<p>Yep, the average value for anything is always the integral divided by the interval.</p>
<p>How bad is it that I totally disregarded the fact that 700 people were initially at the ride in #3? I got the first 2 parts to the question right because that had nothing to do with the initial value. But for the second two parts, I got 800 and the integral - 800r = 0. I can’t believe I just skipped over that part. Both my answers would have been right had there not been a line initially. Can I still get partial credit for the two problems or will it not count at all because I didn’t factor in the initial line?</p>
<p>You should get partial credit for sure</p>
<p>Can we discuss the multiple choice questions as well? Did all of us (who took form A) have the same multiple choice questions?</p>
<p>never^^ can you discuss mc</p>
<p>How did you guys do number 5 on the FRQ</p>
<h1>2</h1>
<h1>3</h1>
<p>a. find area under the graph from 0-3
3200 people arrived.
double check this its kinda hard to look at graph without drawing on it:p
b. its increasing from 2-3 because the rate at which people are arriving is still greater than the rate at which people are entering the ride.
c.not sure about this one but said
time=3 since until that point the line has been increasing because more people are arriving than going on the ride. at time 3 the line is at its longest and after that point the wait time shortens because line shortens.
d. ∫(0-8)(r(t)-800)dt
when the rate arriving is equal to the rate entering so there is no wait time.</p>
<h1>4</h1>
<p>a.∫(0-9)(6-(2x^(1/2)))dx
b.pi∫(0-9)((7-(2x^(1/2))^2-(1)dx
c.y=2x^(1/2)
x=y^2/4
∫(0-6)((y^2/4)<em>(3</em>y^2/4))dy</p>
<h1>5</h1>
<p>a. g(3) and g(-2)
they give you the graph of g’ so you know the area under the graphs will give you the points of g(x).
g(0)=5 so to find g(-2) just find the area from -2 to 5.
g(-2)=5-pi
same with g(3)
g(0)=5
0-2=pi
2-3=3/2
g(3)=5+pi+3/2</p>
<p>i felt like there was no work for one but they gave us a gigantic box.</p>
<p>b. point of inflection is where derivative of g(x) goes from inc to dec or dec to inc.
happens at x=0 x=2 x=3
justify: at these points the deriv of g’ is 0 or undef and crosses the x axis since the graph of g’ goes from inc to dec or dec to inc.
c. find critical points of h(x)
find h’(x) and set equal to 0.
h’(x)=g’(x)-x
find where x=g’(x)
happens at x=3 and x=2^(1/2)
use a number line and find out if these are local max/min.
3 is neither and 2^(1/2) is maximum</p>
<h1>6</h1>
<p>a. find equation of tangent line at (1,2)</p>
<p>use dy/dx=xy^3 since derivatives are the samw.
y-2=8(x-1)</p>
<p>b. y-2=8(1.1-1)
to explain the over/under you need to find out if the graph is concave up/down
use the second derivative equation((y^3)(1+)(3x^2)(y^))) and find what it is at 1.1.
without actually finding the value you know at 1.1 its going to be positive so the graph is concave up at 1.1 and the tangent line will fall under it so it is an under estimate.</p>
<p>c. dy/dx=xy^3.
1/(y^3)dy=xdx
(-1/2)y^-2=1/2x^2+c
plug in 1,2 to find c. c=-5/8
(-1/2)y^-2=1/2x^2-(5/8)
multiply entire thing by 2 to make numbers nicer
switch around with some algebra
y^2=-1/(x^2-10/8)
get y alone
y=±(-1/(x^2-5/4))^(1/2)
when x=1 y is positive so we know equation is positive
y=(-1/(x^2-5/4))^(1/2)</p>
<p>^ I think that’s 6.</p>
<p>noticed and did 5:p</p>
<h1>1.</h1>
<p>a. ∫(0-6)(f(t))dt
=142.275 cubic feet of snow
b.rate of change at 8
f(8)-108
-59.583 cubic feet per hour
c. h(t)= amount of total snow removed between time intervals
0-6=0 cubic feet
6-7=125 cubic feet
7-9=216 cubic feet
d. ∫(0-9)f(t)dt-341
53.334 cubic feet left</p>
<h1>2</h1>
<p>a. find slope at time 6
7,21
5,13
21-13/7-5=8/2=4 hundres of entries per hour
b. l4=(2(0)+3(4)+2(13)+1(21))
r4=(23+2(21)+3(13)+2(4)
(r4+l4)/2=85.5
85.5/8=10.688
this is the average amount of entries enter per hour from time 0-8.
c. did this one wrong but understand now
23-∫(8-12)p(t)dt=700
d. to find max of p(t) find p’(t)
p’(t)=0
t=9.18, t=10.8165
make a table or whatever you use to find the values.
find values of endpoints also since its asking for absol max.
max is at 12.</p>
<p>On 3A, I DID put 3200 in the beginning, but when I was finished with everything and double-checked my answers, I added 700 to that… ***** ***** *****…</p>
<p>Ok, so how exactly were we supposed to know (in 5C) that sqrt 2 is a critical point, if no point was signaled on sqrt 2</p>
<p>Just superimpose the function y = x onto the graph of g’(x). Anytime time the two functions intersect would be a point where h’(x) = 0 and is thus a critical point. You would see two points, one somewhere along the circle and another at x = 3. Now knowing that the circle is of radius 2 and is centered around the origin, you know that x^2 + y^2 = 4 and for x and y to be equal (the point which the circle would intersect y = x), both x and y would have to equal x^(1/2).</p>
<p>Lame 
thats pretty dumb</p>
<p>Okay quick question…on FRQ 6, I accidentally solved for a in box c and c in box a. I put a huge error between them and wrote SWITCH, as well as crossed out the 6a and put 6c, and vice versa. Will they give me credit? I know I got both of them right…</p>
<p>^ They should notice that.</p>
<p>Depends on the grader…</p>