<p>What did you guys get for the approximation of the waste in the landfill?</p>
<p>1411 I think.</p>
<p>Sweeet. I accidentally thought that was asking for the integrated equation (the last part) and had to quickly erase it, rewrite it as the last part, and figure out the approximation.</p>
<p>Nevermind…</p>
<p>Staller, we are twins, did the same thing on 5a). For 5b) I forgot to do implicit (it’s W, not t 
); however, it is concave up regardless, so it does underapproximate. Looks like I got some credit there too, yay!</p>
<p>-121 is the best number.</p>
<p>Someone post solutions to <a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board, I’d like to see how well I did ^.^ .</p>
<p>Coming out of the exam I felt like I preformed at a high 4 level, but given the BC Calc curve, bet I got a 5. Pretty familiar with implicit differentiation but still was dumbfounded by #5C, I know I’ll see the solution and just be embarrassed. 6B and C were tough for me as well, but I’ve always sucked with Taylor series. As a whole, exam was quite what I was expecting and very manageable. No complaints.</p>
<p>The answer to #5 c, assuming I recall correctly, was W=1100e^(t/25)+300</p>
<p>^ I got the same. Woooo!</p>
<p>Who thought the one with the graph with the quarter circles was a ***** due to arithmetic (3 or 4).</p>
<p>As has been stated before, #6c was -121, and 6d required you to use the Lagrange error bound formula, in which case your error ended up to be ~1/4000 depending on what value you used from the graph (I used 30).</p>
<p>@jalmoreno yeah that one was a pain to do i think i got -6-9pi/4 for g(3) </p>
<p>G’(X) being 2 + f(x)?
and g(-3) = 2</p>
<p>i got the absolute maximum being 2.5 (idk if correct)</p>
<p>x=0 being point of inflection</p>
<p>and the avg rate of change being -2/7 but it didn’t apply because x=0 wasn’t differentiable (meaning left hand and right hand limits being unequal, that’s what i wrote)</p>
<p>i hope i’m right >.<</p>
<p>I don’t remember, but I’m unsure about X=0 being a point of inflection.</p>
<p>Everything else seems right.</p>
<p>I think I said X=-4 is a point of inflection because the function isn’t differentiable for X=0. That might be wrong though.</p>
<p>f’(x)=g’‘(x) f’(x)=0 at x=0 and x=-4</p>
<p>I can help with 4.</p>
<p>a) -6 + (9pi)/4
b) max at x = 1.5, b/c equation stops increasing here (change from + to -)
c) I dunno, but I guessed that it was at x = -4, 3. I didn’t think it was at x = 0 because f(x) wasn’t continuous at 0.
d) Mean Value Theorem only works if the equation is continuous on that interval.</p>
<p>What do you guys suppose will be the point-spread on each of these problems?</p>
<p>yeah i was wrong just looking at the graph, x=0 can’t be a point of inflection because the concavity doesn’t change wow i must have blanked out on that part lul</p>
<p>g’(x) = f(x) +2</p>
<p>to find critical points, set g’(x) to 0</p>
<p>0 = f(x) + 2
f(x) = -2 at xcoordinate 2.5, not 1.5 </p>
<p>how to find 2.5 …</p>
<p>slope of the line segment = -2
y - 3 = -2 (x - 0) >>> y = -2x + 3</p>
<p>subsitute y with -2 and solve for x to find where critical point is
-2 = -2x +3
-5 = -2x
x=2.5</p>
<p>These are my solutions to the calculator questions anyone correct if you know</p>
<p>1.)</p>
<p>a - speed = sqr rt ((dx/dt)^2) + ((dy/dt)^2) at t = 3
answer - 13.007</p>
<p>b - slope of tangent line at t =3 (dy/dt)/(dx/dt) = sin(9)/13 =
answer = .032</p>
<p>c - find position of the function at t = 3</p>
<p>FnInt(dx/dt,x,0,3) = X(3) - X(0)
21 = x(3) - 0
FnInt(dy/dt,x,0,3) = X(3) - X(0)
.7735 = x(3) - (-4)
answer = (21,-3.226)</p>
<p>d - total distance is equal to
FnInt((dx/dt)^2 + (dy/dt)^2,x,0,3</p>
<p>answer = 21.091</p>
<p>2.)
a -
F(b) - F(a)/b-a
52 - 60/5-2
answer = -8/3 degrees celsius per min</p>
<p>b - the expression represent the average value of the temperature from 0 minutes to 10 minutes</p>
<p>trapezoidal -
(1/10) (2((60+66)/2) + 3((60+52)/2) + 4((52+44)/2) + 1((44+43)/2)) = 52.95 degrees celsius</p>
<p>c - The integral of H’(t) represent the change of temperature of the tea from 0 minutes to 10 minutes. </p>
<p>H(10) - H(0) = -23 degrees celsius</p>
<p>d - </p>
<p>integral of B’(t) from 0 to 10 will give u the change in temperture</p>
<p>100 - 65ish = 34.813</p>
<p>then u do 43 - 34.813 = 8.817</p>
<p>For number 4(c) x=-4 isn’t a solution because the the derivative doesn’t exist at x=-4. The limit as x approaches -4 from the left does not equal the limit as x approaches -4 from the right.</p>
<p>^ those answers are exactly what i got</p>
<p>for #4c, I put x=-3, and x=0, though my teacher told us it was only x=0 because that was when the function changed from increasing to decreasing, and that it the derivative didn’t exist there.</p>
<p>I’m still a bit iffy about him saying it’s not also x=-3, because there’s a vertical asymptote there?..</p>