![]() ![]() Please view the General Questions Megathread before asking a question. When the AP week comes, please DO NOT discuss the multiple choice section! However, in accordance with the agreements made with the College Board in regards to the release of the FRQ's, you may discuss them on this site when they are released. Everyone has their own specialties! Help with what you know and get help with what you don't is the golden rule. If you would like to share resources for this purpose, message the moderators first.ĭon't try to one-up each other with scores. College Board and many textbook publishers have and continue to send copyright notices when they are shared here.Īdvertising is not allowed without moderator approval. Positive discussion is encouraged.ĭo not ask for or share audit exams or other illegal/copyrighted materials. And we're done.Please try to keep discussions on-topic about AP courses. So this negative 121 mustīe the sixth derivative of f evaluated at 0. Here, x to the sixth over here, you have 6 factorial over ![]() This is going toīe the sixth degree term in Taylor approximation, Of the sixth derivative of f evaluated at 0 times x to Taylor series approximation of f is going to be f prime The Taylor series centered at 0, or at 0, orĪpproximated around 0, the sixth degree term in the Because if you take the derivative, this constant will obviously disappear. Or essentially, we just took the indefinite integral of this side, of 4t plus 1. Out the sixth derivative of f evaluated at 0. Antiderivative of 1 is t, plus t, and then youre going to have a plus a constant. Trying to figure out- so they want us to figure Is the fourth derivative of f evaluated at 0ĭivided by 4 factorial. It's the second derivative of f evaluated at 0ĭivided by 2 factorial. That derivative divided by that degrees, thatĭerivative evaluated at 0 divided by thatĭegrees factorials. Problem 1 BC CALCULATOR a) Speed of the particle at time t3 (x(3))2+(y(3))2(4t+1)2+(sin(t2)) 2 t3 169+(sin(9))213.007 Acceleration vector at t3 x'(3),y'(3)4,2tcos(t2) t3 4,6cos94,5.467 b) dy dx dy dt dx dt sin(t2) 4t+1 sin9 13 0. Is centered around 0, and that's what we care about Term of the Taylor series, its coefficient And if you look at yourĭefinition of the Taylor series right here- and we go intoĭepth on this in another Khan Academy video where we talkĪbout why this makes sense- you see that each degree Non-zero terms of the Taylor series of f. We were able to come up with the first four Opposed to just taking the sixth derivative of Of f about x equals 0 tells us that there Us find the first four terms of the Taylor series The product rule over and over again, and the chain And then to evaluate it at 0,īecause this is x squared here. AP Calculus BC AP Exam Free Response Question Review-Sequences and Series Questions AP Exam 2015 BC 2014 BC 2013 BC 2012 BC 2011 BC 2011 BC Form B 2010 BC 2010 BC Form B 2009 BC 2009 BC Form B 2008 BC 2008 BC Form B Question Statistics Question Mean Score Points Possible Your Score 3.29 3.10 3.34 4.75 3.53 N/A 2.60 N/A 1.79 4.42 N/A Mr. If you just tried to find the sixth derivative ofį, that would take you forever. Of the sixth derivative of f evaluated at 0. ![]() Many mathematicians like to joke that they are lazy, but that laziness is an admirable trait because it is actually an optimization of effort, meaning that you can quickly dispense with one problem so as to move on to another. or AP Calculus AB score of 3 (or equivalent AB subscore on BC exam). Everyone progresses at their own rate, the key is to keep working at it. Past papers are a fantastic way to prepare for an exam as you can practise the. I am also not belittling you either because you may not yet see the implications, or, if you do, chose not to use that knowledge. You need to be an expert at it and do it many times to start to get a feel for the concepts and implications I am talking about. It is one thing to be able to do the "cookbook" calculations, that is, relying on the application of the rules of differentiation and integration that apply to a given situation, and quite another to start to get a feel for the meaning of what you are doing and the implications of those meanings which will help guide you to analytic solutions.ĭon't get me wrong, I am not belittling "Brute Force". In this case, the question, and its several sections, are leading you, step by step, to an analytic solution, which is yours for the taking. Doing it that way is often called the "Brute Force" method. ![]() You could, but you would need to contend with multiple applications of the product rule, things can get messy and that is where errors can be made. ![]()
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