and did you get x-(3/2)x^3 for G3(x) ?
For 6b, did anyone recognize the rational function f’ as 1/(1+3x) for the radius of convergence (R=1/3)?
The original sigma, not the derivative, is (1/3)ln (1+3x).
@reqhuuya , I’m not 100% certain of what I got but think that I got the same answer as you for part 6b.
thanks mathinduction.
@mtxing that’s a clever solution, it should get full credit. There are lots of ways to solve for y and that’s one i wouldn’t have thought of
@noname054 You are just flat out wrong. m=1 and b=1.
Lets go with what you are saying and try to plug it in.
y = 2x - 2 … as u claim m = 2 and b = -2.
Part c) gives us the point (2,3) so lets plug those in!
3 = 2(2) - 2
3 =/= 2 …
Lol you are flat out wrong, and you were never supposed to solve the differential equation. Ever.
The way you were supposed to solve it is as follows:
y = mx + b
dy/dx = m = 2x - y
y = (2x-y)x + b
3 = (2(2) - 3)2 + b
3 = 2 + b
b = 1
y = mx + 1
3 = 2m + 1
2 = 2m
m = 1
@jetynz The given point F(2) = 3 was only applicable for part c) not for part d). There are an unlimited set of solutions to differential equations, and d) asked for one that was linear. The solution through f(2) = 3 is not linear. The two problems are completely independent, and the given point could only be used for the specific solution y = f(x), not the different solution curve y = mx + b given in part d)
@nearea The questions are not completely independent. Part d) refers to the differential equation in part c) and it says " Let y = f(x) be the particular solution to the differential equation with the initial condition f(2) = 3."
And then in part d) Find the values of the constants m and b for which y mx b is a solution to the differential equation.
The differential equation is dy/dx = 2x - y.
And yes there are an unlimited set of solutions to differential equations, obviously, but the point (2, 3) better darn be one of them…
Hey guys I’m really dumb and probably going to get a 1 on this exam but I have to make a powerpoint on how to solve 2015 AP Calc BC6 for my Calc BC final project and it’s like due tomorrow LOL. BUT, huge problem, don’t know how to do BC 6 FRQ. Please help T_T
Hi all,
What they were expecting with 6d was definitely not to solve the differential equation but rather for you to take a derivative/second derivative and use substitutions (which is what I ended up doing also as I wasn’t 100% confident in my ability to solve that differential equation).
However, the differential equation isn’t bad to solve (I tried it afterwards) and isn’t that far beyond the scope of the AP. I guess they would give full credit.
Here’s a sketch of the solution:
Let dy/dx=2x-y . Then substitute in a u for dy/dx such that u=2x-y. Then take du/dx which is equal to 2-dy/dx. But we know that u=dy/dx from the previous step; thus, du/dx=2-u. We can use separation of variables on this and get 1/(2-u) du=dx. Integrating both sides -ln|2-u|=x+C. |2-u|=Ae^(-x). It turns out the absolute values don’t matter here (you can try it both ways; it still works). So 2-u=Ae^(-x). Now, you can back substitute with u=2x-y, so 2-(2x-y)=Ae^(-x). 2-2x+y=Ae^(-x), so y=Ae^(-x)+2x-2. In order to be linear across all x’s, A must equal 0, so y=2x-2. Hence, m=2, b=-2 which is the same solution that we get via the other method.
Thus, the differential equation isn’t terrible to solve (although is slightly beyond AP). I didn’t use it myself but would definitely hope that they’d give full credit for that method!
@jetynz Yeah, but in this case, there is only one solution where the slope is constant, as m is a constant slope. For that reason, you need to be able to set the second derivative equal to 0 since the derivative of a constant is 0. When you plug the point (2,3) into the second derivative you found in part b, y" = y - 2x + 2, you get y" = 1. Since the second derivative is equal to 1 for (2,3) for the solution curve through f(2) = 3, it would mean that the slope of the line is changing, which is not possible since the slope of the solution curve y = mx+b is constant.
@nearea , I think you’re right; there’s only one linear solution (the one they ask for). The other solution doesn’t work as a solution to the differential equation. Through my solution in the post above, the only thing that can change is the A value depending on the initial condition; everything else must remain as is, including the m and b values.
@nearea @Mathinduction Oh well… I guess we’ll see.
Anyways, this is just one subpoint!
Does anyone want to go through and make list of all the answers so we can all see and compare? Maybe we can even break down the scoring?
We should go through them all. Maybe all do a google hangouts or something?
Google doc maybe?
@jetynz , there’s probably no need; somebody always compiles answers on this link each year: http://www.askmrcalculus.com/#2015 . He’ll probably post answers in the next day or two.
@rdeng2614 i posted in the wrong thread… i meant to post in AB sorry hahaha
what did you guys have for the 3rd degree G(x) in the last question? Also, what did you put for the rational function in the second to last?
y=mx+b
dy/dx=m=2x-y
m=2x-(mx+b)
m=2x-mx-b
m=(2-m)x-b
Match up the coefficients
(2-m)=0, m=2
m=-b, b=-2
End of discussion.
How are the BC and AB subscore graded? Is the BC score the entire test while the AB is certain parts, are the scores purely separated by BC level and AB level questions only, or how does it work?
@Dorfdude8 , I believe the AB subscore is based just upon questions pertaining exclusively to AB material; whereas, the BC score is the entire test (I’m not 100% sure though).