<p>can you put +or- for the answer to that last slope field question in front of the radical.</p>
<p>I thought the mean value theorem did guarantee the existance because the function was differentiable and continuous and -5/6 was between 0 and -5/2. not sure on the explanation though.</p>
<p>crypto - for AB 5 i don't think the value is guaranteed because the function was not differentiable (because of the sharp point), i don't know about your AB 4c...i thought the extrema were at x = 1 and x = 3</p>
<p>For 5d since they are asking average "rate of change", isn't it just read from the graph problem, where v(20)-v(8)/(12), is taken. Because he is asking you if there is a v'(x) for this value,and if at all we did take the integral it would give us the average value and not the rate of change.</p>
<p>grumpybear - i see your point on MVT for AB5 - hmmm though...</p>
<p>whoops on AB4 d - you are correct. I had that on my exam - lol. It's not slope of f(t), it IS f(t) thanks ill edit</p>
<p>sbi - oops - i'm wrong, you're right. I have gotten that part wrong.</p>
<p>to all - is my graph similar to the one you all drew??</p>
<p>yeah it looks like mine</p>
<p>My graph had straight lines- is it a major error or not</p>
<p>Also for 4d, isn't it just x=4. Because double derivative of g(x) is f'(x), and the values change only at x=2. Verify this guys?</p>
<p>well i think they told you whether f'' was positive or negative so you could take concavity into account</p>
<p>yeah - straight lines is an error - you have to take concavity into account</p>
<p>Point of inflection is just + to -, or - to + and mention the point right. But they aren't asking concavity here. Any other opinions?</p>
<p>sbi - i meant you have to take concavity into account for graph of f(t)</p>
<p>and now that i think of it, I think points of inflection on g(x) is indeed x = 2; it's when f '(t) changes sign, and that is at x = 2</p>
<p>Here's all of my answers - mega compilation (newly edited):
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 = 2
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 - the avg change in v over 8<t<20 is the integral of v(t) of that (a rect and a trapezoid) times 1/(20-8) = 18.3333 m/s - since v(t) is continuous on that interval, there must be some f(c) that equals the average</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>haha, i do soooo bad</p>
<p>Will they take off points for AB 5a if you didn't use integrals to find the displacement of the car? I just added up the rectangle and two triangles, but i said it was equal to the integral from 0 to 24 of v(t). It didn't say specifically that we had to integrate though, to find the value of that integral.</p>
<p>that test was so basic. SAT I Math basic. and im talking about BC.</p>
<p>Yea, I did horrible... wow I wasn't ready. and to think.. I thought I did awesome on the velocity ?... my only issue being that I wrote my piecewise as as velocity! now of only I had taken their derivatives. Wow, I read that wrong and now Im going to slam my head into a wall.</p>
<p>
[quote]
Will they take off points for AB 5a if you didn't use integrals to find the displacement of the car? I just added up the rectangle and two triangles, but i said it was equal to the integral from 0 to 24 of v(t). It didn't say specifically that we had to integrate though, to find the value of that integral.
[/quote]
</p>
<p>No, that's how you are supposed to do that. It shows you understand the principle of integration, that is, finding the area under a curve. It would have been even easier for you just do the problem as a trapezoid, instead of three figures.</p>
<p>for AB FR 6,
how and what did u get for the approximation of f(1.1)??</p>
<p>and also, isnt the MVT not guaranteed cuz theres a straight line on the graph?? i dont know...</p>
<p>MVT wasnt guaranteed because the graph wasnt differentiable (corner).</p>
<p>the equation of the tangent line at (-1,1) is y = 2x - 3. f(1.1) = 2(1.1) - 3 = -.8</p>
