2010
Question 1
(a) 142.275 cubic feet (must show integral for full credit)
(b) -59.583 cubic feet per hour
0
, 0≤𝑡≤6
(c) ℎ(𝑡) = { 125(𝑡 − 6) , 6 < 𝑡 ≤ 7
125 + 108(𝑡 − 7), 7 < 𝑡 ≤ 9
(d) 26.335 cubic feet (must show integral and h(9) for full credit)
Question 2
(a) 4 hundred entries per hour
(b) Trapezoidal sum: 10.688; the meaning is the average number of hundreds of entries in the box
between noon and 8PM.
(c) 7 hundred entries (must show integral for full credit)
(d) Entries are being processed most quickly at time t = 12 (must set P’(t) = 0 and identify all four
possible candidates – 8, 9.184, 10.816, 12 – for full credit)
Question 3
(a) 3200 people (must show integral for full credit)
(b) Increasing because r(t) > 800 for 2 < t < 3.
(c) t = 3; 1500 people waiting in line
𝑡
(d) 0 = 700 + ∫0 𝑟(𝑠)𝑑𝑠 − 800𝑡
Question 4
(a) 18 (must show integral and antiderivative for full credit)
2
9
(b) 𝜋 ∫0 ((7 − 2√𝑥) − (7 − 6)2 ) 𝑑𝑥
6 3 4
𝑦 𝑑𝑦
16
(c) ∫0
Question 5
(a) 𝑔(3) =
13
+
2
𝜋;
𝑔(−2) = 5 − 𝜋
(b) The graph of y = g(x) has points of inflection at x = 0, x = 2, and x = 3 because g’ changes
from increasing to decreasing at x = 0 and x = 3, and g’ changes from decreasing to
increasing at x = 2.
(c) ℎ′ (𝑥) = 𝑔′ (𝑥) − 𝑥 ; Critical points are 𝑥 = √2 (relative max) and 𝑥 = 3 (neither a minimum
nor a maximum).
Question 6
(a) 𝑓 ′ (1) = 8; 𝑦 = 2 + 8(𝑥 − 1)
(b) 𝑓(1.1) ≈ 2.8; Because f(x) is concave up, the approximation 2.8 is less than f(1.1).
(c) 𝑓(𝑥) =
2
√5−4𝑥 2
2011 (Form B)
Question 1
(a)
(b)
(c)
(d)
171.813 mm (must show integral for full credit)
2.864 mm/day
𝑉(𝑡) = 100𝜋𝑆(𝑡); 𝑉 ′ (7) = 602.218 cubic mm per day
𝐷(0) = −0.675 < 0 and 𝐷(60) = 69.3773 > 0; Because D is continuous, the IVT implies that
there is a time t, 0 < t < 60, at which D(t) = 0. At this time, the heights of water in the two cans
are changing at the same rate.
Question 2
(a) The left-hand and right-hand limits are not equal, so r is not continuous at t = 5.
(b) 258.053 (must show integral for full credit)
(c) 𝑟 ′ (3) = 50; The rate at which water is draining out of the tank at time t = 3 hours is increasing at
50 liters per hour squared.
𝐴
(d) 12,000 − ∫0 𝑟(𝑡)𝑑𝑡 = 9000
Question 3
(a)
22
3
(must show integral for full credit)
2
(b) Volume = ∫0 2𝑦(6 − 𝑦 − 𝑦 2 )𝑑𝑦
(c) 𝑓 ′ (𝑥) = 2
1
√𝑥
;2
1
√𝑥
1
1
1
= 1 → 𝑥 = 4 ; The point P has coordinates (4 , 2).
Question 4
(a) x = 4; It is a relative maximum.
2(𝑥−6)
;
𝑥4
−2
−1
(b) 𝑓 ′′ (𝑥) =
(c) 3 − 2𝑥
+𝑥
The graph of f is concave down on the interval 0 < x < 6.
(must show integral and antiderivative for full credit)
Question 5
(a)
(b)
(c)
(d)
0.03 meters/sec2
The total distance that Ben rides from t = 0 to t = 60 is 139 meters.
Yes, the Mean Value Theorem implies there must be a time where v(t) = 2.
1.5 meters per second
Question 6
(a) 6π2 (must show antiderivative for full credit)
(b) x = 0 and x = π
𝜋
(c) ℎ′ (𝑥) = 𝑔(3𝑥) ∙ 3 ; ℎ′ (− 3 ) = 3𝜋