<p>for fr#5 (d) why you use -1 instead of -4</p>
<p>also for #5 (b) if you plug y=0 in. the derivate is 5x^2+3 which is always large than zero. but in order for the function the tangent x axis the derivate has to be 0 at some point.</p>
<p>youknowme,
you're right I screwed up on 5b.</p>
<p>this is what I did for 5 d
given: f(-1) = -4
f(-.5) = -4 + .5<em>f'(-1,-4) = -4 +.5</em>6 = -1
f(0) = -1 + .5<em>f'(-.5,-1) = -1 + .5(5</em>.5^2 - 6/(-1-2))
f(0) = -1 + .5(5*.25 +2)</p>
<p>wow, can't believe i actually got one question right.</p>
<p>5 b) dy/dx = 0, y= 0</p>
<p>This can never happen. If you do the math, you get the square root of a negative #</p>
<p>bah i am so dumb, haha i only did the euler's question up to f(-.5)... forgot to do the last approximation. how much do you think that questin was worth.. i'm surprised though, overall i got a lot of the same answers.. i'm thinking i got around a 40-45 on the FRQ, which is much, much higher than I expected</p>
<p>(b) it's possible for the x-axis to be tangent to the graph, since the the y position at x=-1 is -4, but since the slope is positive that means the y position will increase from -4. The slope is also decreasing according to the second derivative, so the slope could concievably decrease to zero at the same time the y position hits zero.</p>
<p>i said it was impossible, cuz for x-axis to be tangent y has to be 0, and dy has to be 0 as well, but those two conditions are never satisfied at the same time if u plug it into the equation.</p>
<p>sorry, didn't see bongo's post.</p>
<p>(d)
c = -3 + integral from 2 to infinity of m(t)</p>
<p>could u explain how you got that? that's the one thing i didn't get</p>
<p>Damn, I messed up on 3 (d) as well.
it should be
c = -3 + integral from 2 to infinity of dy/dt</p>
<p>in order for there to be a horizontal asymptote, dy/dx must approach zero, this can be observed visually, since at horizontal asymptotes the graph tends to plateu and level off because the slope is approaching zero.</p>
<p>So if we were to take the indefinite integral of dy/dt from t=2 to infinity that would give us change in y from when t = infinity to when t=2. Remember when t= infinity the slope is zero so the actual y position at t = infinity would be equal to the equation of the asymptote. The reason I add -3 is because at t = 2 y is at -3.
When we take the indefinite integral it is:
(y value evaluated at t= infinity) - (y value evaluated at t = 2)
so when we add -3, what we're doing is
(y value evaluated at t= infinity) - (y value evaluated at t = 2) + (y value evaluated at t = 2)
which gives us our desired value.</p>
<p>dangggg how do you guys remember all these answers from the test?
right after i take the test, I tend to forget almost EVERYTHING.. probabaly my brain's adjusting to make room for the next information to come in...but i mean, i don't remember anything except that the rocket B was going 50 :p</p>
<p>and yea. non-calculator part of the MC was definitely harder than the calculator active one... Im hoping for a 5, but not too sure now. PLZZZZZZ 5555555555!!!!!!!!</p>
<p>for 3d, what if you just put integral from 0 to infinity of dy/dt instead, shouldn't that be the same thing? (theoretically, even though you don't know what's at 0)</p>
<p>wow....I should never have signed up for this class. They mutiple choice killed me and I skipped 10 overall. On the free response I think I might get about 50% of the points. I think I got 51 for rocket b too but who knows. If I get a 3 I'll be happy....sad</p>
<p>back to the problem on the interval of convergence.. are you sure it's (-1,1)? I got (-1,1] or something</p>
<p>i got (-1,1) bc the first condition for the alternating series test of convergence is that it lim as n-> inf must equal 0...but if i recall correctly, it didn't...</p>
<p>I took form B (international)...the non calculator FR was horrible. Just take a look at the last FR on that and you know what I mean.</p>
<p>A question for ya'll...does the curve of 65% = 5 apply for for the WHOLE test or just the BC parts of the test? And how are the AB subscores given?</p>
<p>happyggal,
lol, I didn't remember all these questions, all the free response questions have been availible on college board since yesterday afternoon.</p>
<p>prometheus,
if you took the integral from 0 to infinity it would only be correct if the original function, y(t), passed through (0,0).
Take this example:
lets say at t= infinity y=6. Thus the horizontal asymptote would have the equation of y=6. However, let's also say that at t=0, y=1.
So if you took the indefinite integral of dy/dt from 0 to infinity, you would obtain:
6-1 = 5
y=5 clearly is not the equation of the assymptote.</p>
<p>it's (-1, 1]</p>
<p>-1 works by alt series test</p>
<p>Bongo,
when you let x = 1
yes it does alternate signs, HOWEVER a of n+1 is always greater than a of n (ie 5/6 >4/5) so it violates the alternating series assumption, and like when x=-1, the lim as n--> infinity is not zero therefore it diverges
so the interval is (-1,1)</p>