<p>
</a></p>
<p>Solutions for number 6 :).</p>
<p>ARGHHH....the MC part for Calculus BC was soo hard this year. the curve better be lenient!</p>
<p>even my calc teacher told the class after the exam that the 2007 exam was harder than the previous years.</p>
<p>For 6, I might have put 1/2 instead of -1/2. 11/24 doesn't ring a bell, but I don't think I did it any other way, so I hope I just forgot that I put that answer.</p>
<p>Well, I got all of #6 right, except for the lagrange error bound - I used the one for Maclaurin series rather than The alternating series one. You know, Rn <= max f^(n+1) (x-c) ^n+1 / (n+1)! Which was stupid because it took a lot longer than just looking at the next term.</p>
<p>Would this still be correct though? I got a maximum error of 1/300 for the answer, which is different than 1/320, but could I still get credit for it?</p>
<p>I did an improper intergral.. I guess it doesn't matter though, because doing the limit thing would not affect and answer and neither would it be doing something wrong.. I HOPE</p>
<p>Well since my topic got ignored, here are the answers:</p>
<p>1)
a)37.962
b)1871.190
c)174.268</p>
<p>2)
a)8264 cars
b)[0, 1.617)U(3,5.076)
c) t = 3, 5126.591 gallons</p>
<p>3)
a) 10.371
b) -1.732 or -sqrt(3)
c) 0.5</p>
<p>4)
a)y-2 = e^2(x-e)
b)concave up
c)some random stuff, but i remember that C = 2 - 2e^3/9</p>
<p>5)
a)30.8ft
b)7200pift^3/min
c)19.3
d)underapprox</p>
<p>6)
a) don't remember
b) -1/2
c) 11/24
d) max error = 1/320</p>
<p>Kastsm:
The lagrange error bound can be used on all taylor series, but it simplifies to the next term in the series if the series is alternating.
What you did should still work, but you did way more work than was necessary.</p>
<p>welll
left 7 blank on the mc and got 0 points on #3 and #6 for sure
sweeeeeeeet >_<</p>
<p>:P very pretty answer for number 6. I shall now post the answer to number 3:</p>
<p>My teacher didn't give us back the green sheet, but I will try to remember #3. There are two equations, r=2 and r=3 + 2cos(x). They intersect at 2pi/3</p>
<p>a. find the area bounded by the two equations:
(S(a, b) represents integral from a to b)
the area represented would be 2 times the integral of r=2 from 0 to 2pi/3, all of this added to 2 times the integral of r=3 + 2cos(x)
2<em>(1/2)S(0, 2pi/3) 4 dtheta + 2</em>(1/2)S(2pi/3, pi) (3 + 2cos(x))^2 dtheta = 10.371</p>
<p>b. In this next problem, you need to find dr/dt given dr/dt = dr/dtheta
first find dr/dtheta of r= 3 + 2cos(x) (i'm using x as theta so I don't have to spell it out) so dr/dx= -2sin(x)= -1.73 or -sqrt(3), which is dr/dt.</p>
<p>c. find dy/dt given dy/dt= dy/dtheta.
So, find dy/dtheta. y = r*sin(x), so (3 + 2cos(x))sin(x)
Find the derivative, which is: -2sin(x)^2 + (3+cos(x))cos(x), which when you plug in pi/3, you get dy/dt = .5</p>
<p>For b and c, you also had to provide explanations, but as I am not entirely sure what all they are looking for, I will not try to mislead anyone or perhaps embarass myself :(</p>
<p>Ok thanks goodusername. Yeah it took a good ten minutes by the time I simplified it all down. </p>
<p>On number 3, the polar coordinates question, I somehow got part (a) right even though we never learned how to find the area between two polar curves. </p>
<p>I ended up using three different equations to figure it out - I used the area of a semicircle for the part to the right of the y axis, then the area of a polar graph equation for the area between the intersection points, then figured out the theta measure of the 2 sectors I had left over and used Asector = 1/2r^2 theta and multiplied it by two. Wow, I can't believe that worked!</p>
<p>Wasn't 6 (c) 11/24, not 11/12 ?</p>
<p>Yea, it was 11/24, my bad (typo)</p>
<p>Ok, that's good.</p>
<p>quick question
If for #6 I made a mistake in the original series (part a), do I still get points if I carried down that answer for parts b,c,d?</p>
<p>His (yatta!) is right. The radius is (20/(1 + x^2))/2. Then you square that to get r^2. Then you divide by 2 again because it is a semicircle. (this comes out to be dividing by 8) Then you take the integral and multiply by pi.</p>
<p>Noober: I think you made a mistake:</p>
<p>1) You forgot the -2 factor, so it should have been [(20/(1 + x^2) - 2)/2]^2 as the integrand. Then you do the final integral -> pi({integral from -3 to 3} [(20/(1 + x^2) - 2)/2]^2)/2 = 174.268</p>
<p>Hrm. :/ concerning crosscurrent. I know that for physics you can use a wrong value and when you plug it into a formula, though set up right, and come out with an answer, though wrong, you are able to get full points. But.. I am not entirely sure about this. I would have to say probably not, because someone can just say, for instance, the series is represented by, 1+x, which would make the other answers very easy.</p>
<p>Right?</p>
<p>I believe so.. and might I add, very pretty presentation ;D</p>
<p>There ya go Noober ;), and yea, that is an awesome presentation.</p>