<p>wow xxsteelxx, thanks! I will be using that as well now.</p>
<p>A month and two weeks to go.</p>
<p>How well have you guys memorized stuff? Our teacher lets us use cheat sheets, so I haven’t memorized much…</p>
<p>I memorized most.</p>
<p>Second time reviewing Calc BC right now. All I need to do left is Series (Curse Taylor polynomials).</p>
<p>I am quite surprised, what do you guys have to memorize? I’m guessing power series expansion for certain functions (sin x, cos x, e^x, etc)?</p>
<p>^god, you just remind me of the future tribulation I will have memorizing the power series expansion. I barely memorize those the first time I study BC material.</p>
<p>just remember the short hand formula (series notation) and you can do all sorts of stuff from there</p>
<p>I have a good feeling this year… <a href=“http://www.math.lsu.edu/~neubrand/AP05/AP%20Calculus/apc03_calculus_frq89-97a.pdf[/url]”>http://www.math.lsu.edu/~neubrand/AP05/AP%20Calculus/apc03_calculus_frq89-97a.pdf</a></p>
<p>No need to thank me. :)</p>
<p>dude you are THE man</p>
<p>Thx xxsteelxx, even though you say no thx. Wow, only two-three polars. Maybe there’s not going to be a polar this year.</p>
<p>Hey guys, do you know if there is any way possible to take the derivative of a function of you have ONLY a CHART with numbers on it? my teacher mentioned something…</p>
<p>thanks!</p>
<p>what kind of chart? be more specific…</p>
<p>I studied for less than 2 hours and thought I had no mastery of Calc. I got a 5. If you are good at math and have the fundamentals down for Calc, you will do amazingly well!</p>
<p>@lemone:</p>
<p>a chart with just x and y values.</p>
<p>I think that you use (delta y)/(delta x) (i.e. slope) to find a derivative at a point. If they tell you to find the derivative at say, x=3 and they give you f(3) and f(2.999), you use change in y (f(3)-f(2.999)) divided by change in x (3-2.999) to find your slope at 3.</p>
<p>my advice for people who are stressing: spend less time reading the material, spend more time doing actual problems</p>
<p>definitely correct^^^^
Right now I’m working on FRQ’s from AP central, how’s it coming along?</p>
<p>umm i have a question:
A curve C is defined by the parametric equations x = t^2 - 4t + 1 and y = t^3
What is the equation of the line tangent to the graph at the point (-3,8)?</p>
<p>i’m not sure what i did was correct…but i plugged in -3 for the x equation and 8 for the y equation and got t=2
so i found dx/dt and dy/dt and when i plugged in 2 for dx/dt it was 0, and for dy/dt it was 12
and i thought that when dx/dt is 0 and when dy/dt is not 0, then there must be a vertical tangent at that point because i remember my teacher saying something about that
so then the equation would be x=-3?
correct me if i’m wrong…
thank you!</p>
<p>Yes, that seems to be the right answer. The other thing you could have done is find that t= y^(1/3) and plug that in the other equation and just take the derivative from there. You’ll get an undefined answer which gives you a vertical tangent of x=-3.</p>
<p>oh ok thank you! =)
and one last question…
The position of a particle moving in the xy-plane is given by the parametric equations x= t^3 - 3t^2 and y= 2t^3 - 3t^2 - 12t. For what values of t is the particle at rest?
a) -1 only
b) 0 only
c) 2 only
d) -1 and 2 only
e) -1, 0, 2</p>
<p>it’s at rest when velocity is 0, or when dy/dx is 0
but when i solve for it, i know that 0 would make the denominator undefined, and same for 2. so -1 would work (a)
but my friend says the answer is (c) 2
i dont get how he got that…=(</p>