<p>Questions are Up</p>
<p><a href=“http://media.collegeboard.com/digitalServices/pdf/ap/apcentral/ap13_frq_calculus_bc.pdf[/url]”>http://media.collegeboard.com/digitalServices/pdf/ap/apcentral/ap13_frq_calculus_bc.pdf</a></p>
<p>Okay so when I was doing the exam for 6c I got for the coefficient 4f’’(0) which was equal to -8/3. Did anybody else get this?</p>
<p>I had the other form of the test… will these questions ever be released? Will we be graded on the same curve? I didn’t have any diff eq’s, and my area problem wasn’t a polar area one.</p>
<p>is anyone willing to create the answer sheet???</p>
<p><em>joins the club</em></p>
<p>Ok, this may seem ridiculous, but I barely learned about polar stuff so I struggled a bit for 2. .<strong>. For 2c, do you simply find the equivalent of t for theta (with the equation from 2b), plug it in the derivative and get the coordinates from the graph or something? I`m sooo sorry. My teacher breezed by polars and I had no idea how to do them. .</strong>____.</p>
<p>Would anyone like to explain to me why the absolute minimum for 4b was -8?</p>
<p>I thought that since f(x) = integral of f’(x) from 0 to x, if you plug in 0 for x (where the abs min occurs), then you get f(x) = 0.</p>
<p>IM pretty sure it was 0, cause it was the integral from 0 to 0. If it was -8, damn I guess i didnt learn the fundemental th. right</p>
<p>It’s -8. The condition f(8) = 4 implies the position is at 4 at x = 8, and the total value of the integrated curve is 12. Therefore, just work backwards and at x=0, f(0 = f(8) - 12 = 4 - 12 = -8.</p>
<p>Would this be a valid way to do #6 part a?</p>
<p>Since it says show that f’(0) = 2, i set up:
p(x) = -4 + 2x</p>
<p>Then, I let p(x) = -3 and x = 1/2
Which leads to -3=-3
Therefore, f’(0) = 2</p>
<p>I know I probably should’ve shown it another by initially setting up -3= 4 + (1/2)k, but would my process earn credit?</p>
<p>Anyone going to post a google docs answers?</p>
<p>For #4, part b, I didn’t get x=0, or y=-8, but I think it’s a 3 pt question. I identified x=6 as f’(t) =0 in my sentences, and the endpts as candidates, but got the wrong y value for x=0, so I concluded x=6 was the abs min. How many pts do you think I could get here?</p>
<h1>4 part d, I wrote the tangent line and used the wrong pts since it was a line to g(x) so it had to be using g(x) pts by converting the f(1/2) = -5/2. I took the derivative and got g’(3) = 75. Will they ignore the line even though I boxed it as my final answer? I guess I was tired and didn’t read carefully.</h1>
<p>@shining I did the slope formula from P(1/2) to f(0) and i got the same answer xD</p>
<p>I thought the test was okay, I was confident with 15/28 on the first mc and 13/17 on the second one. Besides FRQ’s 2 and 6, I think I did fine. I didn’t even attempt 6 but that’s my own fault for not learning Taylor series lol. So thinking back about everything, I’m giving myself a lowest possible score of ~58 points. Would that be good enough for a 4?</p>
<p>What do you guys think the pts per part will be for #6? I only got part b, I think, and the first term -4x for part c ( forgot to do chain rule).</p>
<p>Did anyone get the other form for the FRQ? Do you know when/if they’ll post it?</p>
<p>what did you guys get for 4c? my friend told me it was [3,4] but I had [1,4] for some reason 
eff me I’m freaking out now cause I feel like I made so many stupid mistakes</p>
<p>@exterminate
I believe I got (3,4) for the interval man. And I think it could’ve also been (0,1)</p>
<p>k. im stupid. i dont belong here lol. bye</p>
<p>It is [0,1] and [3,4]</p>
<p>You’re looking for when f’’ is negative —> f’ is decreasing and f’ is positive —> f is increasing. </p>
<p>Those are the only intervals tht meet the requirements.</p>
<p>at least I got a 9 on #5. that gives me a little hope for my life. maybe I actually have a chance of making over 20k a year when im older :)</p>