{"id":8458,"date":"2017-01-24t16:28:37","date_gmt":"2017-01-25t00:28:37","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=8458"},"modified":"2022-06-14t09:48:46","modified_gmt":"2022-06-14t16:48:46","slug":"ap-calculus-ab-free-response-practice-problems","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/ap-calculus-ab-free-response-practice-problems\/","title":{"rendered":"ap calculus ab free response practice problems"},"content":{"rendered":"
one of the best ways to study for the ap calculus free response questions is to practice with as many ap exam style problems as you can.<\/p>\n
here is a handful of ap calculus free response style practice problems that you can work through, along with the full solutions. these only represent a small fraction of the types of questions that might come up on an ap calculus ab exam; the more problems you can do before the exam, the better\u00a0prepared you will be.<\/p>\n
solution:
\na) whenever we have a velocity question, we should think back to the following relationships:<\/p>\n
<\/p>\n
the total displacement therefore in the integral of the velocity function between the bounds given<\/p>\n
<\/p>\n
this is an integral that can easily be done by hand. however you should learn how to use you graphing calculator to do this question because many integrals will be far more difficult.<\/p>\n
b) slowing down and speeding up are questions about whether or not the acceleration is positive or negative<\/p>\n
<\/p>\n
cos(t) is negative when t = 10. therefore, the object is slowing down.<\/p>\n
c) now we need to take the indefinite integral to get a function of position. do not forget to solve for our constant.<\/p>\n
<\/p>\n
to solve for c, we need to know one point: at time t=10, the car is at a position x = 5.<\/p>\n
<\/p>\n
the initial time, t=0 gives us a position: x(0) = 0-cos(0)-.839 = -1.839<\/p>\n
d) simply find a root for our initial equation for velocity. t = 5.76 is one of several possibilities.<\/p>\n
<\/p>\n
solution:<\/p>\n
a) this function has a relative minimum at x = 0. we should see that it is a minimum because the derivative goes from negative to positive at this point.
\nb) this function has a relative maximum at x = -2.5 we should see that it is a maximum because the derivative goes from positive to negative at this point.
\nc) the inflection points exist where the second derivative is 0. we should see this in the graph as existing around x= \u2013 1.25, x = 2, and x = 4<\/p>\n
solution:<\/p>\n
this is a quintessential optimization problem. it is often a good idea to sketch the situation given. visualizing the problem often helps in finding the correct governing equations. ap exams sometimes do this for you.<\/p>\n
<\/p>\n
the variables in this problem are the lengths of x and y. these are the only two properties which change.<\/p>\n
optimization problem will almost always have two functions involved; in this case the two functions involve area and perimeter.<\/p>\n
<\/p>\n
we know that perimeter p = 50<\/p>\n
we need to maximize a = xy with the constraint 50 = x + 2y.<\/p>\n
we can rewrite the constraint as one function of another. we put this into the area equation to give:<\/p>\n
<\/p>\n
at this point, we should go back to the constraint and figure out the domain of y. in this case, with 50m of fence, y=25m. we will come back to this later.<\/p>\n
we are asked to find the maximum value of a. as in all maximum\/minimum value problems, we find the derivative.<\/p>\n
<\/p>\n
we set this equal to 0 and solve for y. in this case, y = 12.5. if we put these numbers back into the constraint:<\/p>\n
50 = x + 2y<\/p>\n
x = 25<\/p>\n
answer:
\nwe are\u00a0asked for dimensions — the fence is 25m x 12.5m<\/p>\n
solution:<\/p>\n
this problem is an optimization problem, similar to the last one. if we\u2019re ever not sure, first step is always to draw out the situation:<\/p>\n
<\/p>\n
two governing equations:<\/p>\n
equation 1:<\/p>\n
<\/p>\n
equation 2:<\/p>\n
<\/p>\n
we know w = l, so we can use this to simply the first two equations:<\/p>\n
<\/p>\n
<\/p>\n
we now have a two variable problem:<\/p>\n
<\/p>\n
plug into volume equation<\/p>\n
<\/p>\n
any maximization or minimization, we are going to take the derivative and set to zero. in this case, take derivative of v and set to 0.<\/p>\n
<\/p>\n
we can throw out the negative case because it has no physical meaning in this problem. \u00a0always remember, if a question is asking something about the physical world, ask yourself the question ‘does the answer make sense?’ \u00a0if you get answers that seem ridiculous, go back over your work.<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
the box is a cube.<\/p>\n
these four questions represent just a small fraction of the possible types of questions\u00a0that can be asked on the ap exam. check back later to see additional sample multiple choice questions, related rates questions, as well as\u00a0comprehensive reviews<\/a> of a ton of other topics that you might see on the ap calculus exam.<\/p>\n","protected":false},"excerpt":{"rendered":" one of the best ways to study for the ap calculus free response questions is to practice with as many ap exam style problems as you can. here is a handful of ap calculus free response style practice problems that you can work through, along with the full solutions. these only represent a small fraction […]<\/p>\n","protected":false},"author":224,"featured_media":8469,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24930],"acf":[],"yoast_head":"\n