AP Calculus AB & BC Solutions, 2014 Form A
This guide contains all the solutions to the free-response questions from both the AB and BC exams. Note that some of the problems are common to both tests, but have only been printed once.
Calculus AB-only problems
2(a) We need to find the points of intersectio intersection n so we set y = f (x) = x 4 x = 2.3. So the volume is
⇒ − 2.3x2 + 4 = 4 =⇒ x = 0 and
2.3
V y=−2 = π
[f (x)
0
2
− (−2)]
2.3
dx = π
[f (x) + 2]2 dx = 33.244
0
2(b) 2(b) The length length of a leg is y f (x) = 4 (x4 2.3x3 + 4) = 2.3x3 x4 . Since we are dealing with right triangles, both legs are congruent and the area of each cross section is just 21 leg leg. So the volume is 2.3 1 (2.3x3 x4 )2 dx = 3.574 2 0
−
⇒ −
−
2.3
2(c) The area area of R i s R is
0
[4
−
×
×
−
− (x4 − 2.3x3 + 4)] dx. We simplify and split the integral: k
0
3
2.3x
4
−x
2.3
dx =
k
2.3x3
− x4 dx
Solving the equation above will find the value of k k the problem seeks.
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5(a) f has has a relative minimum whenever the sign of f f changes from a negative number to a positive number. This occurs only at x = 1. 5(b) Since
f (1)
− f (−1) = 0, the mean value value theorem guarantees some c such that f (c) = 0. 1 − (−1)
5(c) We just use the chain-ru chain-rule: le: h (x) =
From the table, we find that h (3) = 5(d) The key is to recognize recognize that easy: 3
−2
1 2
7
=
1 f (x) f (x) = f (x) f (x)
·
1 . 14
f (g (x)) g (x) dx = f (g(x)) + C by the chain-rule. Now the problem is
f (g (x)) g (x) dx = f (g (3))
− f (g(−2)) = f (1) − f (−1) = 2 − 8 = −6
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6(a) I don’t know how to mak makee a graph using this software software,, but just use a graphing graphing calculator calculator to plot the solution to part (c) and see how your sketch compares. 6(b) We find the slope using
dy : dx
dy dx
− 1)cos0 = 2 The equation of the tangent line is y − 1 = 2x =⇒ y = 2x + 1 . (0, (0,1)
= (3
At x = 0.2 we find that y (0.2) =
(0.2) = 2(0.2) + 1 = 1.4 . f (0
6(c) Separation Separation of variables variables and then algebra: algebra: dy = dx dy = 3 y dy = 3 y ln 3 y =
(3
− y)cos x
cos x dx
−
− | − |
cos x dx
sin x + C
Plug in (0, 1) to find that ln 3
| − 1| = sin sin 0 + C = =⇒ C = = ln ln 2. We now deal with the algebra: ln |3 − y| = sin x + ln ln 2 eln |3−y|
3
−y
+ln 2 = esin x+ln = 2esin x
Rearranging we get the final solution to be y = 3
− 2esin x
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Calculus AB & BC Solutions
1(a) The average average rate of change change of A (t) on 0
− A(0) = −0.197 pound/day. ≤ t ≤ 30 is A(30) 30 − 0
1(b) A (15) = 0.164. This means that the amount amount of grass clippings in the bin at the beginning of day 15, in pounds, is decreasing at the rate of 0.164 pound/day.
−
1 1(c) We want want t such that A (t) = 30 Thus, we have A (t) = 2.7526 =
30
1 calculator we find that that A(t) dt. Using our calculator 30 0 t 6.687(0.931) = 2 .7526 = t = 12.415 .
⇒
⇒
30
A(t) dt = 2.7526.
0
1(d) We have A (30) = 0.05598 and A(30) = 0.78293 78293.. Thus, Thus, we have L(t) 0.78293 = 0.05598(t 30) and want to find t such that L (t) = 0.5; so we have 0.5 0.78293 = .05598(t 30) = t = 35.054 .
−
−
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−
−
− − ⇒
−
2(a) The area of the shaded shaded region region is 9π + 4 2(b) x = r cos θ = cos θ(3
−
dx 2sin2θ); dθ
θ= 3
π
=
π/2 π/2
0
1 2 r dθ = 9.708 2
−2.366 .
Solution 1. - Rectan Rectangul gular ar The distance between the two curves is 2(c) Solution (x1 x2 )2 + (y1 y2 )2 . Let cos θ and y 1 = 3 sin sin θ. We have x 2 = cos θ(3 2 sin(2 sin(2θ)) r1 = 3 and r 2 = 3 2sin(2θ). We have x 1 = 3 cos and y2 = sin θ(3 2 sin(2 sin(2θ)). )). So x1 x2 = 2sin2θ cos θ = a and y1 y2 = 2sin θ sin2θ = b. Enter a2 + b2 into Y1 and take the numerical derivative of Y1 at θ = π3 to get that the rate of change is 2.000 .
− −
−
√
−
− −
−
−
− r2 = 2 sin 2θ = distance between the two curves at θ = π3 . The deriv derivati ative ve of 2 sin 2θ at θ = π3 is 4cos 2 · π3 = −2 . Soluti Solution on 2. - Polar Note that r1
2(d) 2(d) Note Note that that dr dθ
θ= 6
π
=
dr dθ
· dθ dt
=
dr dr . Thus Thus,, = dt dt
π − 4 cos(2 cos(2θ) · 3 = − 12cos2θ =⇒ −12 cos cos 2 · 6
−6 .
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=
− 6 =⇒
3
3(a) g(3) =
this is the the area area bounde bounded d by the graph graph of f and the x-axis on [ 3, 3]. 3]. Thus Thus f (t) dt; this
−
−3
g(3) = 21 (5)(4) + 21 (1)( 2) =
−
⇒ g(3) (3) =
9
3(b) g is increasing when f (t) is positive and concave down when the slope of the line tangent to f (t) is negative. This occurs on ( 5, 3) and on (0, 2). (Note: x = 5 meets these conditions, but is excluded because the problem specifies “open intervals”)
− −
−
5xg (x) 5g(x) 4 4 ; g (3) = 9 and g (3) = = 2 25x 4 0
− − −2; so h (3) = 5(3)(−2) − 5(9) = − 1 . − 25(32 ) 3 3(d) p (x) = f (x2 − x) · (2x − 1) =⇒ p (−1) = f (2) · −3 = −2 · −3 = 6 3(c) h (x) =
−
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≤ t ≤ 8 is vA(8)8 −− v2A(2) = − 1103 meters/minute. 4(b) 4(b) Yes – since since vA is continuous, vA (5) = 40 and vA (8) = − 120 it follows from the intermediate value theorem that v A (t) = −100 at some time 5 ≤ t ≤ 8. 4(a) The average average acceleration acceleration of train A on 2
12
4(c) The position position of train train A at t = 12 is given by
vA (t) dt + 300 , where the +300 comes from the fact
2
that V a (2) = 300.
A trapezoidal approximation gives 21 (3)(100 + 40) + 21 (3)(40 120) + 21 (4)( 120 train A is approxi approximatel mately y 450 meters meters west west of its starting starting point point at t = 12.
−
−
− 150) = −450 450..
So,
4(d) 4(d) Let d be the distance between train A and train B . We want want d at t = 2. 2. Let Let dA be the east-west distance of train A from the origin, and let d B be the north-south distance of train B from the origin. By the distance formula d2 = d 2A + d2B Taking the derivative of each term with respect to time t and cancelling factors of 2: + dB dB dd = d A dA
We know that d = 500 because d is the hypotenuse of a right triangle with legs d A = 300 and d B = 400. We know from the table that dA = v A = 100 at t = 2, and we know from the function vB = d B that vB (2) = 125. We just substitute and solve for d : 500d = 300(100) + 400(125) =
⇒ 5d = 300 + 4(125) =⇒ d = 60 + 100 = 160
The rate of change of the distance between between the trains is 160 meters/m meters/minut inutee .
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1
5(a) R =
2
xex
0
1
− (−2x) dx =
1
5(b) V y=−2 =
x2
(xe
+ 2)
2
0
2
xex + 2x dx =
0
− (−2x + 2)
2
ex
2
2
+ x2
dx
1 0
=
e + 1
2
5(c) We can consider consider each of the 3 parts of the perimeter perimeter separately separately..
For y = 2x: The graph forms a right triange with the x -axis with legs 1 and 2; thus the hypotenuse (part of the perimiter) is 5.
−
√
2
For x = 1: 1: At x = 1 the value of xex is e and the value of y = is e ( 2) = e + 2 .
−−
For y = xe : The formula formula for the length length of this curve curve is
1 + f (x)2 dx; so the perimeter is
0
1
Thus Thus the perimet perimeter er
1
x2
2
2
1 + 2x2 ex + ex
0
−2x is −2.
2
dx.
1
So the total perimeter P is P =
2
1 + 2x2 ex + ex
0
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2
2
dx + e + 2 +
√
5.
6(a) The series series converg converges es if lim
−
n→∞
n+2 2
lim ( 1)
n→∞
Thus R =
tn+1 < 1. So we have tn n+1
n + 1
(x
n+1
− 1)
n
· 2n(−1)n+1(x − 1)n
|(−1)(2)(x − 1)n| < 1 =⇒ |(x − 1)| < 12
< 1
1 2
6(b) 6(b) The n th term of f f is ( 1)n+1 2n (x 1)n−1 (just take the derivative of the n th term of f f ). Let T (x)n represent the n term Taylor series for f about x = 1, so we have
−
−
T (x)3 = 2
− 4(x − 1) + 8(x − 1)2 + . . . + (−1)n+12n(x − 1)n−1 + . . .
6(c) T (x)n is a geometric series with first term 2 and ratio 2(x 1). 2 2 This series converges to = . To find f we must integrate this expression (since 1 [ 2(x 1)] 2x 1 it represents f ). So, 2 dx = ln 2x 1 + C 2x 1
− −
−− −
−
| − |
−
So, f (x) = ln 2x 1 + C . We now need to eliminate C . Note that f (1) = C because because ln 1 = 0. 0. If you you write the Taylor series for f (x) using the sum given in the probem, you will find that f (1) = 0 because all the terms have a factor of x 1 and thus go out to 0. Thus C = 0 and f (x) = ln 2x 1 .
| − |
−
| − |
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