<p>@ prometheus</p>
<p>For that problem, I can see either 1 Point for Rocket Choice, 2-3 Point for Reason (ie: Knowing what to do)</p>
<p>Rocket B was def going 50</p>
<p>I got Rocket B going at 14 miles per second or whatever the units were. And I asked some of my friends (who are good at Calculus), and they agreed. Oh well.</p>
<p>nope i got 51 and im pretty sure me and promethus are right..14 is way off from rocket A so u know that wouldn't be the right answer lol</p>
<p>Let's compare free response answers
This is what I got for free response 1
(a) find area of region R
1.949
(b) volume of R rotated about y=-3
34.198
(C) expression of R rotated about y-axis
integral from .158 to 3.146 of (2pi(x)(lnx -x +2))dx</p>
<p>yea u got #1 right</p>
<p>i got the same numbers for part a and b tamirms but for c</p>
<p>got</p>
<p>th int from -1.84xxx to 1.14xxx of (y+2)^2 - (e^y)^2 dy</p>
<p>free response 2
(a)
1657.82
(b)
all t that is in [12.428, 16.121]
avg val is 199.426 left turning cars per hour
(c)
A signal should be installed.
Consider t from 14 to 16.
integral of 14 to 16 of L(t) is = 412.265
421.265 * 500 = 206132.87 > 200,000, QED.</p>
<p>hguh1130, that's probably because I used the shells method. Anybody else that used cylindrical shells for part c of 1 that can verify my answer?</p>
<p>can anyone solve 3, 5 and 6?</p>
<p>same answers for the car problem though, thougm my intergral for part C was +/- 1 from the maximum of the function and i think par A said round to the nearest Car..</p>
<p>fyi i took the Ab exam, these question were shared</p>
<p>i got B with 50. just redid the problem and got B > A by 50>49 again. how are you guys getting 51?</p>
<p>for 6, just recalling from the test after having seen the problem once again (but not having redone it)...i remember to have gotten...</p>
<p>a) -1<x<1 b)="" y'(0)="0" y''(0)="15/12" ==""> relative minimum</x<1></p>
<p>fr # 3
(a)
speed = 1.2075, acceleration vector <.3956, -.7407>
(b)
Vertical tangent line when t = .693
(c)
m(t) = dy/dx = (dy/dt)/(dx/dt) = 4t/( (1+t^3)( arcsin((1-2e^(-t)) ) )
lim t-> infinity of m(t) is zero, since denom aproaches to infinity
(d)
c = -3 + integral from 2 to infinity of m(t)</p>
<p>I thought for a) that it could x could equal -1 too?</p>
<p>I think I agree with everything else though</p>
<p>hugh, yeah I overlooked the round to the nearest whole number part for 2(A)</p>
<p>dont feel bad.. i took the intergral of du/sqrt of U as .5* the sqrt of u instead of 2* the sqrt of u :(</p>
<p>besides those 2 what was the third shared question?</p>
<p>fr # 4
(a)
.55 ft/s^2
(b)
integral of v(t) from 10 to 70 is total distance traveled from time interval of 10 to 70 seconds
using subintervals of 10-30, 30-50, 50-70
(20)(22) + (20)(35) + (20)(44) = 2020 feet
(c)
v(t) for rocket b is = 6sqrt(t+1) -4
v(80) = 6*9 -4 = 54 -4 = 50
Rocket a's velocity is 49
therefore, Rocket b is faster</p>
<p>for fr #5
(a) dy/dx = 6, d^2 y/dx^2 = -9
(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.
(c)
-4 + 6(x-1) + -4.5(x-1)^2
(d)
f(0) = -1 + .5(5(.25) +2)</p>
<p>fr # 6
(a) most people probably got radius of convergence = 1 about x=0, the question is whether you can include -1 and/or 1 in the interval.
let x = -1
thus f(x) reduces to an infinite sume of n/n+1, if we take lim as n-> infinity, we find the limit does not equal zero therefore it diverges</p>
<p>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)
(b)
y = f(x) - g(x)
y'(0) = f'(0) - g'(0) = -.5 - -.5 = -.5+.5=0
y''(0)= f"(0) - g"(0)= 4/3 - 1/12, clearly it's greater than zero,
thus the second derivative tell's us it's concave up, the first derivative tell's us it's zero, by definition it's a relative min.</p>