Standard Level 2 – pg 110 – answers?
1) a) m = ½
b) m= 3/2
c) m = 2
2) a) m = -1/2 b) m = -1
c) m = -3/2
3) dy/dx = 0
4) dy/dx = 0
5) dy/dx = 1
9) y’ = -4t+3
10) y’ = 3x
14) y’ = cos(x) 15) y’ =
−1
2
2
11) y’ = 3t -2
−1
-3cos(x)
16) y’ =
𝑥2
−2𝜋
−1
21)
9𝑥 3
2√𝑋 3
19) y’ = 9𝑥 4
20) y’ =
25) f’(0) = 0
26) y’(2) = 36
27) y’(0) = 4
d) m = 3
d) m = -2
6) dy/dx = 3
7) y’= 2x
8) y’ = 2t+2
1
12) dy/dx = 6x2-2x+3
13) dy/dx = 2x+2sin(x)
−1
𝑥4
-𝜋sin(x)
17) Dxy =
22) 12𝑥 2
23) f’(1) = -1
6
18)
−4
3𝑥 3
3
1
32) f’(x) = 2 x- 3 + 𝑥 3
5
24) f’(5) = 3
36) h’(x) = 2 - 𝑥 2 =
2𝑥 2 −1
𝑥2
43) y’ = 4x3-6x
y’(1) = -2
2x+y=2
47) y’ = 4x3 -16x
y’ = 0 when 4x3-16x=0
4x(x2-4)=0
4x(x-2)(x+2)=0
X=0,2,-2
Horizontal tangents at (0,0), (2,-14), (-2,-14)
60) a) point (1,1) and nearby point could be (1.0073138, 1.0221024)
m=
1.0221024−1
1.0073138−1
≈3.022 (answers vary)
y=3.022(x-1)+1
b) f’(x)=3x2
T(x) = 3(x-1)+1
c) The accuracy worsens as you move away from (1,1)
d)
-3
-2
-1
-0.5
-0.1
0
∆𝑥
f(x)
-8
-1
0
0.125 0.729
1
T(x)
-8
-5
-2
-0.5
0.7
1
0.1
1.331
1.3
0.5
3.375
2.5
1
8
4
2
27
7
3
64
10
69) a) The slope appears to be greatest between A and B.
b) The average rate of change between A and B is greater than the instantaneous rate of change at
c)
d) The average rate change are approximately equal between B and C, and between D and E.
71) a) s(t) = -16t2 + 1362
s’(t) = v(t) = -32t
b)
𝑠(2) − 𝑠(1)
2−1
= 1298 – 1346 = -48 ft/sec
c) v(t) = s’(t) = -32t, v(1) = -32 ft/sec, v(2) = -64 ft/sec
d) -16t2 + 1362 = 0
√1362
≈ 9.226 sec
4
√1362
-32( 4 ) = -8√1362 ≈
t2 = 1362/16 → t =
√1362
)=
4
e) v(
-295.242 ft/sec.
72) s(t) = -16t2 -22t + 220, v(T) = -32T -22, v(3) -118 ft/sec
s(t) = -16t2 -22t +220 = 112 (after falling 108 ft)
-16t2 -22t + 108 = 0
-2(8t2 + 11t - 54) = 0
-2(t – 2)(8t + 27) = 0
T = 2, so v(2) = -32(2) – 32 = -86 ft/sec.
73) s(t) -4.9t2 + 120t
v(t) = -9.8t + 120
v(5) = 71 m/sec
v(10) = 22 m/sec
74) s(t) = -4.9t2 + so when t = 6.8, s0 =226.6 m
75) aver vel. = 2 miles / 4min = 30 miles/hr. over 0<t<4,
0 miles / min over 4 < t < 6,
1 mile/min over 6< t < 10.
76) 5/6 miles/min = 50 miles/hr over 0<t<6 min,
0 miles/min over 6<t<8 min,
2 miles/min = 30 miles/hr
77) (0, 0) (6, 4), (8, 4), and 10, 6)
78) (0, 0), (4, 2), (6, 2), (10,6)
81) A = s2,
86)
𝑑𝑇
𝑑𝑠
𝑑𝐴
𝑑𝑠
𝑑𝐴
= 2𝑠 , when s = 4 m, 𝑑𝑠 = 8 m2
= K(T – Ta)
87) y = 2x2 – 3x + 1
89) y = x3 – 9x, y’ = 3x2 – 9,
Points of tangency are (0, 0) and (3/2, -81/8)
At (0, 0) the slope = -9, at 3/2, -81/8) the slope is -9/4.
The equations are 9x + y = 0 and 9x + 4y + 27 = 0
91) f(x) must be continuous at x = 2 to be differentiable at x = 2.
So 8a = 4 + b
Also, f’(x) must be the same from both sides, so 3ax2 = 2x at x = 2.
This means 3a(2)2= 2(2), so 12a = 2 and a = 1/3. Therefore 8(1/3) = 4 + b, so b = -4/3.