<p>48 hours have passed, let's discuss the questions. CB hasn't put the FRQ's online, though.</p>
<p>Here's what I remember of FRQ's:
1. Area between curves/ rotation. It was not an improper integral, you had to change viewing window. Pretty easy.
2. Was this the water one? If so, it was ok. The most water was at t=3, right?
3. Polars and parametrics. Area between polar curves, motion along curve, parametrics stuff. I had no idea what to do for the last 2 parts :(.
4-5. Don't remember. They were alright, I think.
6. Taylor series thing, error bound. Good thing I learned them 2 days before the exam :D.</p>
<p>4 was a differentiation thing and 5 was the radius problem.</p>
<p>and for 3, i didnt put t=3 because the graph was not differentiable at t=3. even if u said it was, at that point, it could have either been 250 or 2000. so, i took the derivative and got 6.067. not sure if im right tho, i have to look at the problm once again.</p>
<p>and for the convergence divergence problem, it would go to zero, because 1/ infinity approaches zero. u had to do the limit, u cant just take the integral of infinity.</p>
<p>and for the polar/parametric, for part c, it said that dr/dt was equal to dr/dtheta right? the eqn was r=3 + 2cosx(somthing like that), so i took the derivative, which would equal dr/dt. and then plug in pi/3.</p>
<p>for part d of that problem, basically the same thing, but more painful. it said that dy/dt was equal to dy/dtheta. so, u had to convert the polar graph to linear, so u use y=rsinx, which would equal y=(3+2cosx)sinx. after foiling all that crap, u derive it and plug the number, which i think was again pi/3. i hope im right.</p>
<p>hey for number 2., i am sure that t=3 is right... cuz at the two critical points where the derivative equals 0. those are minimums points... so u have to check the end points of the interval,,, check 0, 3, 7, ,,,,, at 3,,, the amount of water is the greatest...</p>
<p>for #1,,, for teh semicircles part,,, i used the equation pi/8*(y-2)^2/4 =pi/8(y-2)^2</p>
<p>That's all i remebered... and for te series one.. it was easy ,,r giht... for teh last part,, i asid that thsi was an alternatign series,,, so error less than 1st term ommitted.. which equaled to 1/320</p>
<p>uch i messed up on the limit of the series question. I assumed u treat it as an integral and use Lhopitals rule. there goes a few points. The rest of the ? i got right though.
On the polar ?, waht did people get (esp for first part).
I also got 1/320</p>
<p>series of e^x is 1 + x + x^2/2 + x^3/(3<em>2)
so plugging in -x^2 gives: 1 - x^2 + x^4/2 - x^6/(3</em>2).. ((-1)^n)x^(2n)/n!</p>
<p>b. It then asked for lim x-> 0 of (1- x^2 - 1 +x^2- x^4/2 + x^6/6)/x^4, which comes out to be (-X^4/2 + x^6/6)/x^4, which when x approaches zero, comes to be -1/2.</p>
<p>c. It then asked for the integral from 0 to 1/2 of the first two terms of the function, so the integral is x - x^3/3, which is 1/2 - 1/24, which is 11/24.</p>
<p>d. It then asks why is it less than 1/200, so I explained that it was an alternating series approaching zero with the next term's absolute value less than the previous one's (not sure if it really meant to ask for this previous explanation, but I put it just in case), so the next term was determined by x^5/10, which is 1/(32*10), which is 1/320</p>