<p>I can't believe that horizontal asymptote one stumbled me...ekkk</p>
<p>I'm good at work, here's my take on it:</p>
<p>Work is defined as force times distance: W = F X d. When a variable Force, F, moves an object along the x-axis from a to b, we approximate an element of work done by the force over a short distance change-in-x by
W = F * change-in-x</p>
<p>FTC yields, W = lim [n approcahes infinity]sigma (k=0, infinity) F * x = Integration [F(x) dx] limits: a, b</p>
<p>^thats AP Physics C [i got a 5 last year on both parts of the AP C Tests] </p>
<p>This problem is a little bit tricky, and partially incomplete, F in this case can equal weight of the water (weight meaning mass X gravity), he failed to gives us weight so the problem is IMPOSSIBLE as it stands...had he give us weight, then the integrad would have been:</p>
<p>W = wight * pi INTEGRATION {shell method}</p>
<p>OH AND BTW 1 and -1 is the correct answer for the asymptotes :)</p>
<p>Ok, RES, here's another absolute value thing for you:</p>
<p>Integration [lower limit -1, upper limit 1] {1 - |x|} dx = ? </p>
<p>NO CALCULATOR, and EVALUATE THE INTEGRAL ALL THE WAY, the answer is a number...:)</p>
<p>This killed me on a practice thing we had today in class; I don't know how to integrate absolute functions. What do you do?</p>
<p>I googled around and found a formula:</p>
<p>Integration of l x l = ( x * l x l ) / 2, and so:</p>
<p>x - xlxl/2 taken from -1 to 1 = 1?</p>
<p>all right, you don't have to integrate this one...
think about it, integration is defined by area under the curve...
so if you draw the function out on paper, you will get a upsidedown "v" shifted up 1 unit on the y-axis.
This will make 2 triangles bounded with the x axis, each with base of 1, and height of 1, so the area is the sum of the areas of the 2 triangles, so .5 (.5 * base * height) +.5 = 1</p>
<p>hope this helps...
let me know if you need more explanation, I have to go to drivers ed now</p>
<p>^ Gotcha, thanks man.</p>
<p>I'm pretty sure it is calc BC... I know we did it in class. Hmm.. maybe it's just not on the exam? Who knows, sorry, then ):</p>
<p>Yeah, definitely not on exam. It contains density and gravity.<br>
Unless that information is given or explained (which is high unlikely), it won't be on the BC test.</p>
<p>edit = Theres no need to say sorry! You just taught be something new, lol.</p>
<p>The Work problem was in my class textbook, so I understand why WantIvy put it on there. OTOH, I haven't seen it in any BC practice exam, so I guess (thank god) that it's safe to say that we don't need to know that for the exam.</p>
<p>NEW PROBLEM!!!</p>
<p>No calculator:
1)A particle moves along a line with velocity v(t) = t^2 - t. The total distance that it covers from t=0 to t=2 equals: ?</p>
<p>This is probably an "AB" question, but whatever...</p>
<p>I'm going to post 2 q's just to move things along...</p>
<p>2)The general solution to the differential equation y' = [(1 - 2x) / y] is a family of what? (circles, squares, hyperbolas, etc?)</p>
<p>They're both AB questions, although they'll never ask #2 on the AB test, as best I can tell.</p>
<p>Question:</p>
<p>in a taylor series for sinx, lets say the function is sin(3x^2), can you just write out the taylor series for sinx and plug in the (3x^2) for X?</p>
<p>if so, why isnt this done for BC 2004 form A question 6?</p>
<p>For the last question on the previous page:</p>
<ol>
<li><p>I got 2.</p></li>
<li><p>x^2 + y^2 = 2x + C .....so I'm guessing ellipse? Or some sort of curved structure?</p></li>
</ol>
<p>For thorps:</p>
<p>When you do something like that, you must multiply the new plugged-in series by the derivative of 3x^2, and when that's done, I just take the initiative to just multiple the sine polynomials by a negative one to mess up their alternation.</p>
<p>Let me check the BC 2004 real quick. I'll explain it in a more serious tone, then :)</p>
<p>Thorps, no, you cannot do that. Think about it. Because of the chain rule, all of your derivatives will be incorrect. And the alternating thing usually doesn't work unless you are dealing with a Maclaurin sin or cos function.</p>
<p>Thorps:</p>
<p>I assume you came up with a simpler version of the last page of this:</p>
<p>Why you can't just throw in 3x^2 inside:</p>
<p>x - (x^3)/31 + (x^5)/5! .....?</p>
<p>You can, but there is an additional step. The derivatives are no longer then same when you have this new center.</p>
<p>Recall that a polynomial approx is basically:</p>
<p>f^n(center) * (x - a)^n all over n!</p>
<p>where n is your degree. For sinx, it's always either 1 or 0.
For sin(3x^2) it's a whole another ball game. You have to derive that to whatever times the degree you have, plug in zero, and try to find a value that doesn't come out to 0.</p>
<p>I'm having a little problem with this, trying to find out why my calculator is giving me something different for the first two terms of the polynomial approx of sin(3x^2)</p>
<p>I got:</p>
<p>27x^6 as the first term because of</p>
<p>sin(3x^2)'s 2nd derivative value at 0 = 6;</p>
<p>6(3x^2)^3 all over 3! gives 27x^6. </p>
<p>The calculator's first two terms are:</p>
<p>3x^2 - (9x^6)/2</p>
<p>^Why?</p>
<p>^^^</p>
<p>Never mind. You only account for derivatives/chainrule when it's off center from zero, or not a maclaurin. </p>
<p>For sin(3x^2) 's taylor poly. centered at zero, you can go ahead and just plug it in.</p>
<p>Ah...spent like 30 minutes on that.</p>
<p>RESmonkey:</p>
<p>On the 2004 BC question you cant plug it in even though its centered at 0 because theres an additional +pi/4 , which even with 0 will mess everything up, correct?</p>
<p>thanks...</p>
<p>^^^^^</p>
<p>I dont get it. take a look at my review book (2007-08 Petersons)</p>
<p>This says that you would be able to just plug it right in!</p>
<p>so can you or can you not?!</p>
<p>You can plug it in. As far is I know, if you know the Maclaurin series for f(x), then to find the Maclaurin series for f(3x^2), you simply plug it in. </p>
<p>I think that things get messed up when you have the Taylor Series centered at a different x, but because the Maclaurin Series for sin(x) converges for all real x, we don't need to make a Taylor Series to mess with any sin(x) type problem because the Maclaurin Series works just fine at any point on a sin(x) curve.</p>
<p>The main point is that we have a Maclaurin series on our hands so we can plug it in. </p>
<p>Just my take on this - I didn't actually read this anywhere and I'm hoping I'm right.</p>