Exam 1 - Part I – 28 questions – No Calculator Allowed ...
Transcript of Exam 1 - Part I – 28 questions – No Calculator Allowed ...
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Exam 1 - Part I – 28 questions – No Calculator Allowed - Solutions
1. Find
�
limx→0
6x5 − 8x3
9x3 − 6x5
A.
�
23
B.
�
−89
C.
�
43
D.
�
−83
E. nonexistent
�
B. limx→0
6x5 − 8x3
9x3 − 6x5 = limx→0
2x3 3x2 − 4( )3x3 3− 2x2( ) =
−89
2. Let f be a function such that
�
limx→4
f x( ) − f 4( )x − 4
= 2 . Which of the following must be true?
I. f is continuous at x = 4. II. f is differentiable at x = 4. III. The derivative of
�
′ f is continuous at x = 4.
A. I only B. II only C. I and II only D. I and III only E. I, II and III
�
C. Students should recognize that the statement tells them that the derivative of f exists at x = 4 and thus it must be true that f is continuous at x = 4 as well. The question is, what about ′ ′ f ? It is easy to construct a piecewise function for ′ f whose left and right limits are 2 at x = 4.
′ f x( ) =2x − 6,x ≤ 42,x > 4⎧ ⎨ ⎩
and thus ′ ′ f x( ) =2,x ≤ 40,x > 4⎧ ⎨ ⎩
so ′ ′ f is not continuous at x = 4.
And it is easy to generate constants so that f is continuous at x = 4 : f x( ) =x2 − 6x,x ≤ 42x −16,x > 4⎧ ⎨ ⎩
3. If
�
f x( ) = 2x + 1( ) x 2 − 3( )4, then ′ f x( ) =
A.
�
2 x 2 − 3( )3 x 2 + 4x −1( ) B.
�
4 2x +1( ) x 2 − 3( )3 C.
�
8x 2x +1( ) x 2 − 3( )3 D.
�
2 x 2 − 3( )3 3x 2 + x − 3( ) E.
�
2 x 2 − 3( )3 9x 2 + 4x − 3( )
�
E. Product/chain rule : ′ f x( ) = 2x + 1( )4 x 2 − 3( )32x( ) + 2 x 2 − 3( )4
= 2 x 2 − 3( )34x 2x + 1( ) + x 2 − 3[ ]
′ f x( ) = 2 x 2 − 3( )39x 2 + 4x − 3( )
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4.
�
x 4 − x( )2∫ dx =
A.
�
x 9
9+ x 3
3+ C B.
�
x 9
9− x
6
6+ x 3
3+ C C.
�
x 9
9− x
6
3+ x 3
3+ C
D.
�
x 4 − x 2( )33
+ C E.
�
x 4 − x 2( )33 4x 3 − 2x( ) + C
�
C. x 8 − 2x 5 + x 2( )∫ dx = x 9
9− x
6
3+ x 3
3+ C
5.
�
ddx
tan t4 −1( ) dt1
x 3
∫ =
A.
�
sec2 x12 −1( ) B.
�
tan x 4 −1( ) C.
�
tan x12 −1( )
D.
�
3x 2 tan x12 −1( ) E.
�
12x11 tan x12 −1( )
�
D. ddx
tan t4 −1( ) dt1
x 3
∫ = tan x3( )4−1⎡
⎣ ⎢ ⎤ ⎦ ⎥ 3x2( ) = 3x2 tan x12 −1( )
6. Newton the cat begins to walk along a ledge at time t = 0. His velocity at
time t, 0 ≤ t ≤ 8, is given by the function whose graph is given in the figure to the right. What is Newton’s average speed from t = 0 to t = 8?
A. 0 B. 2 C. 3
D.
�
94
E. 5
�
C. Average Speed =v t( ) dt
0
8
∫8 − 0
=v t( ) dt + v t( ) dt + v t( ) dt − v t( ) dt
6
8
∫4
6
∫2
4
∫0
2
∫8 − 0
Average Speed =6 + 10 + 5 + 3
8=
248
= 3
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7. The table below gives selected values of
�
v t( ) , of a particle moving along the x-axis. At time t = 0, the particle is at the origin. Which of the following could be the graph of the position,
�
x t( ) , of the particle for 0 ≤ t ≤ 4 ?
t 0 1 2 3 4
�
v t( ) 3 0 -1 1 3
A. B. C.
D. E.
D. Velocity is zero at t =1 and at 2 < t < 3 so the graph of x t( ) must have horizontal tangents at t =1 and at 2 < t < 3.
8. The graph of a twice-differentiable function f is shown in the figure
to the right. Which of the following is true? A.
�
f −2( ) < ′ f −2( ) < ′ ′ f −2( ) B.
�
f −2( ) < ′ ′ f −2( ) < ′ f −2( ) C.
�
′ ′ f −2( ) < ′ f −2( ) < f −2( ) D.
�
′ f −2( ) < f −2( ) < ′ ′ f −2( ) E.
�
′ ′ f −2( ) < f −2( ) < ′ f −2( )
�
D. Since f is decreasing, ′ f −2( ) < 0 and since f is concave up, ′ ′ f −2( ) > 0. Since f −2( ) = 0, the largest is ′ ′ f −2( ) and the smallest is ′ f −2( ).
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9. What is the slope of the line tangent to the graph of
�
y = 1e2x x − 2( )
at x = 1 ?
A.
�
−2e
B.
�
−1e2
C. 0 D.
�
1e2
E.
�
−2e2
�
D. y = 1e2x x − 2( )
= e−2x
x − 2
′ y =x − 2( ) e−2x( ) −2( ) − e−2x
x − 2( )2 ⇒ ′ y 1( ) =2 e−2( ) − e−2
12 = 1e2
10. What are all the horizontal asymptotes of
�
f x( ) = 6 + 3ex
3− ex in the xy-plane?
A.
�
y = 3 only B.
�
y = −3 only C.
�
y = 2 only D.
�
y = −3 and y = 0 E.
�
y = −3 and y = 2
E. limx→∞
6 + 3ex
3− ex= −3 lim
x→−∞
6 + 03− 0
= 2
11. If f is continuous for all real numbers x and
�
f x( )1
4
∫ dx = 10, then f x − 2( ) + 2x[ ]3
6
∫ dx =
A. 37 B. 39 C. 35 D. 57 E. 25
�
A. f x − 2( ) + 2[ ]3
6
∫ dx = f x − 2( ) dx3
6
∫ + 2 dx3
6
∫ u = x − 2,du = dx x = 3,u = 1, x = 6,u = 4
f x − 2( ) dx3
6
∫ = f x( ) dx1
4
∫ = 10 2x dx3
6
∫ = x23
6= 36 − 9 = 27 ⇒10 + 27 = 37
12. If
�
f x( ) = x 3 + x 2 + x + 1 and g x( ) = f −1 x( ), what is the value of
�
′ g 4( ) ?
A.
�
185
B. 1 C.
�
157
D.
�
16
E. 24
�
D : Inverse : x = y 3 + y 2 + y + 1 = 4 ⇒ y = 1.
1 = 3y 2 + 2y + 1( ) dydx ⇒ dydx
= 13y 2 + 2y + 1
dydx y=1( )
= 13y 2 + 2y + 1
= 16
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13. Find the y-intercept of the tangent line to
�
4 x + 2 y = x + y + 3 at the point (9, 4).
A. -2 B. 10 C.
�
52
D. 15 E.
�
49
�
B. 2x
+ 1ydydx
= 1+ dydx
⇒ dydx
1y−1
⎛
⎝ ⎜
⎞
⎠ ⎟ = 1− 2
x
dydx
=1− 2
x1y−1
⇒ dydx 9,4( )
=1− 2
312−1
= −23
Tangent line : y − 4 = −23
x − 9( )
y − intercept : y − 4 = −23
−9( ) ⇒ y = 10
14. The function f is continuous and non-linear for
�
−3 ≤ x ≤ 7 and
�
f −3( ) = 5 and f 7( ) = −5. If there is no value c, where
�
−3 < c < 7 , for which
�
′ f c( ) = −1, which of the following statements must be true?
A. For all k, where − 3< k < 7, ′f k( ) < −1. B. For all k, where − 3< k < 7, ′f k( ) > −1. C.
�
For some k, where − 3 < k < 7, ′ f k( ) = 0. D.
�
For − 3 < k < 7, ′ f k( ) exists. E.
�
For some k, where − 3 < k < 7, ′ f k( ) does not exist.
�
E. This is the Mean - Value Theorem which states that there must be some value k.
− 3 < k < 7 such that f 7( ) − f −3( )
7 + 3= −5 − 5
10= −1. Since there is no such value k,
then there must be a value k on − 3 < k < 7 where f is not differentiable.
15. If
�
x 2 + 2y 2 = 34 , find the behavior of the curve at (-4, 3). A. Increasing, concave up B. Increasing, concave down C. Decreasing, concave up D. Decreasing, concave down E. Decreasing, inflection point
�
B. 2x + 4y dydx
= 0 ⇒ dydx
= −2x4y
= −x2y
dydx −4,3( )
= 46
> 0 Curve is increasing.
d2ydx 2 =
−2y + x dydx( )
4y 2 d2ydx 2
−4,3( )=−6 − 4 2
3( )36
< 0 Curve is concave down.
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16. A large block of ice in the shape of a cube is melting. All sides of the cube melt at the same rate. At the time that the block is s feet on each side, its surface area is decreasing at the rate of
�
24 ft2 hr . At what rate is the volume of the block decreasing at that time?
A.
�
12s ft 3 hr B.
�
6s ft 3 hr C.
�
4s ft 3 hr D.
�
2s ft 3 hr E.
�
s ft 3 hr
�
B. A = 6s2 ⇒ dAdt
= 12s dsdt
⇒−24 = 12s dsdt
⇒ dsdt
= −2s
ft hr
V = s3 ⇒ dVdt
= 3s2 dsdt
= 3s2 −2s
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = −6s ft 3 hr
17. A squirrel climbs a telephone pole and starts to walk along the
telephone wire. Its velocity v of the squirrel at time t, 0 ≤ t ≤ 6 is given by the function whose graph is to the right. At what value of t does the squirrel change direction?
A. 1 and 5 only B. 2 only C. 2 and 4 only D. 1, 3, and 5 only E. 1, 3, 4 and 5 only
�
B. Direction changes ocurs when velocity switches from positive to negative or from negative to positive. This occurs at time t = 2 only.
18. The graph of
�
′ f x( ), the derivative of f , is shown to the right. Which of the following statements is not true?
A. f is increasing on 2 ≤ x ≤ 3. B. f has a local minimum at x = 1. C. f has a local maximum at x = 0. D. f is differentiable at x = 3. E. f is concave down on -2 ≤ x ≤ 1.
�
B. Local minima occur when ′ f x( ) switches from negative to positive. This occurs at x = 2. A. is true as f is increasing when ′ f x( ) > 0 C. is true as local maxima occur when ′ f x( ) switches from postitive to negative. D. is true as ′ f 3( ) exists. ′ ′ f 3( ) does not exist. E. is true as f is concave down when ′ f x( )is decreasing.
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19. The difference in maximum acceleration and minimum acceleration attained on the interval 0 ≤ t ≤ 3 by the particle whose velocity is given by
�
v t( ) = 2t 3 −12t 2 +18t −1 is A. 4 B. 6 C. 18 D. 21 E. 24
�
E. a t( ) = 6t 2 − 24t + 18 ⇒ ′ a t( ) = 12t − 24 = 0 ⇒ t = 2
a 0( ) = 18, a 2( ) = −6, a 3( ) = 0 Maximum acceleration = 18 Minimum acceleration = −6 Range = 24
20. The function f is continuous on the closed interval [0, 8] and has the values given in the table below.
The trapezoidal approximation for
�
f x( ) dx0
8
∫ found with 3 subintervals is 20k. What is the value of k?
�
x 0 3 5 8f x( ) 5 k 2 7 10
A. 4 B.
�
±4 C. 8 D. -8 E. No values of k
A. 32
5+ k2( )+ 22k2 + 7( )+ 3
27+10( ) = 20k
15+ 3k2 + 2k2 +14 + 51= 40k⇒ 5k2 − 40k + 80 = 0
5 k2 −8k +16( ) = 0 ⇒ k − 4( )2 = 0 ⇒ k = 4
21.
�
sin2x1− sin2 2x
∫ dx =
A.
�
2ln cos2x( ) + C B.
�
12ln cos2x( ) + C C.
�
12sec2x tan2x + C
D.
�
12sec2x + C E.
�
2tan2x + C
�
D. sin2x1− sin2 2x
∫ dx = sin2xcos2 2x
∫ dx = tan2x sec2x dx =∫ 12
sec2x + C
22. The line
�
x + y = k , where k is a constant, is tangent to the graph of
�
y = 2x 3 − 9x 2 − x +1. What are the only possible values of k?
A. 1 only B. 0 and - 29 C. 1 and -29 D. 0 and 3 E. 1 and -26
�
E. y = −x + k ⇒ ′ y = −1 y = 2x 3 − 9x 2 − x + 1⇒ ′ y = 6x 2 −18x −1 = −1⇒ 6x x − 3( ) = 0 x = 0,y = 1 so x + y = 1 x = 3,y = −29 so x + y = −26
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23. The graph of
�
y = ′ f x( ) , the derivative of f , is shown in the figure to the right. If
�
f 0( ) = −1, then f 1( ) =
A. 1 B. 2 C. 3 D. 4 E. 5
�
B. Method 1: ′ f x( )0
1
∫ dx = f 1( ) − f 0( ) ⇒ f 1( ) = f 0( ) + ′ f x( )0
1
∫ dx
′ f x( )0
1
∫ dx = area of trapezoid =12
5 + 1( ) = 3⇒ f 1( ) = −1 + 3 = 2
Method 2 : ′ f x( ) = −4x + 5 ⇒ f x( ) = −2x2 + 5x + C
f 0( ) = 0 + C = −1⇒ f x( ) = −2x2 + 5x −1⇒ f x( ) = −2 + 5 −1 = 2
24. The average value of
�
sin2 x cos x on the interval
�
π2, 3π2
⎡ ⎣ ⎢
⎤ ⎦ ⎥ is
A.
�
−23π
B.
�
23π
C. 0 D. -1 E. 1
�
A. favg =
sin2 x cos xπ 2
3π 2
∫ dx
3π2
−π2
u = sin x,du = cos x dx x =π2
,u = 1 x =3π2
,u = −1
favg =u2
1
-1
∫ du
π=u3
31
−1
=−23π
25. The functions f and g are differentiable and
�
f g x( )( ) = x 2 for all x. If
�
f 4( ) = 8, g 4( ) = 8, ′ f 8( ) = −2, what is the value of ′ g 4( )?
A.
�
−18
B.
�
−12
C.
�
−2 D.
�
−4 E. Insufficient data
�
D. ′ f g x( )( ) ′ g x( ) = 2x ⇒ ′ g x( ) = 2x′ f g x( )( )
′ g x( ) = 8′ f 8( )
= 8−2
= −4
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26. Let f be a twice-differentiable function such that
�
f a( ) = b and f b( ) = a for two unknown constants a and b, a < b. Let
�
g x( ) = f f x( )( ). The Mean Value Theorem applied to
�
′ g on [a, b] guarantees a value k such that a < k < b such that
A.
�
′ g k( ) = 0 B.
�
′ ′ g k( ) = 0 C.
�
′ g k( ) =1 D.
�
′ ′ g k( ) = 1 E.
�
′ g k( ) = b − a
�
B. ′ g x( ) = ′ f f x( )( ) ⋅ ′ f x( ) ′ g a( ) = ′ f f a( )( ) ⋅ ′ f a( ) = ′ f b( ) ⋅ ′ f a( ) ′ g b( ) = ′ f f a( )( ) ⋅ ′ f b( ) = ′ f a( ) ⋅ ′ f b( ) so ′ g a( ) = ′ g b( ) Since f is twice differentiable, ′ g is differentiable so the MVT guarantees that
′ ′ g k( ) = ′ g b( ) − ′ g a( )b − a
= 0
27. The graph of
�
f x( ) = x 2 + 0.0001 − 0.01 is shown in the graph to the right. Which of the following statements are true?
�
I. limx→0
f x( ) = 0.
II. f is continuous at x = 0.III. f is differentiable at x = 0.
A. I only B. II only C. I and II only D. I, II, and III E. None are true
�
D. A trap problem. This looks like x which is not differentiable at x = 0.
But the function is given and ′ f x( ) =x
x2 + 0.0001 and ′ f 0( ) = 0.
28. What is the region enclosed by the graphs of
�
y = x − 4x 2 and y = −7x ?
A.
�
43
B.
�
163
C. 8 D.
�
683
E.
�
803
�
B. Even without a calculator, students should be able to sketch the line and parabola.
x − 4x2 = −7x⇒ 4x2 − 8x = 0 ⇒ 4x x − 2( ) = 0 ⇒ x = 0,x = 2
A = x − 4x2 + 7x( ) dx0
2
∫ = 8x − 4x2( ) dx0
2
∫ = 4x2 −4x3
3⎡
⎣ ⎢
⎤
⎦ ⎥
0
2
= 16 − 323
=163
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Exam 1 - Part II – 17 questions – Calculators Allowed - Solutions
29. The slope field for the equation in the figure to the right could be
A.
�
dydx
= x + y 2 B.
�
dydx
= x − y 2 C.
�
dydx
= xy
D.
�
dydx
= x + y E.
�
dydx
= x 2 − y
�
E. Since the slope is 0 at 1,1( ), choices A and C are eliminated
Since the slope is 0 at −1,1( ), choice B is eliminated The slope field cannot be D because along the line y = −x, all slopes should be zero.
30. Let R be the region between the graphs of
�
y = 2sin x and y = cos x as shown in the figure to the right. The region R is the base of a solid with cross sections perpendicular to the x-axis as rectangles that are twice as high as wide. Find the volume of the solid.
A. 4.472 B. 7.854 C. 8.944 D. 15.708 E. 31.416
�
D. 2sin x = cos x⇒ x = 0.464,3.605
s = 2sin x − cos x. Area = 2s s( ) = 2s2. V = 2 2sin x − cos x( )2
0.464
3.605
∫ dx = 15.708
31. The first derivative of a function f is given by
�
′ f x( ) = x − 2( )esin 2x . How many points of inflection does the graph of f have on the interval 0 < x < 2π?
A. Two B. Three C. Four D. Five E. Six
�
B. ′ ′ f x( ) = 2cos2x x − 2( )esin 2x + esin 2x
As shown on the graph of ′ ′ f x( ), there are 3 values of x where it changes sign.
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32. A conical tank is leaking water at the rate of
�
75in3 min. At the same time, water is being pumped into the tank at a constant rate. The tank’s height is 60 in while its top diameter is 20 inches. If the water level is rising at the rate of
�
5in min when the height of the water is 10 inches high, find the rate in which water is being pumped into the tank to the
nearest
�
in3 min . (The volume of a cone is given by
�
V = 13πr2h )
A.
�
44 in3 min B.
�
175 in3 min C.
�
250 in3 min D.
�
119 in3 min E.
�
31 in3 min
�
D. V = 13πr2h r
h= 10
60⇒ r = h
6
V = π3h2
36⎛
⎝ ⎜
⎞
⎠ ⎟ h = πh3
108
dVdt
− 75 = πh2
36dhdt
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 100π
365( ) = 43.633
dVdt
= 75 + 43.633 ≈119 in3 min.
33. At which points is the tangent line to the curve
�
8x 2 + 2y 2 = 6xy +14 vertical? I. (-2, -3) II (3, 8) III. (4, 6) A. I only B. II only C. III only D. I and II only E. I and III only
�
A. 8x 2 + 2y 2 = 6xy + 14
16x + 4y dydx
= 6x dydx
+ 6y⇒ dydx
= 6y −16x4y − 6x
= 3y − 8x2y − 3x
Tangent line vertical when dydx
is undefined so 2y = 3x.
−2,−3( ) is on the curve but 4,6( ) is not.
34. For the function f ,
�
′ f x( ) = 4 x − 3 and f 2( ) = 4. What is the approximation for
�
f 2.1( ) found by using the tangent line to the graph of f at x = 2.
A.
�
−2.6 B. 4.5 C. 4.8 D. 5.4 E. 9.4
�
B. Tangent line : ′ f 2( ) = 4 2( ) − 3 = 5 ⇒ y − 4 = 5 x − 2( ) ⇒ y 2.1( ) ≈ 5 2.1( ) − 6 = 4.5. Be careful of the trap answer (D). You aren't given the tangent line equation, just the formula for the slope of the tangent line.
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35. Let
�
F x( ) be an antiderivative of
�
x 3 + x +1 . If
�
F 1( ) = −2.125, then F 4( ) =
A. -15.879 B. -11.629 C. 7.274 D. 15.879 E. 11.629
�
E. x3 + x + 1 dx1
4
∫ = F 4( ) − F 1( ) ⇒ F 4( ) = F 1( ) + x3 + x + 1 dx1
4
∫
F 4( ) = −2.125 + x3 + x + 1 dx1
4
∫ = −2.125 + 13.754 = 11.629
36. What is the area of the region in the first quadrant enclosed by the graph of
�
y = 2cos x,y = x, and the x-axis?
A. 0.816 B. 1.184 C. 1.529 D. 1.794 E. 1.999
�
A. A = x dx0
1.03
∫ + 2cos x dx1.03
π 2
∫ or A = cos−1 y2
⎛ ⎝
⎞ ⎠ − y
⎡ ⎣ ⎢
⎤ ⎦ ⎥ dx
0
1.03
∫ = 0.816
37. A particle moves along the x-axis so that at any time t > 0, its acceleration is given by
�
a t( ) = cos 1− 2t( ) . If the velocity of the particle is -2 at time t = 0, then the speed of the particle at t = 2 is
A. 0.613 B. 0.669 C. 1.331 D. 1.387 E. 1.796
�
D. v 2( ) = v 0( ) + a t( ) dt = −2 + 0.613 = −1.387 ⇒ Speed = 1.3870
2
∫
38. A right triangle has legs a and b and hypotenuse c. The lengths of leg a and leg b are changing but at a certain instant, the area of the triangle is not changing. Which statement must be true?
A. a = b B.
�
dadt
= dbdt
C.
�
dadt
= − dbdt
D.
�
a dadt
= −b dbdt
E.
�
a dbdt
= −b dadt
�
E. A = 12ab⇒ dA
dt= 1
2a dbdt
+ b dadt
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 0
a dbdt
= −b dadt
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39. A particle moves along a straight line with velocity given by
�
v t( ) = t − 2.5cos2t . What is the acceleration of the particle at t = 2 ?
A. 0.168 B. 0.238 C. 0.451 D. 0.584 E. 1.450
�
B. a t( ) = 1− 2.5cos2t ln2.5( ) −sin2t( ) 2( ) ⇒ a t( ) = 1− 2.5cos4 ln2.5( ) −sin4( ) 2( ) = 0.238
40. If the graph of
�
′ f x( ) is shown to the right, which of the following could be the graph of
�
y = f x( )?
I. II. III. A. I only B. II only C. III only D. I and III only E. II and III only
�
A. Since ′ f x( ) > 0,x ≠ 0, f x( ) is increasing. But since ′ f 0( ) = 0
f x( ) has a horizontal tangent at x = 0. For III, ′ f 0( ) does not exist.
41. The rate at which the gasoline is changing in the tank of a hybrid car is modeled by
�
f t( ) = t + .5sin t − 2.5 gallons per hour, t hours after a 6-hour trip starts. At what time during the 6-hour trip was the gasoline in the tank going down most rapidly?
A. 0 B. 2.292 C. 3.228 D. 4.203 E. 6
�
A. ′ f t( ) = 12 t
+ .5cos t = 0. ⇒ t = 2.292 or t = 4.203.
Since the change is negative, we want the value of t when f t( ) is a minimum.
f 0( ) = −2.5 f 2.292( ) = −0.611 f 4.203( ) = −0.886 f 6( ) = −0.190
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42. The expression
�
175
ln 7675
+ ln 7775
+ ln 7875
+ ...+ ln2⎛ ⎝ ⎜
⎞ ⎠ ⎟ is a Riemann sum approximation for
A.
�
ln x75
⎛ ⎝
⎞ ⎠ dx
1
2
∫ B.
�
ln x75
⎛ ⎝
⎞ ⎠ dx
76
150
∫ C.
�
175
ln x dx76
150
∫ D.
�
ln x dx1
2
∫ E.
�
175
ln x dx1
2
∫
�
D. This is a right Riemann sum. The base is 1
75 and the heights are ln 1 +
175
⎛ ⎝
⎞ ⎠ ,ln 1 +
275
⎛ ⎝
⎞ ⎠ ...ln 1 +
7575
⎛ ⎝
⎞ ⎠
So this is ln x dx1
2
∫ .
43. Let f be the function given by
�
f x( ) = cos t2 + t( ) dt0
x
∫ for − 2 ≤ x ≤ 2. Approximately, for what
percentage of values of x
�
for − 2 ≤ x ≤ 2 is
�
f x( ) decreasing?
A. 30% B. 26% C. 44% D. 50% E. 59%
�
B. ′ f x( ) =ddx
cos t2 + t( ) dt0
x
∫ = cos x2 + x( ) By the graphs below, ′ f x( ) < 0 for − 2 < x < 1.849 and 0.849 < x < 1.728 So ′ f x( ) < 0 for 1.849 + 2 + 1.728 − 0.849 = 1.030 values out of 4 = 26%
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44. The rate of change of people waiting in line to buy tickets to a concert is given by
�
w t( ) = 100 t 3 − 4t 2 − t + 7( ) for 0 ≤ t ≤ 4. 800 people are already waiting in line when the box office opens at t = 0. Which of the following expressions give the change in people waiting in line when the line is getting shorter?
A.
�
′ w t( ) dt1.480
3.773
∫ B.
�
w t( ) dt1.480
3.773
∫ C.
�
800 − ′ w t( ) dt1.480
3.773
∫
D.
�
′ w t( ) dt0
2.786
∫ E.
�
w t( ) dt0
2.786
∫
�
B. Since w t( ) represents the rate of people waiting, w t( )t=a
t=b
∫ dt represents the change of
the number of people in line between 2 times. The line is getting shorter when its rate of change is negative. The 800 people already in line doesn't enter into it.
45. A particle moves along the x-axis so that its velocity
�
v t( ) =12te−2t − t +1. At t = 0, the particle is at position
�
x = 0.5. What is the total distance that the particle traveled from t = 0 to t = 3 ?
A. 1.448 B. 1.948 C. 2.911 D. 4.181 E. 4.681
�
D. Distance = v t( ) dt −0
1.69
∫ v t( ) dt or 1.69
3
∫ v t( ) dt = 4.181.0
3
∫ The starting position has no bearing on the answer.
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Exam 1 – Answers and Description
# Ans. Topic # Ans. Topic1 B Limits/Horizontal Asymptotes 29 E Differential Equations2 C Definition of Derivative 30 D Area/Volume3 E Derivatives of Basic Functions 31 B Function Analysis4 C Integration Techniques 32 D Related Rates5 D Derivative of Accumulation Function 33 A Horizontal/Vertical Tangent Lines6 C Average Value of a Function 34 B Tangent Lines and Local Linear Approx.7 D Straight-Line Motion - Integrals 35 E Fundamental Thm of Calculus8 D Function Analysis 36 A Area/Volume9 D Tangent Lines and Local Linear Approx. 37 D Straight-Line Motion - Integrals
10 E Limits/Horizontal Asymptotes 38 E Related Rates11 A Fundamental Thm of Calculus 39 B Straight-Line Motion - Derivatives12 D Derivative of Inverse Function 40 A Function Analysis13 B Implicit Differentiation 41 A Absolute Extrema14 E Intermediate/Mean Value Theorem 42 D Riemann Sums15 B Implicit Differentiation 43 B Derivative of Accumulation Function16 B Related Rates 44 B Derivative as a Rate of Change17 B Straight-Line Motion - Derivatives 45 D Straight-Line Motion - Integrals18 B Function Analysis19 E Absolute Extrema20 A Riemann Sums21 D Integration Techniques22 E Tangent Lines and Local Linear Approx.23 B Derivative as a Rate of Change24 A Average Value of a Function25 D Derivatives of Basic Functions26 B Intermediate/Mean Value Theorem27 D Continuity/Differentiability28 B Area/Volume