0<\/sub> to y<\/em>1<\/sub> in the time interval from t<\/em>0<\/sub> to t<\/em>1<\/sub>, then we can express this fact using the derivative:<\/p>\n <\/p>\n letting the difference in time be h<\/em> = t<\/em>1<\/sub> – t<\/em>0<\/sub>, then the formula becomes:<\/p>\n <\/p>\n the final piece of the puzzle is to use the given information that y<\/em>\u00a0‘ = f<\/em>(t<\/em>, y<\/em>).<\/p>\n that means that you can replace y<\/em>\u00a0‘ by its current value as determined by the function f<\/em>. if you’re only going from t<\/em>0<\/sub> to t<\/em>1<\/sub> (a single step), then you get:<\/p>\n <\/p>\n on the other hand, why stop at just one step? once you know the value of y<\/em>1<\/sub>, then you can use that value in a similar way to find out y<\/em>2<\/sub>. then use y<\/em>2<\/sub> to find y<\/em>3<\/sub>, etc.<\/p>\nwhy does it work?<\/h3>\n