<p>I've recently started looking over some of the AP free-response questions from previous tests and I simply have massive trouble understanding the questions. I've had great understanding of my course's concepts all year (100 the last three quarters), but I simply haven't been exposed to the types of tedious questions the College Board placed on the past exams. Does anyone have any suggestions? I really don't want to wind up with a 3 or 4 on this exam.</p>
<p>I can’t think of anything more than going through the questions and looking at the solutions.</p>
<p>Practice through many FR questions and understand and analyze the steps that lead to the answer.</p>
<p>Can you post a link to one of the questions you don’t understand and explain what you don’t get about it? I (or somebody else) could help explain what you don’t understand.</p>
<p>Seems to me like your class isn’t as rigorous as it should be then.</p>
<p>AP Central has a long list of old AP test free responses that you can practice with.</p>
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<p>Most of the questions pretty much. I’m going over them now and trying to do them on my own, but it can be quite confusing and challenging.</p>
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<p>Although my textbook covers everything (I think), it’s around eleven years old. Most of the folks in my class are actually struggling to pass (and even unjustly whining about the teacher). Then again, I’m not in a school with a majority of very strong math students in general.</p>
<p>Off the top of my head, you should be familiar with:</p>
<p>Limits, functions, continuity, derivatives (all types and application of), antiderivatives (and some techniques, like substitution), definite integrals, FTC, related rates, volume (washer, disc, etc.), differential equations, numerical approximations of integrals, and some other miscellaneous stuff.</p>
<p>I had the same issue with the FRQs at first. I have since gone through every FRQ on AP Central and I now know what patterns exist from year to year. I know exactly what the readers are looking for on each question, and even how many points to expect for certain types of questions i.e. 6 whole points for one differential equation. I would recommend doing this so you feel much more comfortable with the types of questions that will be on there. Fortunately, there are always area/volume questions that are an easy 9, and hopefully there will be a slope field this year, which is another easy 9. Those high earning questions are good to have if you get some hard ones that you can only manage about 3 or 4 points on.</p>
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<p>My book has all of those; and I’m going to do enclosed areas and volumes this week.</p>
<p>Basically, I’ve done a lot more past questions for experience; and I’ve also gone through my current textbook. The problem with my current book is that it does not have examples of the types of questions that the College Board would put on the exam nowadays. What I really need help with (as far I remember), is volume. I’m having trouble with washer problems that do not involve the x-axis and instead involve lines such as x=-2 or y=1 as axes of revolution. Can anyone explain that type of problem to me?</p>
<p>Are you familiar with big R and little r? If you are, just find big R and little r with respect to the off-center axis of revolution.</p>
<p>I understand the big R and little R, but I don’t know what you mean by “respect with”. Is there anything numerical or algebraic that changes with the integral?</p>
<p>Big R = distance from ‘top’ to axis of revolution</p>
<p>Little R = distance from ‘bottom’ to axis of revolution</p>
<p>You’ve apparantly only had an axis of revolution at y=0, so you’re distance is the functions minus 0, which is just your function.</p>
<p>Say I have f(x) and g(x), f(x) > g(x) > 0 (meaning f is the “top”), and I want to rotate around y = 1.
R = distance from ‘top’ to axis of revolution = distance from f(x) to y=1 = f(x) - 1.
r = distance from ‘bottom’ to axis of revolution = distance from g(x) to y=1 = g(x) - 1.</p>
<p>Hopefully now you really understand R and r.</p>
<p>Sorry to jump in here, but @ChemE14:</p>
<p>What if the f(x) function is cut off at say y=8 and it goes above it a small amount. If it asks you to find the volume around the x-axis, how would you do that?</p>
<p>Ex: imagine an upside down parabola in the first quadrant</p>
<p>I’m not sure I understand the question. Could you post an example of a function/functions? ( f(x) = )</p>
<p>Also, there was a reason I put “top” and “bottom” in quotation marks in quotations.</p>
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Nothing is different. You just take the top function minus the bottom function to give you the “height” of the rectangle and dx is your width.</p>
<p>So, ChemE, basically, if I wanted to find the area of (-x^2)+3 (from -2 to 0) at an axis of y=2, I would have to take into account the (x^2)+5 as well as 2^2 since there is no other “bottom” function?</p>
<p>Okay, first I want to make sure I get this question right.</p>
<p>f(x) = -x^2 + 3, rotated around axis of y=2.</p>
<p>So what we have is a single function where part of it is below the axis of revolution, the interval (-2, -1), and part of it is above the axis of revolution, the interval (-1, 0).</p>
<p>First, because this is just one function, we use the disc method, which only concerns big R. See, my explanation about big R and little r was for washer method (when there’s a gap in the solid), and it was based on the assumption that we were dealing with the first quadrant with two functions that never crossed each other and never crossed the axis of revolution. That was why I put the “top” and “bottom” in quotes - for this case, big R is the top and little r is the bottom. But now I’ll do what I probably should have done and define big R and little r much more clearly.</p>
<p>Big R is the larger distance from the solid to the axis of revolution, while little r is the smaller distance from the solid to the axis of revolution. So say the function(s) is/are underneath the axis of revolution, then big R isn’t the function on top - it’s the function on the bottom, since that’s where the greater distance is. And if there is any intersecting going on, you have to change R and r for each interval. It’s easier to show with an example, and maybe I or someone else will post an example using washer method. But for your current example, we stick with the disk method, which just deals with big R.</p>
<p>Back to the problem:</p>
<p>1) Graph the function and axis of revolution.
Graphing these two makes it easier to visualize the problem. Because this problem is odd in that the function crosses over the axis of revolution, we must first identify the point of intersection*. y=2 and f(x) = -x^2 + 3 intersect when x = -1, so we will split the problem into two parts,</p>
<p>V<em>total = V</em>left + V_right</p>
<p>Where V<em>left is on the interval (-2, -1) and V</em>right is on the interval (-1, 0). Looking at the graph, it’s pretty clear why we should do this.</p>
<p>*Actually, we don’t have to split this up for this example, but let’s just continue with what we have.</p>
<p>2) Identify R for each part.
Remember, R is the distance from the axis of revolution to ANY point on the function. If you wanted to know the distance from y=2 to f(-2), you would subtract -1 from 2 to get 3. If you wanted to know the distance from y=2 to f(-1.5), you would subtract 0.75 from 2 to get 1.25. So hopefully you can see that our R, the distance from y=2 to any point on f(x), is </p>
<p>R = 2 - f(x) <---- for V_left</p>
<p>Now let’s move on to our other part of the volume. Hopefully after working out the previous part, you can see that the distance from axis of revolution y=2 to the function at any point is</p>
<p>R = f(x) - 2 <---- for V_right</p>
<p>3) Set up the integral for rotational volume
You should know that the equation for rotational volume using the disk method is</p>
<p>V = pi * [integral from a to b] [R(x)]^2 dx</p>
<p>But we have V<em>tot = V</em>left + V_right, so let’s find out what our V’s are</p>
<p>V<em>left = pi * [integral from -2 to -1] [2 - f(x)]^2 dx
V</em>right = pi * [integral from -1 to 0] [f(x) - 2]^2 dx</p>
<p>I think you can take it from here. Find V<em>left, V</em>right, then add together to get V_tot.</p>
<p>EDIT: Just read what you wrote again, and saw you said “I wanted to find the area…” hopefully that was just a typo and I didn’t type out step-by-step for volume for nothing!</p>
<p>I guess -x^2+4 at y=-2 from -2 to 0 would be better for washer. How would I do that?</p>
<p>Well, washer method is when you have two functions. For example, say you had f(x) = x and g(x) = x + 1, and you wanted to find the volume when the region between f(x) and g(x) is rotated around y = -1. That’s washer method - when you rotate that region, there is a gap in the middle of the solid. Washer method basically takes the volume if the solid were filled and subtracts the volume of the empty space.</p>