DescentBB

Need one, especially if one isn't familiar with the BC-exclusive topics on the AP test.

I've taken practice tests for AB and BC. The AB is cake, but BC-- .

I wouldn't even be in this mess if I wasn't pressured into it by my math teacher, but I might as well make the best of it. So any good resources/books? I know there must be thousands out there to choose from.

edit:



I've managed to figure it all out, except this. Help? TIA.

Not really an AP BC resource, but this is the best math site ever:
http://www.quickmath.com

Free Mathematica on the web. Great to check most algebra/calculus questions.

Stewart's "Early Transcendentals" is considered a very good calculus book. People also like Thomas&Finney. I don't know how well these books prepare you for the BC exam, though -- I don't remember what was even on it.

Check out http://www.powellsbooks.com/

They have a god selection of technical books.

Please see above.

Ok: here is how you want to go about solving this question. You are given the graph of a function, y=F(x). You are also given x=G(t) and y=H(t) (G(t) is known (integral of x'(t)), H(t) is not). So, they are asking you questions conerning y(t) given a position on a y(x) graph. Approach for the first question: You know that dy/dx is positive. You can also say that dy/dx = (dy/dt)/(dx/dt) (dt's cancel) Therefore, if dx/dt at that point is positive, dy/dt must also be positive, and vice versa. You can find dx/dt simply by integrating the equation (in terms of t) they give you, solve for t given C's x, and then plugging that t into the dx/dt equation. I think if you grasped that, you should be able to figure out the rest. The key to it lies in the fact that dy/dx = (dy/dt)/(dx/dt).

Thanks, here's what I've got:

A. dy/dx is visually positive at C, and dx/dt is negative for all values between T = 0 and T = 9, except T = 3, which is B, so dx/dt can be assumed negative. Therefore, dy/dt must be negative, since dy/dx=(dy/dt)/(dx/dt).
B. Because dy/dx=(dy/dt)/(dx/dt), for dy/dx to be undefined, dx/dt must equal zero. Since the only point between T = 0 and T = 9 where dx/dt equals zero is T = 3, dy/dx is undefined at T = 3.

C. Since at T = 8, the tangent line is y = (5/9)x-2, dy/dx = 5/9 at T = 8. x'(8) = dx/dt at T = 8. Therefore, 5/9 = (dy/dt)/x'(8). Then (for whatever dx/dt and dy/dt come out to equal) f'(t)= (dx/dt)i + (dy/dt)j. The speed I'm not sure about, but I'd hazard to guess that it's sqrt((dx/dt)^2+(dy/dt)^2)?

D. Since the change in Y from A to D is visibly zero, change in position from A to D (T = 0 to T = 9) can be summarized as change in X. Therefore, the change in position is the integral from T = 0 to T = 9 of x'(t)dt, which can be calculated on a calculator.

I think I'm finally understanding this. Any thoughts?

Oh, BTW, I actually ended up with a Cliffs AP Calculus book. But, before anyone criticizes, it actually was the only book I could find that actually discussed the more advanced BC topics. And it even pointed them out as such too. It doesn't sound that impressive, but the others just dwelled on the most mundane crap or were 90% practice tests. (edit: BTW, I also have an (8.5 x 11)in. x 600 page Princeton Review book sitting on the floor from school.)