AP CALCULUS AB FRQ:
2005 #6
Solved by: Brandon Borko
AP Calculus AB- Ms. Zhao Pd. 4
2005 FRQ #6
NO CALCULATOR
THE QUESTION
Question has 9 points total
2 points for Part A
2 points for Part B
5 points for Part C
Part A deals with slope fields
Part B is approximating a yvalue
Finally, Part C is finding the
particular solution to the
given differential equation
• On the axes provided, sketch a slope
field for the given differential equation
𝑑𝑦
2𝑥
( = − ) At the twelve points
𝑑𝑥
𝑦
indicated {x-values -1,0, and 1; y-values
2,1,-1, and -2}.
PART A (2 POINTS)
The graph for this question should be as
the one on the left.
(1 pt) for correct points with 0 slope {points
with x-value 0}
(1 pt) for correct points with correct nonzero slopes
(-1,2) and (1,-2) with a slope of 1
(1,2) and (-1,-2) with a slope of -1
(-1,1) and (1,-1) with a slope of 2
(1,1) and (-1,-1) with a slope of -2
• Let 𝑦 = 𝑓 𝑥 be the particular solution to the differential
equation with the initial condition 𝑓 1 = −1. Write an
equation for the line tangent to the graph of 𝑓 𝑎𝑡 (1, −1)
and use it to approximate 𝑓(1.1)
Using the differential equation, we can find the slope to f(x) at
(1,-1)
𝑑𝑦 −2𝑥
−2(1)
=
=
=2
𝑑𝑥
𝑦
(−1)
After finding the slope, use the initial condition of (1,-1) to find
the tangent line.
𝑦 + 1 = 2(𝑥 − 1)
Then just plug in 1.1 to approximate.
𝑦 = 2 1.1 − 3 = −0.8
f(1.1)≈ -.8
PART B(2 PTS)
1 pt for finding
the tangent
equation:
𝑦+1=2 𝑥−1
1 pt for
approximating
f(1.1) ≈ −.8
• Find the particular solution 𝑦 = 𝑓(𝑥) to the
𝑑𝑦
−2𝑥
given differential equation ( =
)with
𝑑𝑥
𝑦
the initial condition 𝑓 1 = −1.
PART C(FIRST 2 POINTS)
The very first thing you must do to solve this problem is
separate the variables. If you don’t separate the variables
you will get zero points on this part, maximizing your FRQ with
4/9 points.
To do this you must multiply both sides by dx and then by y,
thus separating the variables as so:
𝑑𝑦 −2𝑥
−2𝑥𝑑𝑥
=
; dy =
;
𝑑𝑥
𝑦
𝑦
𝑦 𝑑𝑦 = −2𝑥 𝑑𝑥
After separating the variables, you must then find the
antiderivatives of each side:
Always
𝑦 𝑑𝑦 = −2𝑥 𝑑𝑥
remember
to add C
𝑦2
= −𝑥 2 + 𝐶
2
1 pt for separating
variables:
𝑦 𝑑𝑦 = −2𝑥 𝑑𝑥
(EXTRA IMPORTANT, if NOT
DONE 0/5 points on entire
part C)
1 pt for finding the anti
derivatives of each side:
𝑦2
= −𝑥 2 + 𝐶
2
• Find the particular solution 𝑦 = 𝑓(𝑥) to the given
𝑑𝑦
−2𝑥
differential equation ( =
)with the initial
𝑑𝑥
𝑦
condition 𝑓 1 = −1.
PART C (REMAINING 3 PTS)
After finding the antiderivatives of each side, you must
then find the constant of integration, the variable C, for the
equation. To find this you must substitute the initial
condition into your equation:
(−1)2
2
1
3
= −(1)2 + 𝐶; 2 = 𝐶 − 1; 𝐶 = 2 𝑜𝑟 1.5
C is equal to 1.5 or 3/2
If C is not present in the equation, then you can only get a
maximum of 2/5 on the part.
After finding C, you must then solve for y, using algebra:
𝑦2
3
= −𝑥 2 + ; 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 2 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑦 2 = −2𝑥 2 + 3
2
2
Because the solution goes through (1,-1) y is negative.
𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑜𝑜𝑡 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 − 𝑦 = 3 − 2𝑥 2 ;
𝑑𝑖𝑣𝑖𝑑𝑒 − 1 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠
𝑦 = − 3 − 2𝑥 2
1 pt for there being and finding
the constant of integration,
C=3/2
(If not present in work, maximum
of 2/5 for part)
1 pt for using the initial condition
(to solve for C)
1 pt for correctly solving for y
𝑦 = − 3 − 2𝑥 2
OVERVIEW 9/9
• A) If you find the graph below as your answer that is 2 pts, one for the points
with 0 slope, and 1 for the points with a non-zero slope.
• B) If your tangent line is 𝑦 + 1 = 2 𝑥 − 1 then that’s 1pt
• If your approximation for f(1.1) is -0.8 then that is another pt
• C) If you separate your variable with y on one side and x
on the other whereas 𝑦 𝑑𝑦 = −2𝑥 𝑑𝑥 that is 1pt.
•
•
•
•
𝑦2
2
The antiderivatives are = −𝑥 2 + 𝐶, which is another point
C=3/2, another point
If you used the initial condition of (1,-1), that is another point
If your equation for the particular solution is 𝑦 = − 3 − 2𝑥 2 , that is the final point
You got 9/9 on this question
CITATIONS
• Images found on Google Images
• http://apcentral.collegeboard.com/apc/public/repository/_ap05_sg_calcul
us_ab_46569.pdf for answer sheet