@Sanders99 I believe so.
Can someone tell me what I would get if I get 30/45 on the multiple choice, and a 6 on each FRQ?
@Sanders99 based on the 2008 curve, you’d get a 5.
1a) int R(t) from 0 to 8 =76.570
1b) D(3)>R(3), so amount of water is decreasing
1c) A(t)= amount at time t has a min at 0, 8, or where R(t)=D(t) at t=3.271. A(0)=30, A(3.271)=27.964, A(30)=48.543. Abs min at t=3.271
1d) 30+intR(t)-D(t) from 0 to t = 50
2a) Int g-f from 0 to 1.032 + int f-g from 1.032 to 2 = 2.004
2b) Int (f - g)^2 frin 1.032 to 2 = 1.283
2c) dh/dx = df/dx - dg/dx = -3.811 at x = 1.8
3a) v’(16)=5 meters/min^2
3b) Total distance traveled, in meters, from t=0 to 40. = 7600.
3c) B’(t) = 15 m/min^2
3d) 350 m/min
4a) slopes
4b) d2y/dx2= 2-dy/dx = 2-2x+y. In Q2, x<0 and y>0, so d2y/dx2>0, thus concave up
4c) f(2)=3; slope at (2,3)=1, so neither a rel min nor rel max at (2,3)
4d) dy/dx at (2,3) =1, so m=1. Plug in point and slope to find b=1
5a) Max at x=-2, f’=0 ad changes from + to -
5b) (-2,-1) and (1,4) b/c f’’<0 and f’<0 on these intervals
5c) f’’ changes sign at x=-1, 1, 3
5d) Int f’(t) from 1 to x = f(x) - 3. f(4)=15, f(-2)=6
6a) y-1=(1/4)(x+1)
6b) dy/dx is undefined where 3y^2=x. Substituting into original equation,
y^3 - (3y^2)y =2 so y = -1. x=3. (3, -1)
6c) d2y/dx2 = [(3y-x)(dy/dx) - (y)(6y*dy/dx - 1)]/(3y^2-x)^2 = 1/32 at (3, -1)
I would look at 5B. I’m pretty sure that it is (1,3), not (1,4)
Yer right! (1,3).
And 5d) is f(-2) = -6, not 6.
I’m pretty sure 5d is 12 and -9 because the first derivative is all negative there, so the function has to be decreasing. It can’t go from -6 to 15, that would be increasing
Dang, I think you’re right! f(4)=3 + -12 = -9 and f(-2) = 3+(-9). Geez, how many other mistakes?!?
I mean 3-(-9), so 12
@Bugiardi I disagree with 4d. y = x + 1 doesn’t satisfy dy/dx = 2x - y for all point on y = x + 1. dy/dx should equal 1 for all points since m = 1, but, for example, at (0, 1), dy/dx = -1.
Hahah I think the rest are right! Except 4d I don’t think was talking about the specific point (2, 3), I think it was completely seperate from 4c. I left it blank so I don’t know for sure, but I heard it was m = 2 and b = -2. These values work because for any points along this line, if you plug the x and y values into dy/dx, you’ll always get 2
@zach1198 do you think I’m still going to get credit for 6c, I just flipped the 1/2/16 should have been 1/32 all my steps were right just untill the final answer which I know I would not get the 1 point.
@Hark123 Definitely. You’ll probably get around 2 points just for getting the expression correct for d^2y/dx^2.
I was pretty unclear about part d) of this question in general – do you have a proposed solution? (I haven’t read through this whole thread). By solutions to the differential equation, did they mean ANY line satisfying ANY valid point in the slope field? In which case m=1, b=1 would work. But also m=0, b=2 for (1,2). Smarter people than me will have to share… Clearly (look at the slope field) the function y is nonlinear. Or maybe I just missed the point entirely. I’d love to hear thoughts…
@Zach1198 but don’t you think that part would be a little more worth it since there are only 3 parts and the 1st part was so easy…I would guess that part (a) would be 2 points, part (b) would be 3 and part © 4 points so at least 3 points?
lol @dogeater
@Bugiardi I posted this earlier:
dy/dx = 2x - y
d^2y/dx^2 = 2 - 2x + y
Since the second derivative of a line is 0, to find the equation of the line, we need to find a general solution to when d^2y/dx^2 = 0.
d^2y/dx^2 = 2 - 2x + y = 0
y = 2x - 2
@Hark123 I’m sure you will get a good amount of credit for that problem since you did the right work, you just made a small error at the end. If you set it up correctly you will probably get most of the points.
@Zach1198 that’s good to hear at least.