<ol>
<li><p>Contribute.
Here's the link:</p></li>
<li><p>Only for internationals: The general consensus is that, this exam, was much harder than Form A.</p></li>
<li><p>Thinking of curve of 62? Yes.</p></li>
</ol>
<p>Please help!</p>
<p>AP:</a> Calculus AB</p>
<p>click 2010, form B.</p>
<p>Thanks!</p>
<p>Did anyone get number 2 part c??</p>
<p>More like how the heck do you solve all of number 2? I bsed the whole thing… I mean like in part a, don’t you solve that by graphing the second function and seeing when the lines intersect the x axis? It intersects at like more 100 points lol.</p>
<p>For 2a, I think all you do is just graph g’(x), which is given, and find the zeroes within the given interval</p>
<p>That’s what I’m saying. The thing is, the graph intersects the x axis at A LOT of points, like at least 25 points… Try graphing the function yourself.</p>
<p>Also, for 4d:</p>
<p>(d) Write expressions for the squirrel’s acceleration a(t), velocity v(t), and distance x(t) from
building A that are valid for the time interval 7 < t < 10.</p>
<p>I thought this question was so vague… I expressed v(t) as the integral of a(t) and s(t) as integral of v(t), but I didn’t know what to do with a(t)…so I just but a bunch of different answers haha I think I wrote down v’(t) and even included some kind of equation that I solved. They should have been more specific about the “expression” lol</p>
<p>Malfunction, are you sure you set your window to [0.12,1]? I remember getting 2 or 3…</p>
<p>holy crap. form b looks like the easiest **** ever. that ****es me off so freaking much.</p>
<p>dayadam: That’s what I thought when I saw the Form A questions.</p>
<p>I didn’t do the AB but I can help you with #2.</p>
<p>For 2. a) you just set g’(x) = 0 and solve. You’ll get x + 1/x = pi or x + 1/x = 2pi or x + 1/x = 3pi (the rest are irrelevant in the given domain). Each should give you one solution for x.
b) concave downwards => g’‘(x) < 0. Solve the inequality.
c) first you need to say g(0.3) = g(1) + integral (g’(x) dx) from 1 to 0.3, and find that with the calculator. Then find g’(0.3), and the equation is y - g(0.3) = g’(0.3)(x - 0.3)
d) you need to show that for 0.3 < x < 1, g’(x) is either increasing or decreasing (and this, I suspect, you get from b), but I don’t know without actually solving it). From that, you can say that if g’(x) is decreasing, the tangent is above, or the other way around.</p>
<p>Feel free to correct me, though. =)</p>
<p>@oceanangel
We had exactly (I think so, at least) the same question in Calc BC. You’re supposed to find the slope between 7 and 10 for a(t), and you get a(t) = -10. For v(t), you just write the equation of a straight line passing through either point with slope = a(t). For x(t), you need to first find the area under the graph from 0 to 7, then express the area under the graph from 7 to t in terms of t (which would be integral of v(t) dt from 7 to t). Add the 2 expressions and you get x(t).</p>
<p>hmmm I don’t remember what I got specifically for 2c but I did it without a calculator (ran out of time) and got this huge thing because I didn’t simplify it lol</p>
<p>I think that’s what I did for 2c. But for some reason I kept getting some pretty ■■■■■■■■ answers lol. I’ll try going over this again some time.
EDIT: UGH I was looking at the interval starting from 0.012 not 0.12 pffffff. There’s only two points. 
Careless errors FTW! Oh well, I just hope I get some partial credit on this question…</p>
<p>@estrat: Yup I remember getting -10 for a(t). For v(t) I didn’t know what they wanted, so I just wrote integral of a(t) AND the equation of the line haha I don’t know what the graders will do with that >.< But for s(t), I think I just put the integral, but I don’t remember adding 2 integrals… I think I just put integral of v(t) from 7 to 10…AHH!! =(</p>
<p>Oh, did anybody get 660 on that midpoint approximation thingy? Then something like 255, then subtracting them, right?</p>
<p>I remember 660, but not 255…is that the same question?</p>
<p>Yeah, part b of the same question. Then part three asks for the difference or something (water going in and water going out).</p>