Maths Quest Maths C Year 12 for Queensland 2e 1 WorkSHEET 8.2 Advanced exponential functions Name: ___________________________ 1 The vertical motion of the suspension system of a motor vehicle is given by ( ) t e y t 4 sin 2 - = where y is the displacement from the centre of oscillation at time, t. For this system when , p = t calculate (a) the displacement (b) the velocity (c) the acceleration of the system. (a) ( ) ( ) 0 4 sin 2 = = - p p p e y (b) ( ) ( ) ( ) ( ) p p p p p - - - - - = + - = + - = e e e t e t e y t t 8 4 cos 8 4 sin 2 4 cos 8 4 sin 2 ! (c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) p p p - - - - - - - - - = ´ - = + - = - - = - - - = + - - = - - = e e y t t e t t e t t t t e t t e t t e y t t e y t t t t t t 16 8 2 4 cos 8 4 sin 15 2 4 cos 8 4 sin 15 2 4 sin 16 4 cos 4 4 cos 4 4 sin 2 4 sin 16 4 cos 4 2 4 cos 4 4 sin 2 4 cos 4 4 sin 2 ! ! ! ! ! 2 Solve for A and B in the following equation, given that ≠0 and ≠0. & + & − 2 + + =0 If & ≠ 0, clearly BOTH & + & − 2 = 0 and + = 0 Solve simultaneously; take equation2: = − = −1 now, 1+ & − 2 = 0 +1 −1 =0 so, = ±1 Hence, = 1 = ±1 *** This is an IMPORTANT concept and relatively “different” to the use of the Null Factor Law
Transcript
Maths Quest Maths C Year 12 for Queensland 2e 1
WorkSHEET 8.2 Advanced exponential functions
Name: ___________________________ 1 The vertical motion of the suspension
system of a motor vehicle is given by ( )tey t 4sin2 -= where y is the
displacement from the centre of oscillation at time, t. For this system when ,p=t calculate (a) the displacement (b) the velocity (c) the acceleration of the system.
(a) ( ) ( )
04sin2
== - pp pey
(b)
( ) ( )( ) ( )
p
pp pp-
--
--
=
+-=
+-=
eee
tetey tt
84cos84sin24cos84sin2!
(c)
( )( )
( )( )( )( )
( )p
pp-
-
-
-
-
-
-
-
-=
´-=
+-=
--=
---=
+-
-=
--=
eey
ttette
ttttette
tteyttey
t
t
t
t
t
t
1682
4cos84sin1524cos84sin152
4sin164cos44cos44sin24sin164cos42
4cos44sin24cos44sin2
!!
!!
!
2 Solve for A and B in the following equation, given that 𝐶 ≠ 0 and 𝐷 ≠ 0. 𝐴& + 𝐵& − 2𝐵 𝐶 + 𝐴𝐵 + 𝐵 𝐷 = 0
If 𝐶&𝐷 ≠ 0, clearly BOTH 𝐴& + 𝐵& − 2𝐵 = 0
and 𝐴𝐵 + 𝐵 = 0
Solve simultaneously; take equation2:
𝐴𝐵 = −𝐵 𝐴 = −1
now, 1 + 𝐵& − 2𝐵 = 0 𝐵 + 1 𝐵 − 1 = 0
so, 𝐵 = ±1
Hence, 𝐴 = 1𝑎𝑛𝑑𝐵 = ±1
*** This is an IMPORTANT concept and relatively “different” to the use of the Null Factor Law
Maths Quest Maths C Year 12 for Queensland 2e 2
3 In a particular electrical circuit at time t, the current, I, is given by
tQIdd
=
and the charge, Q, is ( )tAeQ t pcos-= . Initially at time t = 0, the charge in the system is 10. (a) What is the charge when t = 5? (b) What is the current in the system
when t = 5?
(c) How does the charge change over
time?
(a) Note that ( ) 100 =Q
( ) ( )
( ) ( )( ) ( )( ) 5
5
0
1055cos105
cos1010
0cos0
-
-
-
-
-=\===\==
eQeQ
tetQA
AAeQ
t
pp
(b)
( ) ( )( )( )
( )5
5
105cos105
sincos10sin10cos10
dd
-
-
-
--
=-=\
+-=\--=
=
eeI
tteItete
tQI
t
tt
ppp
ppp
(c) As ( ) 0, ®¥® tQt because
0cos10lim =-
¥®te t
tp .
Maths Quest Maths C Year 12 for Queensland 2e 3
4 The motion of a pendulum subject to damping has a solution of the form
÷øö
çèæ += -
4sin2 ptAex t
where x cm is the displacement of the pendulum at time t. If x = 4 cm when t = 0 seconds, find: (a) the value of the constant A (b) the velocity of the pendulum
when it first passes through its central position
(c) the first value of t for which the
pendulum has zero velocity.
(a)
242
14
4sin4
4sin4
0 when 4
0
=\
´=
÷øö
çèæ=
÷øö
çèæ=
==
-
A
A
A
Ae
tx
p
p
(b)
4
4
04
sin
4sin240
,0When 4
sin24
2
2
pp
pp
p
p
p
-=
=+
=÷ø
öçè
æ +\
÷ø
öçè
æ +=
=
÷ø
öçè
æ +=
-
-
nt
nt
t
te
x
tex
t
t
for integer values of n. When n = 1, t is first positive.
p43
=t
When the pendulum passes through its central position
cm/s. 24
cos244
3
,4
3en Wh
4sin2
4cos24
4cos24
4sin28
23
23
2
2
2
p
p
pp
p
pp
p
p
-
-
-
-
-
-=
=÷øö
çèæ
=
úû
ùêë
é÷øö
çèæ +-÷
øö
çèæ +=
÷øö
çèæ ++
÷øö
çèæ +-=
e
ex
t
tte
te
tex
t
t
t
!
!
(cont.)
Maths Quest Maths C Year 12 for Queensland 2e 4
3 (cont)
(c) When does ( ) ?0=tx!
04
sin24
cos
04
sin24
cos24 Solve 2
=÷øö
çèæ +-÷
øö
çèæ +\
=úû
ùêë
é÷øö
çèæ +-÷
øö
çèæ +-
pp
pp
tt
tte t
÷øö
çèæ+-=
÷øö
çèæ+=+\
=÷øö
çèæ +
-
-
21tan
4
21tan
4
21
4tan So
1
1
pp
pp
p
nt
nt
t
for integer values of n. Select n = 1
seconds. 820.2»\ t
Maths Quest Maths C Year 12 for Queensland 2e 5
5 A new worker on an assembly line can perform a particular task in such a way that if x units are completed in a day, then after t days on the assembly line
( )xktx
-= 100dd
for some positive constant k and 100£x for all t. The worker produces
20 units on the first day and 30 units per day after being on the job 10 days. (a) How many units per day will the
person produce after being on the job 15 days?
(b) How long will it take for the
person to be working at 90% of their full potential?
(a)
( )( )
kt
kt
ckt
ckt
exxxeee
excktxcktx
tkx
x
tkx
x
-
-
--
--
-=\
=
=
=-
--=-+=--
=-
=-
òò
0
0
100..
100100ln100ln
d001d
d001d
ktexx
xxt
--=\
=-=\==
801008010020
20,0When
0
0
( )
( )units. 521.34
8010015days 15When
801000134.0
87ln
10187ln10
87
801003080100
30 days, 10When
150134.0
0134.0
10
10
»-=
=\-=\
»
÷øö
çèæ-=
÷øö
çèæ=-\
=\
-=\-=
==
´-
-
-
-
-
ext
etx
k
k
e
eexxt
t
k
k
t
(b) At 90% of full potential, the person would produce 90% of 100 units per day.
days. 727.15581ln
0134.01
811080
8010090 Then
0134.0
0134.0
0134.0
»
÷øö
çèæ-=\
=
=
-=
-
-
-
t
e
ee
t
t
t
That is, 156 days will be needed.
Maths Quest Maths C Year 12 for Queensland 2e 6
6 In a wasp population, the rate of growth of the number of reproductive wasps at time t days after 100 days is given by ( ) ( ) ( ) 2500,1002.0 1.0 =-= rtetr t!
(a) Show that
( ) ( ) .301002.0 1.0 +-= tetr t (b) What is the number of
reproductives when t = 30 days? (c) Graph ( ) ( )0.12 100 30.tr t e t= - + (d) Use a graphics calculator to find
when the number of reproductive flies is:
(i) 200 000 (ii) At least 200 000
(iii) Greatest (iv) ( )200 000 r 400 000t< <
(a) ( )
( )( )
( )( )( )
( ) ( )
( ) ( ) 30110230220250
250,0When 110211010
10010010
d1010010
d10010dd
d100 Consider
1.0
1.0
1.0
1.01.0
1.01.0
1.0
1.0
+-=\
=+===
+-=\
-=
+-=
+-=
-=
-
òòò
tetrc
crt
ctetrte
ete
tete
ttet
tte
t
t
t
tt
tt
t
t
(b)
( )
( ) 32443030160
3080230,30When
3
3
»+=
+´=
=
re
ert
(c) (Hint: Graph using the graphics calculator)
(d) (Hint: use the graphics calculator with appropriate
calculator functions.)
(i) 81.7 or 108.0 (ii) 81.7 < t < 108.0
(iii) t = 100 (iv) 81.7 < t < 94.9 and 103.8 < t < 108.0
Maths Quest Maths C Year 12 for Queensland 2e 7
7 A population, N, of mice at time t exhibits logistic growth such that
( ) 1000,004.02dd 2 =-= NNNtN
(a) Solve for ( )tN by using the
logistic equation directly. (b) Show that your solution for ( )tN
satisfies the logistic differential equation.
(c) Using appropriate software, graph
( )tN . (d) Determine when ( )tN is maximal. (e) Calculate:
(i) the average rate of change in ( )tN across the interval
8.05.0 ££ t (ii) the instantaneous rate of
change when 1=t .
( )
( )
( ) t
t
at
etN
etN
Nba
eNbaN
Nba
tN
2
2
0
00
0
41500
40010000050
100004.02
(a)
-
-
-
+=\
+=\
===
÷øö
çèæ -+
=
( ) ( )( ) ( ) ( )( )
( )
( ) ( )( )( )
( )22
2
22
2
22
2
2
2
22
2
222
12
41
400041
100041100041
500004.0415002
produces 004.02 and41
4000 so
841500
41500 (b)
t
t
t
t
tt
t
t
tt
t
e
ee
eee
NNe
etN
eetN
etN
-
-
-
-
--
-
-
---
--
+=
+
-+=
+
´-
+
´-
+=¢
-´+-=¢
+=
And this agrees with tNdd above.
( )
( ) 1000
004.02dd
on tosoluti a is41
500
2
2
=
-=
+=\
-
N
NNtN
etN t
(c) .
(cont.)
Maths Quest Maths C Year 12 for Queensland 2e 8
6 (Cont.)
(d) N(t) is maximal
( )
250
5002121When
=
´=
÷øö
çèæ=batN
Hint: When is N = 250? Use the graphics calculator to find where N(t) = 250 meets the logistic curve.
693.0=\ t (e) (i)
( ) ( )
unit time.per population of head 7.2473.0
5080
»
- .N.N
(ii) Hint: use a graphics calculator feature
which determines the derivative at a point. 227.9 head of population per unit time.
Maths Quest Maths C Year 12 for Queensland 2e 9
8 In a community of 10 000 people, a person who has been exposed to a virus has contracted the virus. If we assume that all people in the community may be susceptible to this virus, and if the epidemic spreads at a rate in direct proportion to the number who have it, N, and to the number who have not contracted it, 10 000–N, then we may model the spread of the virus as follows:
( ) ( ) 10,00010dd
=-µ NNNtN
(a) Using the integral calculus, show
that at time t weeks the number of people who have contracted the virus is given by
( ) ktetN 0001099991
00010-+
=
where k is a constant of proportionality.
(b) If after 3 weeks 400 people have
contracted the virus, determine the value of the proportionality constant k correct to three decimal places.
