14 1.m707 .m843 .cm^422 .cm562 .cmtdV
dt 80-1250O496 Answersf
dP
dt̄
̄
̄
t=4=4 Profit is increasing at $ 4000 per year
after 4 years.
dP
dt̄
̄
̄
t=10=28 Profit is increasing at $28 000per
year after 10 years.
dP
dt̄
̄
̄
t=25=88 Profit is increasing at $88 000per
year after 25 years.
2a iQ(0) = 100 iiQ(25) = 50 iii Q(100) = 0
bidecr. 1 unit per year iidecr.p^12 units per year
c Q^0 (t)=¡
5
p
t
< 0
3a 0 : 5 m
bt=4: 9 : 17 m, t=8: 12 : 5 m, t=12: 14 : 3 m
c t=0: 3 : 9 m year¡^1 , t=5: 0 : 975 m year¡^1 ,
t=10: 0 : 433 m year¡^1
dAs
dH
dt
=
97 : 5
(t+5)^2
> 0 for allt> 0 , the tree is always
growing.
4aC^0 (x)=0: 0009 x^2 +0: 04 x+4dollars per pair
bC^0 (220) =$ 56 : 36 per pair. This estimates the additional
cost of making one more pair of jeans if 220 pairs are
currently being made.
c $ 56 : 58 This is the actual increase in cost to make an extra
pair of jeans ( 221 rather than 220 ).
dC^00 (x)=0: 0018 x+0: 04
C^00 (x)=0when x=¡ 22 : 2. This is where the rate of
change is a minimum, however it is out of the bounds of the
model (you cannot make< 0 jeans!).
5a iE 4500 iiE 4000
bidecrease ofE 210 : 22 per km h¡^1
iiincrease ofE 11 : 31 per km h¡^1
c dC
dv
=0 at v=^3
p
500 000¼ 79 : 4 km h¡^16a
dV
dt
=¡ 1250³
1 ¡
t
80́
L min¡^1
batt=0when the tap
was first opened
c d(^2) V
dt^2
=^125
8
L min¡^2
This shows that the rate of change ofV is constantly
increasing, so the outflow is decreasing at a constant rate.
7aThe near part of the lake is 2 km from the sea, the furthest
part is 3 km.
b
dy
dx=
3
10 x
(^2) ¡x+^3
5
dy
dx
̄
̄
̄
x=^12
=0: 175 , height of hill is increasing as gradient
is positive.
dy
dx
̄
̄ ̄
x=1^12
=¡ 0 : 225 , height of hill is decreasing as
gradient is negative.
) top of the hill is betweenx=^12 andx=1^12.
c 2 : 55 km from the sea, 63 : 1 m deep
8ak= 501 ln 2¼ 0 : 0139
bi 20 grams ii 14 : 3 grams iii 1 : 95 grams
c 9 days and 6 minutes ( 216 hours)
di¡ 0 : 0693 gh¡^1 ii¡ 2 : 64 £ 10 ¡^7 gh¡^1
eHint: You should find
dW
dt
=¡ 501 ln 2£ 20 e¡
501 ln 2t
9ak= 151 ln
¡ 19
3
¢
¼ 0 : 123 b 100 ±C
c c=¡k¼¡ 0 : 123
didecreasing at 11 : 7 ±C min¡^1
iidecreasing at 3 : 42 ±C min¡^1
iii decreasing at 0 : 998 ±C min¡^1
10 a 43 : 9 cm b 10 : 4 years
cigrowing at 5 : 45 cm per year
iigrowing at 1 : 88 cm per year
11 a A(0) = 0
bik=
ln 2
3
(¼ 0 :231)
ii 0 : 728 litres of alcohol produced per hour
12 p^212 cm^2 per radian
13 a rising at 2 : 73 m per hour brising
14 b i 0 ii 1 iii ¼ 1 : 11
EXERCISE 14E
1 250 items
2bLmin¼ 28 : 3 m,
x¼ 7 : 07 m
c
3 10 blankets 414 : 8 km h¡^15 at 4 : 41 months old
6aHint: V= 200 = 2x£x£h
b Hint: Showh=
100
x^2
and substitute into the surface area
equation.
c SAmin¼ 213 cm^2 ,
x¼ 4 : 22 cm
d
7 20 kettles 8 C
³
p^1
2 ,e
¡^12
́
9aRecall that Vcylinder=¼r^2 h and that 1 L= 1000cm^3.
b Recall that SAcylinder=2¼r^2 +2¼rh.
c radius¼ 5 : 42 cm, height¼ 10 : 8 cm
10 b μ¼ 1 : 91 , A¼ 237 cm^211 b 6 cm£ 6 cm
12 a 06 x 663 : 7
b l= 100m,x=^100 ¼ ¼ 31 : 83 m,A=20 000¼ ¼ 6366 m^2
13 after 13 : 8 weeks 14 after 40 minutes
15 c μ=30±,A¼ 130 cm^2
16 a Hint: Show that AC= 360 μ £ 2 ¼£ 10
b Hint: Show that 2 ¼r=AC
c Hint: Use the result fromband Pythagoras’ theorem.
d V=^13 ¼
¡μ
36
¢ 2 q
100 ¡
¡μ
36
¢ 2
eμ¼ 294 ±
17 1 hour 34 min 53 s whenμ¼ 36 : 9 ± 189 : 87 m
EXERCISE 14F
1 ais decreasing at 7 : 5 units per second
2 increasing at 1 cm per minute
3a 4 ¼m^2 per second b 8 ¼m^2 per second
4 increasing at 6 ¼m^2 per minute
cyan magenta yellow black
(^05255075950525507595)
100 100
(^05255075950525507595)
100 100 IB HL OPT
Sets Relations Groups
Y:\HAESE\CAM4037\CamAdd_AN\496CamAdd_AN.cdr Tuesday, 8 April 2014 8:39:50 AM BRIAN