AP Calculus Midterm Review
Name:
No Calculators Allowed
For questions 1 and 2, a particle is moving along the x-axis with position given by
x(t) = t3 – 7t2 + 8t + 5 for times on [0, 10].
1. During what time intervals is the speed of the particle increasing?
(A) (7/3, 10]
(B) (4 , 10]
(C) [0, 2/3) and (7/3, 4)
(D) [0, 2/3) and (4, 10]
(E) [2/3, 7/3) and (4, 10]
2. What is the position of the particle when it is farthest to the left?
(A) -14
(B) -11
(C) -47/27
(D) 5
(E) 203/27
3. Find dy/dx for y=4sin2(3x).
(A) 8 sin(3x)
(B) 24 sin(3x)
(C) 8 sin(3x) cos(3x)
(D) 12 sin(3x) cos(3x)
(E) 24 sin(3x) cos(3x)
4.
(A) 5
(B) 3
(C) -3
(D) -5
(E) The limit does not exist.
5. The graph of f’(x) is given below on the domain[-3, 3]. On which interval(s) is the function f(x)
both increasing and concave up?
(A) (-2, 2)
(B) (-2, 0) and (0,2)
(C) (-3,2)
(D) (-2,-1) and (0,1)
(E) none of these
x
f(x) g(x) f’(x) g’(x)
1
3
1
-2
4
2
5
3
1
-4
3
2
1
-2
1
4
4
-3
2
-1
6. If p(x) = xf(x) – g(3x-2), then use the table above to evaluate p’(2).
(A) 11
(B) 10
(C) 8
(D) 6
(E) 4
7. The graph of f(x) consists of lines and quarter circles. What is the value of
(A) (10 -5π)/4
(B) (10 + 5π)/4
(C) (12 + 5π)/4
(D) (12 – 5π)/4
(E) none of these
?
8. A rectangle is drawn in the first quadrant so it has two sides on the coordinate axes and one
vertex on the curve y= - ln(x). Find the x coordinate of the vertex for which the rectangle’s area is a
maximum.
(A) ½
(B) – ln(1/2)
(C) 1/e
(D) e
(E) 1/e2
9. Based on the graph of d2(f(x))/dx2 above, how many points of inflection exist on f(x) on (-4,8)?
(A) 5
(B) 4
(C) 3
(D) 2
(E) 1
10. When the height of a cylinder is 12 com and the radius is 4 cm, the circumference of the
cylinder is increasing at a rate of π/4 cm/min, and the height of the cylinder is increasing four
times faster than the radius. How fast is the volume of the cylinder increasing?
(A) π/2 cm3/min
(B) 4π cm3/min
(C) 12π cm3/min
(D) 20π cm3/min
(E) 80π cm3/min
Calculators Allowed
11. What is the average value of (sin x)4cos x on the interval 0≤ x ≤ π/2?
(A) 3/ ln 4
(B) 6 / (π ln 4)
(C) 4 / ln 4
(D) 8 / ( π ln 4)
(E) 3π / ( 2 ln 4)
12. Let f(x) be a function such that f’(x)= ln x(cos x) + (sin x)/x. On the interval (0,3) f(x) has a
point of inflection nearest x=
(A) 0.352
(B) 1.101
(C) 2.128
(D) 2.259
(E) 2.901
x
-3
-2
-1
0
1
2
3
f(x)
1.25 1.07 0.53 0.27 1.04 3.56 2.18
13. Using the table above and three left-hand rectangles of equal width, approximate
(A) 9.900
(B) 7.720
(C) 5.640
(D) 4.900
(E) 2.820
.
14. In right triangle ∆ABC, <C = 90, point A is moving along AC at a rate of .5 cm/sec and point B is
moving along BC at a rate of 1/3 cm/sec. What is the rate of change in the area of ∆ABC at the
instant AC = 15 cm and BC = 20 cm?
(A) -0.0833 cm2/sec
(B) -0.4167 cm2/sec
(C) -0.833 cm2/sec
(D) -7.5 cm2/sec
(E) -15 cm2/sec
15. If
and
(A) k+4
(B) k -4
(C) 16 – k
(D) k – 16
(E) -16 - k
, what is the value of
?
Free Response Questions:
16. No Calculator Allowed
17. Calculator Allowed
REVIEW ANSWERS
1E
2B
3E
4D
5D
6B
7A
8C
9D
10 D
11 B
12 B
13 C
14 D
15 A
x(t) is distance, v(t)=x‘ (t) = 3t2 – 14t + 8 = (3t -2)(t -4).
min is at critical point or endpoint. Critical pts x=2/3, 4. Check x(t) values at t=0, 4, 10.
4*2sin(3x)*cos(3x)*3
limit as x->2 of (x-2)(x+3)/-(x-2) = limit as x->2 of -1(x+3)
f ‘ (x) >0 AND f “(x) >0
p’(x)=xf ‘(x)+f(x)-g ‘(3x-2)*3 so p’(2)= 2*1 + 5 –(-1*3)
-1/4π*4 + 2.5 – 1/4π
A=x(-ln x) A’=x(-1/x) + -ln x = -1 – ln x
Min at A’=0 so x=e-1
when g” changes sign or is undefined
Use C=2πr to get dr/dt=1/8. Thus dh/dt=1/2.
f ‘ (x) = max means x is inflection point.
when f” changes sign or is undefined OR f ‘ is a max or min value
2( 1.25 + .53 + 1.04)
A=(1/2)xy dA/dt = .5 ( y dx/dt + x dy/dt) = .5 (15 • -1/3 + 20 • -1/2)
- (Integral from 1 to 7 – integral from 1 to 3) =
-(Integral from 1 to 7 + integral from 3 to 1) = -(-4 +k)
16 a) f’(x) = 0 at x=-3, 1, 4. f’(x) changes sign from pos to neg at -3 and -4. Thus f’(x) has a max at
x=-3 and x=-4.
b) f’(x) changes sign at x=-4, -1 and 2. Thus f(x) has inflection pts at x=-4, -1, and 2
c) f(x) is concave up with pos slope where f’(x) is increasing and positive: (-5,-4) and (1,2)
d) Candidates are where f’(x) changes sign from neg to positive and the enpoints. Check f(1), f(5) and f(5). Absolut min on [-5,5] is f(1)=3
17 a) r(5.4) approx r(5) + r ’(5)∆t= 30+ 2(.4) = 30.8 ft . Since r is concave down on (5, 5.4), the
estimate is greater than r(5.4)
b) dV/dt = 4πr2 dr/dt. At t=5, dV/dt = 7200π ft3/min
c) 2(4) + 3(2) + 2(1.2) + 4(.6) + 1(.5) = 19.3 ft. The integral is the change in radius, in feet, from
0 to 12 mins
d) Since r(x) is concave down, r’(x) is dec on (0,12). The estimate of 19.3 ft is less than the
integral’s value