@HelixKriby Half the people say it’s 1,1 and another half say it’s 2,-2.
anyone have a link to the free response questions?
(it’s been 48 hrs)
They usually post around 4 p.m.
Ha ha - Stand and Deliver.
Since the FRQ’s are up, allow me to shed some light:
I got this wrong on the test either way ._.
We know that we need to find a value for m and b for mx + b= y to be a solution to the differential equation 2x -y.
To begin let us substitute the “y” for “mx + b” and we get: 2x - mx - b.
As many of you pointed out, it is a linear equation, so there cannot be an x term in dy/dx. The only way for x to be canceled out is for m = 2. Therefore, we see m is 2.
Our new dy/dx = -b. Since we know that m = dy/dx and that m = 2, then we get:
2 = -b, b= -2.
Thus, m = 2, b= -2.
^shoot I solved for the tangent line
Was that Form O? I had form G. Just by skimming through it, it seem like my version was “harder”. If so, hopefully the curve will be good for me.
For the y= mx+b one, I just set the second derivative that you found earlier in the problem to 0 since the second derivative has to be zero where the solution curve is linear, as the first derivative would just be a constant.
I believe that the second derivative was y"= y - 2x + 2, so after setting it equal to 0 = y - 2x + 2, you could just solve for y to get y = 2x - 2. Therefore, m = 2, and b = -2.
Just went through all the questions. We reviewed all of these types of FRQs in class. I’m upset I didn’t have this version. Did anybody with Form G feel the same way I do?
FRQ’s are finally up, so I think it is safe to talk about it a little bit now. (lol, I know no one cares about the rules, but I really do trying to obey it)
For the differential equation one, we are given the differential equation dy/dx = 2x-y, and we are asked to find the value of m and b for which y=mx+b is a solution to that differential equation.
I think everyone need to realize that any particular function y=f(x) is a solution to a differential equation if and only if when you differential y=f(x) you get exactly the derivative function the differential equation tells you for ANY VALUE of x. And so laGrangeError is right, the above condition holds only for when m=2 and b=-2.
There are actually quite a few different methods to obtain this result from different aspects of this differential equation:
(1)
If you try to solve dy/dx = 2x-y by something like http://www.wolframalpha.com/ or by hand if you know how to deal with “integrating factors” or some other even more advanced methods, you will obtain the general solution of this differential equation to be y(x) = Ce^-x+2x-2. Then by letting the integration constant C to be 0, you obtain the linear solution y(x)=2x-2, so m=2, b=-2
(2)
You can also solve differential equations by “guessing”. You guess a solution for the equation, then you plug that back in and see under what condition the solution hold. So the AP provide you the guess “y=mx+b”. You plug that in for dy/dx and y, you get m=2x-mx-b. Then you put everything on one side, 0=2x-mx-b-m. The sum of all the x-term and constant-term need to be 0 and for b and m to be a constant, b=-m and 2=m, therefore, m=2, b=-2
(3)
Like someone else said a while ago, you can also use the fact that a line have no second derivative (d^2y/dx^2 = 0), you differentiate 2x-y and get the second derivative, set that equal to 0 and substitute m and whatever back in just like method (2), with some algebra you will surely get m=2, b=-2
(4)
Even if you don’t know how to solve the differential equation for this particular solution. You can work out the slope field for a bigger value of x and y, then you will recognize a pattern that you start at any point in the plane, whenever you go right and up two you will get to a point with exactly the same slope. This pattern implies that the only possible linear solution is the one that with slope 2, because this is the only way you can start at some point and follow the slope and get to another point that give you the same slope and which make it a line. Then you connect the points with slope 2 and you get the line y=2x-2.
I feel so nerdy after writing all these math BS.
Sad that I was too focused on the fact that the general solution of this diff eqn can’t be a linear one, and didn’t think of any of these thing I just wrote. And the worst part is I suddenly figured out all these crap the night right after I took the exam.
I set f’(x) = f’(x+1), since if it’s a linear equation, the slope should be the same everywhere. I solved and got y=2x-2. Any ideas as to whether or not that would get credit?
Hi all,
Yes, I also got m=2 and b=-2 via the taking derivatives and substitution method. I did know how to solve this type of equation (it’s not particularly complicated and isn’t that far beyond the scope of the AP) but wasn’t sure that I’d be able to recall it so just went with the other method instead.
For 4.b…
Can you just do 2-dy/dx as an answer to d^2y/dx^2? And… How detailed should the justification be? I just said all values of d^2y/dx^2 > 0 for all values in Quadrant II; therefore, they are all concave up.
On # 1 a and b, did I have to factor in the D(t) at all? Is it right if I just integrated the R(t) equation from 1 to 8 for part a and for part b just took the first derivative of R and plugged in 3? Did the the D(t) equation factor in only on part c and d?
Did anyone else get 8 for 6c?
@Oblivion96 For 1 a I did not factor in D(t), for b, I think I did. Not sure though, since I thought since the water is leaking, you should also factor that in.
Can anyone explain how to do part c. I could’ve sworn when I plugged in the equation and tried to find the 0, it didn’t give me one.
Guys, also for speed, were you supposed to take the absolute value of the velocity vectors, that’s what I did, not sure about it though.
@goblazers How did you get 8? I thought you were supposed to give a Taylor Polynomial for 6c.
what did you guys say for the “rational function”?