<p>When are the scoring guidelines released??</p>
<p>wouldnt the answer for 6c be 119. because you needed to square your negatives in sin(x^2).</p>
<p>you can confirm it on your calculator</p>
<p>type in d(sin(x^2)+cos(x),x,6)</p>
<p>then plug in 0, the answer is -121</p>
<p>@anotherindiankid.</p>
<p>i’m pretty sure it IS x, and it isn’t k…</p>
<p>@theyellowboss i guess thats right. so does that mean I would only get that part of the question wrong or would I get part b and c wrong as well because its based upon my wrong answer in part a.</p>
<p>1d. I think I may have done it incorrectly, based on what was posted above, but what I did was… </p>
<p>abs(FnInt(dy/dx),x,0,sqr root of pi) + abs(FnInt(dy/dx),x,sqr root of pi, sqr root of 2pi) + abs(FnInt(dy/dx),x,sqr root of 2pi, 3)</p>
<p>Basically, I found the zeros of the velocity graph as it would change from positive to negative or vice versa at those points and took the absolute value of each part to guarantee that I would get the total area/total distance. Is this right or wrong?</p>
<p>By the way, here is a post of all of the solutions…</p>
<p>[2011</a> AP CALCULUS AB and BC Free Response Solutions [DRAFT]](<a href=“http://www.ilearnmath.net/help/index.php?topic=1611.msg5441#msg5441]2011”>http://www.ilearnmath.net/help/index.php?topic=1611.msg5441#msg5441)</p>
<p>The solutions are on a pdf file attached to the post on this page.</p>
<p>Did anyone take BC Form B? I felt that the FRQ was harder than the past released ones.</p>
<p>I would like to beg that there is an error in the unofficial answer key. For 4b, they did NOT consider the end points. And if you did, three was actually the maximum.</p>
<p>is the test different for internationals??
Cuz I can’t recall any of these questions.</p>
<p>@thechampanon yeah, it’s true they do have to check endpoints, but the answer was definitely x = 5/2. not x = 3. [AP</a> Test 2011 AB/BC4](<a href=“http://home.roadrunner.com/~askmrcalculus/11ABBC4.html]AP”>http://home.roadrunner.com/~askmrcalculus/11ABBC4.html)</p>
<p>Dangit… I must’ve screwed up my algebra.</p>
<p>the max was 3, not 5/2, g(5/2) = 5.25 (2(1.5) + .5(1.5)(2.25)) = 5.25, at 3, g(3) = 6 + 0</p>
<p>does anyone know how to do the polar curve multiple choice quesiton?</p>
<p>Yes! :D</p>
<p>That’s what I thought…</p>
<p>
No, that’s not right. Did you see immadinosaur’s link? You made a mistake in your calculations. Besides, the max at three from a logical point of view wouldn’t even make sense anyway because the relative max of the graph is at 5/2, indicating that g(x) is decreasing from the interval 5/2 < x < 3. So there is no way that the y value at 3 can be greater than the y value at 5/2.</p>
<p>This isn’t even hard, guys. g(x) = 2x + ∫f(x)dx on [0,x]</p>
<p>That last line segment: y = 3 - 2x</p>
<p>g(-4) = something extremely negative that I don’t feel like calculating</p>
<p>g(5/2) = 2(5/2) + (1/2)(3/2)(3) - (1/2)(3/2 + 5/2)|3 - 2(5/2)| = 13/4 = 3.25</p>
<p>g(3) = 2(3) + (1/2)(3/2)(3) - (1/2)(3/2 + 3)(3) = 3/2 = 1.5</p>
<p>Argument settled, let’s move on.</p>
<p>$@#%</p>
<p>I think I wrote that 1/10∫H(t) represents the average rate of change (thinking H(t) was the rate of change function). I thought I was clever because I did a similar problem the night before. Don’t ask why I would think that… considering the data given. I do crap like this all the time.</p>
<p>And I forgot to evaluate the acceleration vector at t=3… I thought they just asked us to find the general vector. </p>
<p>For the diff. equations problem… I think I forgot to bring the whole right side to base e and wrote e^(1/25 t) + ln1100 + 300. </p>
<p>I don’t understand why they have to crunch us for time so much… just so that people who get the material make stupid mistakes and they don’t have to bother coming up with more difficult problems. >_></p>
<p>Other than that I did well on them all, except #6. I didn’t get parts B or C right. =/</p>
<p>(and to the argument about the maximum value… you can use logic to figure out that it is 2.5 and not 3, which is what I did. Didn’t have time to crunch the numbers)</p>
<p>hey guys, is it alright, if for a few problems, such as 6b), i didn’t write an actual numerical value. For 6b, as one of the terms, i wrote (1/3! + 1/6!)*x^6 instead of the actual value.
Same for a few other problems. I had all the other steps right though. Would I get points taken off? My teacher told me I wouldn’t… So just to save time, i didn’t.</p>
<p>Thoughts on the curve?</p>