{"id":8914,"date":"2017-02-07t22:18:56","date_gmt":"2017-02-08t06:18:56","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=8914"},"modified":"2017-02-07t20:29:47","modified_gmt":"2017-02-08t04:29:47","slug":"ap-calculus-bc-exam-free-response-practice-problems","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/ap-calculus-bc-exam-free-response-practice-problems\/","title":{"rendered":"ap calculus bc exam free response practice problems"},"content":{"rendered":"
there are only six free response problems on the ap calculus exam, and yet we all dread these questions the most! but as long as you know your calculus, these problems are no more difficult than the multiple choice questions.<\/p>\n
in this review, you’ll get an overview of the free response section and a few practice problems to work out. you can find the solutions by scrolling all the way down.<\/p>\n
the ap calculus bc exam<\/a> consists of two main sections,<\/p>\n just like in the multiple choice section, the free response section is divided into two parts. part a consists of two questions and requires the use of a (graphing) calculator. however, no calculator is allowed on part b, which has four questions.<\/p>\n you are required you to write out all of your steps for each free response question, even if you used a calculator program or special feature to find the answer. correct answers without supporting work may not receive credit.<\/strong><\/p>\n another major difference is that the free response questions usually consist of 2\u20144 subproblems. so while there are only 6 problems total, there may be closer to 20 problems if all parts are counted.<\/p>\n an important strategy to keep in mind: if you get stuck, move on! all of the problems are weighted equally and there is the possibility of earning partial credit.<\/p>\n the following four problems are modeled after actual ap calculus bc free response questions, two in each part. on the ap exam, there will be two more questions in part b. for this practice set, give yourself 30 minutes for each part. then, later on, you can go back and finish the parts that you didn’t get to for practice.<\/p>\n time \u2014 30 minutes. a graphing calculator is required for some problems or parts of problems.<\/strong><\/p>\n (c) find the point(s) of inflection. verify your answer.<\/li>\n (a) find the speed of the other car in miles per hour.<\/p>\n (b) at what rate is the angle \u03b8<\/em> increasing in radians per second?<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n time \u2014 30 minutes. no calculator is allowed for these problems.<\/strong><\/p>\n (note, on the actual exam, this part consists of four question in 60 minutes.)<\/em><\/p>\n (c) set up, but do not integrate, an integral expression, in terms of a single variable, for the volume of the solid generated when r is revolved about the line x<\/em> = -2.<\/li>\n (c) find the value of .<\/li>\n<\/ol>\n the solutions below represent what might be acceptable work for a student to show in order to earn full credit. the purpose here is not to provide detailed explanations, but to demonstrate correct mathematical mastery. in fact, i’ve written a little bit more than a typical student might do on the exam so that you can follow my work easier.<\/p>\n use the second derivative test for relative extrema.<\/p>\n g<\/em>\u00a0”(x<\/em>) = 6x<\/em>.<\/p>\n g<\/em>\u00a0”(-1.291) = -7.746 < 0, so there is a relative maximum at this point.<\/p>\n g<\/em>\u00a0”(1.291) = 7.746 > 0, so there is a relative minimum at this point.<\/p>\n plug each critical number into the original function g<\/em> to get the y<\/em>-coordinates: g<\/em>(-1.291) = 9.303, and g<\/em>(1.291) = 0.697.<\/p>\n there is a relative minimum with coordinates (1.291, 0.697) and a relative maximum with coordinates (-1.291, 9.303).<\/p>\n (b) to get the slope of the tangent line at (-1, 9), evaluate g<\/em>\u00a0‘(-1) = 3(-1)2<\/sup> – 5 = -2.<\/p>\n using point-slope form, y<\/em> – 9 = -2(x<\/em> – (-1)), or:<\/p>\n y<\/em> = -2x<\/em> + 7.<\/p>\n (c) set the second derivative equal to 0 to find possible points of inflection.<\/p>\n g<\/em>\u00a0”(x<\/em>) = 6x<\/em> = 0 \u2192 x<\/em> = 0.<\/p>\n since g<\/em>\u00a0” < 0 for x<\/em> < 0 and g<\/em>\u00a0” > 0 for x<\/em> > 0, there is a change in concavity at x<\/em> = 0. this proves that there is an inflection point at x<\/em> = 0.<\/p>\n now g<\/em>(0) = 5, so the inflection point is (0, 5).<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n now taking derivatives with respect to time, we get: .<\/p>\n next, the given information tells us that dy<\/em>\/dt<\/em> = -25, and dz<\/em>\/dt<\/em> = 15. furthermore, we know that x<\/em> = 1\/2 and y<\/em> = 1\/2. therefore, by pythagorean theorem,<\/p>\n <\/p>\n now plug in to the related rate and solve for dx<\/em>\/dt<\/em>:<\/p>\n <\/p>\n thus, the other car is going roughly 46.2\u00a0 mph.<\/p>\n (b) this time, we use trigonometry:<\/p>\n <\/p>\n the rate is in terms of radians per hour, so we must convert to radians per second by dividing by 3600.<\/p>\n final answer: 0.01978 rad\/sec.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n the area is 45\/2, or 22.5 square units.<\/p>\n (b) for a solid of revolution about the x<\/em>-axis, it’s best to use the washer method<\/a>.<\/p>\n <\/p>\n (c) this time we have to revolve around a vertical line, x<\/em> = -2. the most straightforward setup involves the shell method<\/a>.<\/p>\n the radius of a typical shell is x<\/em> – (-2) = x<\/em> + 2.<\/p>\n <\/li>\n<\/ol>\n<\/li>\n<\/ol>\n\n
practice problems<\/h2>\n
free response section, part a<\/h3>\n
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free response section, part b<\/h3>\n
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solutions<\/h2>\n
problem 1<\/h3>\n
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problem 2<\/h3>\n
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problem 3<\/h3>\n
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problem 4<\/h3>\n