# EXAM 10, 2012

```ID. nui~ber~
Math 2250-1
FINAL EXAM
December 10, 2012
This exam is closed-book and closed-note. You may use a scientific calculator, but not one which is
capable of graphing or of solving differential or linear algebra equations. Laplace Transform tables are
included with this exam. In order to receive full or partial credit on any problem, you must show all
been written so that there are 150 points possible, and the point values for each problem are indicated in
the right-hand margin. Good Luck!
problem
score
1
2
possible
20
______
10
3
20
4
15
5
10
6
15
7
20
8
15
9
25
total
150
at rest at time t =jj4ec, is accelerated in a straight path
provides a consta~force, and which is also subject to drag forces
velocity. The boat
Ic).
value problem
engine that
2ONfor each
mas 400 k
ling abili
to show that the boat velocity v(t) (in
per second) satisfies the initial
v’(t)2—.05v
v(0)0
t\J&amp;141k 2~
~-
f~as
,&lt;.“
+uov
/
~C•CD -ZOv
‘2-
1k).
(5 points)
c
—
2
—.
oSv
Solve the initial value problem in i&amp;
V j- .o5y
&amp;oSt(vt&divide; .o3v)
e
.o5t
C
(10 points)
2
•j 2-o5t44
v
2
e
4-~
4cc
v-t
~ Ct.t0
Ic).
How far does the boat travel in the first 20 seconds?
12.0
~((t0’)
—
C
-Jo
0
9ot
i-
qv—4te
--oct
4
(5 points)
-~o5t-1zo
~ooe
-Jo
ttc\$D€
2
,&ccedil;
Here is a matrix and its reduced row echelon form:
3 9 -5
1 0
—6
1
2 -2
3 -2
0
tzt;f.r
13020
0 reduced row echelon form of A:
0
Find the general solution to the homogeneous matrix equation A&amp;=
combination form.
~..
001
1o
00000
(6points)
)f2t.
~tit
r
r
-4:
4:
k—~
4EN
I~
I ‘~3
+
r
1-3
‘I
~0
t1pa.
V
Let b be the first colunm ofA, i.e.
3
-2
1
I
What is the general solution to the non-homogeneous equation A x= b? Hint: since you afready know the
homogeneous solution from part g, you only need to find a particular solution and superposition in order to
deduce your answer. Since a matrix times a vector is just a linear combination of the columns, and since
you want A 410 be the first column of A, there is a natural choice for 4.
r3
, -s
2
~
I
-z
~es%tJL
o
(3,
L’J
-J
S.o
X
1’
Jo
(4 points)
P Ia
0
r€k~
3
3). Consider the following initial value problem, which could arise from Newtons second law in a forced
mass-spring oscillation problem:
x
~
~f ‘h
x~t1
x
x(0)=0
x’ (0) = 6.
—
—
3~).
If the applied force in this problem is 100 N, then what are the corr sponding values for the Hookes
constant, mass, and damping coefficients? Include correct units.
~ CX’4- k~t
~ t~”~
F~ Ic-p
4-ti-.
(5points)
~.xf4~g=5~
≤O
4
c~
~i2~ ~
32
K’.
3
ktso
3~ Solve this initial problem using the methods of Chapter 5, based on particular and homogeneous
solutions.
2
C)
(15 points)
‘1
rti—G~- i-25 t(r+s)
4-i~
nt.4t (rf3)t__IG -~) r
=)
-h~
,&ccedil;~(4’~t
f1€~c.S4&amp;
tt4t~ r~-it4c
4~cii3t≤is.4faiJ
=Ot-OI-25C.
;c7tc
tSO
~)Ct2.
Xj)t X)j
I
1e~~os4f
X(o)tOt2\ti~ c~=—2.
~,=
o~3c1t4c~
~)
&ccedil;~-o
4
41
Re-solve the initial value problem in
a using Laplace transfonns:
x’’(t) +6x’(t) +25x(t)=50
x(O)=O
x’ (0) = 6.
(15 points)
?~X(s’i
sKL5)-o)
—&oslash;s--~ -I-
1~
S
So+~s
S
Soi-GS
≤
+2SX(C~= ~P.
Esi-C
(s1.4~sns~
F’(st+c,stis’) i-(\$~pi-c)s
5o
zs~~
SD+Gst 2(s~4-~s42S)#(&amp;S~-)S
t StUi~B) 4+ 1. (cc)
~) Ot24~3
d-~—
2
—
4-
~
-
~?
B,—’~
..__—
2i3tc-os4E
5
Use Lapiace transforms to solve the initial value problem
x” (t) + w~x(t)
=
~cos(w0 t)
x(O) =x0
x’ (0) =
2)
0
‘F
c.~X(s)
S
ff_
t
n
(7points)
ftc
4-
‘i”
N(s~t
f?~
S
S
St0
-
‘4
-f4-
V0
‘2.
u.
What physical phenomenon is exhibited by solutions to
(3 points)
kas a- 4eAIl~.
t.aft.
“4-i’wj.
I,
“~“fl ~“&amp;~k -i-LI-
4~
e-~’~s
/ U
DCLI.trs
\
Cr.
•swLLo.%eb.
4t~
cy.ki..~ik
~f ,4-c
42o~ct~c
6
~i
Find the general solution to the first order system of differential equations
x1’(t)
_3
I
xI
x21(t)
2 —2
x2
(10 points)
Hint: The eigenvalues of the matrix are negative integers.
rCA)
(~+~)Cafz~-a
‘I
‘),t-cf,~
-2~
2-
~io
I
—I 1.0
t&gt;~+S~~+c~
to
F ~‘1(t~
L’iii
ate
[!j
b1rLt
tnj
(A4~1)~3’
—t
-it
Ltj
L1e
L~]
~ Sketch the phase portrait for this linear hoi ogeneous first order system, based on your work and
eigendata in part a. Classify the equilbrium point at the origin. S~hu
(0
X~ ,‘X~
3
s44(t
ki.4.j
2
3
2
1~~tt~
—11fl
i~c~ I
Lti
1-’
~ce I
61~
)O 4asJac.s
7
I-, oo,
So
5d4~4 appw~cL v~jii
l3~~ ita fmsL’a”s t-) —02
≤.
.a~
fl
Consider a general input-output model with two compartments as indicated below. The compartments
contain volumes V~, V~ and solute amounts x1 (t), x2 (t) respectively. The flow rates (volume per time) are
indicated by p., 1
1 .6. The two input concentrations (solute amount per volume) are c1, c5.
.
~
140
X~t~)
~ Supposer2
=
p3
=
=
100, r~
=
r4
200, r5
=
‘~a~4I
0
.
Explain whythevolumes V~(t), V2(t)
remain constant.
\(1’(4’j~zou +Ifl-jfl--~j~- a
V1t~c.s~~4—
V~canf.
≤o
kD
2~~Using the flow rates above, c1
=
0.6, c5
=
_%~y,
0
V~
=
=
(4points)
100 gal, show that the amounts of
solutex1(t) intankl andx2(r) intank2satis~’
x1’(t)
x2’(t)
-3
=
1
2 —2
X1
x2 +
120
0
(6 points)
/
—
r0 L~,
‘~n•(.~.)&divide;I&oslash;~cs —3it~j
V7.
V1
~‘LOj- ~z
r~Lc—r0co
Zn ‘5&plusmn; .-2.OO ‘&frac12;
..
V1
So,
~-
VT
iz.c
V
8
2k). Solve the initial value problem for the system
~10~t~&deg;
[x~’(t)
=
-3
1
X1
+
2 -2
assuming there is initi 1 no solute in either
vector, and then use x =
+
o so
e
hj
120
0
Hint: Find a particular solution which is a constant
P. Notice that you have already found
iii problem 6.
(10 points)
—J
Xp-
&ccedil;c~
3
~0
L&ccedil;
101.: [-s
La] 1z
o
o
CI ~Cz
E~- E,-~7 2ztl.o
fliti
4
-tj[c
[no
Lo
2c~ i-Ct
0
a)c1
.a~
~&ccedil;t
S e,
-~
_,
t VT,
#~xIi
_,
—91
t&ccedil;so
4-st
LGo
—
1_I-I
boi-c.14z
~ 4Zc1~Lt
[Go
L&amp;o
i
Qt
~‘-
aS~tojb~c~~
&ccedil;x1w~
qo .4
L’o
c.c.
C
I
~-qt
2
-t
-4ff
~O ~Ljoe -2-at
~c,.L+) ~6o _to~;.t t20t~~
~j4’~t.
Qt ~
9
Find the general solution to this second order system of differential equations, which could arise when
modeling a coupled mass-spring system. Hint: You have already computed eigendata for the relevant
matrix in problem
x1’’(t) =—3x1 +x2
~..
‘‘(1)
=
2
At 1-~
—
2
(c~ctf&ccedil;s.k~)
-t
‘~
C1t..s(+-.ej
1- .1
~.kl
•
~———~
II
(6 points)
~
L’J
+ (L3L~ctb+%5(~Lt)f
L
‘
Describe the two fundamental modes of vibration for this spring system.
5jew- 1%~444
k~as&ccedil;ec
(C3t LqtD~
~ac~
• fotsf
Cctc~toN ~
L’iwA.L
~-
si~n~p(s.
k~asy,
r~”~
~
Lan~esa~ ~ i~,
(4 points)
fwcc.cq4f ~
s~\$ ‘j, pkas.c, tj~4 n~pLi~.4~, 6ot2
~ For Hooke’s constants and masses as shown below, show that the displacement functions
x1 (t), x2 (t) of the two masses below (from their equilibrium hanging positions) satisfy the system of
differential equations in this problem.
I~_,
I
—
2x~”:
I
..i
rr
t
—
Mx)#2Cs~x%)_
2.
(5 points)
‘~1~2r1
t~v: 2
—ts
(.Kz~Xi’I
N’—
I,
~
10
Ii
We have studied and returned to the rigid rod pendulum several times in this course. This is the freelyrotating configuration indicated in the diagram below:
6’
L Lc~ 6
Using conservation of energy we have derived the autonomous second order differential equation that
describes the angle e ( t) of the mass from the vertical reference line, at time 4 arriving at
0’’(r) + fsin(oQ)) =0.
(i~
That second order DE above is equivalent to the autonomous first order system of two differential
equations
x’(t) 1
U)
y’ (t)
=-
fsirqx)
Explain this equivalence.
~W)
&pound;i€!rf(()~
~u4~’~.a
tkt~1..
(4points)
-=
e’ (4)
i.4 {sIjn~c 4L~4
(a)
C
2tOk1V~\$tI~)
4 x(t),
&pound;oive (a’),
TL,
e’~~1
€≠‘~
So
s~
j~i~i.*.
9”•+
sCk&amp;to
(z~
Find all equilibrium (i.e. constant) solutions to the first order system of bE’s above. Explain how
these equilibrium solutions are related to the constant solutions of the second order differential equation for
rigid-rod pendulum.
2U
(x,,y~’l
tthAt~•)
~-
(6 points)
~
3~’~z _tS(k1~x
-~
~‘≤CktxtO
cc
is
C~ ~
ki-
•
fl~’~,Lt
~rL
sA~-
44
k4Sh
4+ E’o+4-m—
LA)”4b~-
h.~s
-
044
is.
h.mcj
.r~
~sbh4\$5
(0, hit) I kt&amp;~
“S
% lit
-fn--
~4K’th7t
(.R,p)
~fr~4fl~
(Sal t3{ a4
t
“C
(2ti,o)
11
e e:~rr, h
x’(t)
y
y’(t) =-fsin(x)
~ Use linearization and the Jacobian matrix to classit~’ the equilbriurn solutions to the first order system
above. Indicate what you can deduce about the stability of these equilibrium solutions based only on the
linearization.
E(g,
.
[o
-.~_
I
Ht”~”
(lOpoints)
a].
tthei~t~..S)r(h%7t,o)t
r~
Lt
~
oJ
2
-t
.
“4
noab, ~+~ta~I=t4
hoet4
KtktTt,
~mn
tID
~tsj ~v~ai~
s1-o4~9..
I
Lto
lEXT~t~~
)
~
Ii €~
~(et
t44&amp;
(n.4 Ci~Aa
‘lr~~;”t~
(kn,&amp;~
~ 4~
)~&plusmn;y~
_____
9~). Use the sep~ah&ouml;n of v iables method we discussed in class (or if you prefer, use conservation of
energy), to show that the parametric solution curves (x(t), y(1)) to the first order system above lie on the
level curves for a certain function. Explain why this shows that the equilibrium solutions that are
borderline based on the linearization work in ~, are actually stable for the original non-linear system.
(5 points)
1k s~fk
~t
Lx
‘:2s 0~12~~
.1.
Lcd
~a
So
7
~
L’o-uts
cos~c .1-
C-
_tcesxtc,(n4
vsot4n.. Cu.n’t5 a..a
(ik~
f&amp;ci&ccedil;
~?,
Lt&amp;nts~
a≠,~&ccedil;~
ELx,&amp;
-~
—
i~c a “~S+*n+ ‘kh.fl 44t
I&lt;C t~PE f,t
‘144
t
Tkas EL~e.a) ka&ccedil;5LoloJwAi~t.
—
‘n4j.,*
~r
~
~
ct~r yi5
~
(km,
h
12
€bfini ~e
&pound;fre’IftS~W ,41E.
~
So k%%L%S+
a.-
-
Table of Laplace Transforms
This table summarizes the general properties of Laplace transforms and the Laplace transforms of particular functions
derived in Chapter 10.
fQ)
F(s)
af(t) &plusmn; bgQ)
aF(s) + hG(s)
f’Q)
sF(s)—f(O)
f”U)
s2F(s)
—
sf(O)
?F(s)
—
s”~f(0)
f
(s
—
[(0)
— .. . —
sink:
f~”(O)
coshkt
s
sinhk:
F(s
—
a)
(s —a)2&plusmn;k2
(s
a)JQ
—
a)
r&deg;’F(s)
e&deg;’ sink?
~~r~ga
—
r) di
F(s)G(s)
~\$sinkt
—
)n+l
coskt
f(r)dr
e~’f(:)
uQ
:fQ)
—F’(s)
:~fQ)
(—iyF’~~(s)
f
fit), periodp
—
ki coskt)
sink:
u(t—a)
“
a
a)2 &plusmn;k2
(2
k2)2
(82
_1_ k2)2
~(sinkr&plusmn;krcoskt)
F(a)diy
1
—
l—e~
—
—
-
8Q —a)
e~’fQ)d1
-
1:/i
(_j)L
,a,
C”
(square wave)
5
lFt1l
— II
(staircase)
lLaJj
I tanh as
—
s
2
—
C”
s(l
—
(‘I
I
I
r(a &plusmn; I)
50+1
13
```