<p>
Yeah I did!</p>
<p>Oh gosh, I botched FRQ #5 SO badly. :(</p>
<p>Someone answer my question about the rotated area question please…</p>
<p>Ooohh, gotcha. Sorry!
Haha thanks :)</p>
<p>Ajeck, I’m not certain but I think it was around the y?
I think I did shell method on the test somewhere, and I usually never use shell unless it’s rotated around the y axis.
But I’m not 100% certain!</p>
<p>i dont really want to discuss yet but i didnt do y…</p>
<p>My friend and I both did x, but another person I talked to did y. I’m now very confused.</p>
<p>Was the volume one a FR question? I’m pretty sure it was around the x because they don’t even require you to know shells for the AP exam anymore… I guess you could have changed the variables if it was around the y-axis but I don’t remember what the equation was.</p>
<p>I think that one was a multiple choice, and I’m pretty sure it was around the y.</p>
<p>the MC one was easy because 4/5 choices used an incorrect formula but i thought that the FRQ one was around the x axis</p>
<p>FRQ was around the x.</p>
<p>yah it was around the x axis</p>
<p>Form 4HBP: <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;
Form B: <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;
<p>wow, so 6c did require A LOT of calculation. I just looked at sin(x^2) and thought that it’s sixth derivative as 0 must be 0 :/.</p>
<p>Wouldn’t 6C be simply taking the derivative of the supplied graph? The graph is f^5 so wouldn’t the answer be non-existent because there’s a vertical tangent at 0, or is that only because the absolute value was taken for the graph?</p>
<p>6C required barely any computation at all. Just multiply the coefficient of the x^6 term in the taylor series by 6!, I think it was -121 or something.</p>
<p>the answer was -121 if you were stupid like me and actually did repeated differentiations instead of the shortcut thats what you got after forever ~_~</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>mianahrs, I have the same answers as you do for #1 and #2. Are you going to work out the rest?</p>
<p>this is what i got for 3</p>
<p>3a ) perimeter = 1 + e^2k + k + integral ( sqroot ( 1 + 4e^4k) ) , limits of integration being 0 to k</p>
<p>3b ) discs method : pi * integral ( r ^2 )
radius in this case was simply e^2k
pi * integral (e^2x)^2 , x , 0 , k ) = (pi/4)(e^4k - 1)</p>
<p>3c ) (pi * e^2) / 3</p>
<p>i got what theyellowboss did, but do you think they’ll take off points for using x in the arc length formula instead of k, cuz i forgot to change it</p>