<p>I'm so mad at myself. I had that answer but after I finished I went back and redid it because I thought it was wrong. I'm so stupid. Hopefully I still got enough points to get a 5.</p>
<p>how did everyone do on the acceleration piecewise equation one?</p>
<p>I think I did okay on it. I'm pretty sure i got a but i'm not 100% sure about the others but I think i got partial credit on them</p>
<p>I think i got all of the acceleration piece-wise right. I missed the negative on the function for part c for the slope field problem. I'm sure I got full credit for the 1st, 2nd and 3rd one. maybe a little more than half for the max and min one.</p>
<p>i thought that was pretty easy i think i got:</p>
<p>0<=0 <8 a(x)= 2;
8<=x<16 a(x)=0;
16<=x<=24 a(x)=-5/6 (i think thats what i put)</p>
<p>isn't t=4 to t=16 zero acceleration? not t=8 to t=16?</p>
<ol>
<li>for piecewise function, we didn't have to put undefined value, right?</li>
<li>For sand at minimum, was it just "Rate coming in- Rate going out = 0" ?</li>
</ol>
<p>Also, for volume, did you guys do it in two steps?</p>
<p>I did </p>
<p>pi Int( (whatever+1)^2-(whatever+1)^2)
for two pieces. i guess that should work?</p>
<p>Also, for inflection point at g(x), was it 2? or was it undefined?</p>
<p>i got 0 < x < 4 = 5 for acceleration, not 2. I think it's -5/2 not -5/6. -5/6 was the answer to the average rate of change.</p>
<p>I agree with Alex Martin. y = 5x and y = -5/2x + c for the acceleration.</p>
<p>piecewise acceleration: 5, 0, -5/2</p>
<p>also, where is this 2492 for total sand coming from? was that even a question?</p>
<p>Do we have to simplify our fractions to receive full credit on this exam?</p>
<p>grumpybear, I think it's coming from part d of the question, which asked to get the minimum amount of sand on the beach on the interval 0 < t < 6.</p>
<p>Nope, no need to simplify arithmetic.</p>
<br>
<p>Nope, no need to simplify arithmetic.</p>
<br>
<p>Phew... That's what I thought... just checking :)</p>
<p>thanks setzwxman!
what was the answer to the part of the acceleration problem which asked what v'(8) and v'(20) were? I thought v'(8) was not defined because it was undifferentiable and v'(20) was -5/2. does that sound right?</p>
<p>How many points do you get right for just drawing the right slopefield?</p>
<p>When you had to integrate the velocity graph for the acceleration problem, did you have to set up the integral, or was it ok to add up the areas by dividing it into triangles and rectangles and finding the areas of those?</p>
<p>I'm quite happy--considering for 2003 you needed 67/108 to get a 5, I reckon I accumulated 72-75.
As for Part II, not a lot of time for it, but I guess I got 42-45/54. What do you guys think?</p>
<p>Here's all of my answers - mega compilation:
AB1:
a) .065,
b) .410
c) 4.559 (or 1.451*pi)</p>
<p>AB2:
a) 31.81593 cubic yards (units needed),
b) Y(t) = 2500 + fnInt(S(t) - R(t))
c) S(4) - R(4) = -1.90875 cubic yards/minute
d) I had minimum time at 0, and minimum sand = 2,500 cubic yards, but I don't think that is correct</p>
<p>AB3:
a) T'(7) = f(x)-f(c)/x-c = f(8)-f(6)/8-6 = -3.5 C/cm
b) integral expression is fnInt(T(x), X, 0, 8) - of course written as a math expression, not a calc expression; avg temp was 75.625
c) T(8) - T(0) = -45 degrees C; change in temp across the wire from 0 cm to 8 cm
d) No - T'' < 0 for all 0 in the interval 0 < x < 8; for proof, plot points TO SCALE or show that the temp is decreasing at a non-constant rate (compare temp change from 5 to 6 cm and 6 to 8 cm); since it is decreasing at a non-constant rate, T(x) is concave down, and thus T'' is negative.</p>
<p>AB4:
a) Extrema at x = 2, rel. max because f '(x) is pos and then neg before and after x = 2, respectively
b) Graph looks like this:
<a href="http://www.teamtwin.com/AB3Graph.gif%5B/url%5D">http://www.teamtwin.com/AB3Graph.gif</a>
c) g(x) attains relative extremum at x = 3; x = 3 is relative max (all area is above x-axis) - note - it cannot be determined if x = 4 is an extremum, since the graph is not completely discernable after x = 3. If the graph swoops out more negative area, than x = 4 is an relative minimum; however, if the area is less than x = 3, the x = 1 would be the relative minimum of g(x), being 0. I'm not 100% sure on this explination, but it's what I put)
d) Point of inflection is when g''(x) = 0 or DNE - this occurs when the graph of f(t) changes sign, which occurs at x = 1 and x = 3
AB5:
a) area = 360 meters; value of displacement of car
b) v'(4) DNE (sharp turn in curve at point - is derivative 5 m/s or 0 m/s?? we cannot know) - v'(20) is the slope of the line at that point - (20-0)/(16-24) = -5/2 m/s^2
c) a(t) = {5 m/s^2 for 0<t<4; 0 m/s^2 for 4<t<16; -5/2 m/s^2 for 16<t<24}
d) I said the value is guaranteed - V(20)-V(8)/12 = -5/6 - since v(t) is continuous on that interval, there must be some f(c) that equals the average; however, there is debate about this part of the question</p>
<p>AB6:
a) I'm not gonna use paint for this - I'll just describe: horizonal line for all x = 0; (-1,1) = 2; (-1,2) = 1; (1,2) = -1; (1,1) = -2; (-1,-1) = -2; (-1,-2) = -1; (1,-1) = 2; (1,-2) = 1
b) slope at (1,-1) = 2; y = mx + b; -1 = 2(1) + b; b = -3;
y = 2x - 3
c) dy/dx = -2x/y
y dy = -2x dx
y^2/2 = -x^2 + C ; <a href="-1">f(1) = -1 particular solution</a>^2/2 = -(1)^2 + C
1/2 = -1 + C
C = 3/2
y^2/2 = -x^2 + 3/2 OR y^2 = -2x^2 + 3</p>
<p>fin!</p>
<p>Any comments, questions, other answers, just post away!</p>
<p>Collegeboard has already put up the Frqs online
1) For 4) b did you guys get a triangle above the x-axis, and a positive slope below it connecting the left vertex.
2) and for the -5/6 The Mean value theorem does not guarantee the existence of the point right.
3) 3c was just read from the graph problem and subtract right. I got -45
3) Last answer for slope field was (-2x^2+3)^(1/2) right
Verify the answers guys.</p>