severdevon_900012_53808844_HW 10.
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Arizona State University *
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Course
93344
Subject
Mathematics
Date
Nov 24, 2024
Type
Pages
12
Uploaded by BailiffPelican2900
10.1
Application:
Finite
Difference
Rules
306
3
pts
Go
back
to
problem
9.4
(from
your
last
HW)
which
had
data
for
the
f
height
y;
(m)
of
a
pumpkin
as
a
function
of
time
t;
(sec).
S
3
T
a)
Use
the
3-point
centered
difference
rule
to
calculate
the
acceleration
(d?y/dt?)
of
the
pumpkin
at
t
=
1.5
seconds.
(Compare
0
'
-
o)
this
value
to
what
you
already
know
the
acceleration
due
to
gravity
should
be
in
m/s2.)
b)
Use
the
4-point
backward
difference
rule
to
calculate
the
acceleration
(d?y/dt?)
of
the
pumpkin
at
t
=
3.0
seconds.
Hint:
you
can
find
this
rule
in
Table
8-1
(pg
318)
of
the
textbook,
or
the
back
of
your
“Class
30/31”
handout.
c)
IF
you
obtained
and
used
measurements
every
0.1
seconds
(instead
of
0.3
seconds,
like
in
the
table
of
data),
quantify
how
much
you’d
expect
the
error
in
your
acceleration
calculation
from
part
(a)
to
change.
Time,
t;
o)
S
Coreed
Differevce
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10.2
Fundamentals:
Taylor
Series
2
pts
This
is
the
equation
for
the
Taylor
Series
expansion
of
f(x+0),
as
a
function
of
f(x)
and
all
its
derivatives
at
x,
that
will
be
given
to
you
on
your
midterm
and
exam
cheat-sheets:
!
62
r
63
"’
64
nr
f(x+0)=f(x)+df
(x)+7f
(x)+§f
(x)+Zf
(x)+--
Start
with
that
equation
and
use
it
to
derive
the
expression
for
f(x;
—
5Ax)
(i.e.
f(x)
evaluated
five
nodes
to
the
left
of
x;)
as
a
function
of
f{x;)
and
all
the
derivatives
at
x;
(up
to
the
FIFTH
derivative).
{=
Sax
o°
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fx+8)=f(x)+6f'(x)+
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10.3
Application:
Error
Order
and
Precision
7pts
The
following
is
a
5-point
difference
scheme,
over
equally-spaced
x;,
for
d3f/dx>
at
x
=
x;:
1
f(x)
=
E(f”
=6,
+12f,
=10£,+3f,,)
+
Error
Write
out
Taylor
Series
expressions
for
each
of
the
four
fi.3,
fio,
fi1,
fi1
to
the
FIFTH
derivative,
like
you
did
in
8.4,
and
then
combine
them
using
the
given
difference
scheme
above
to
...
a)
Calculate
the
discretization
error
order
(i.e.
write
the
error
=
O(Ax?)
for
some
integer
p).
b)
Calculate
the
precision
of
the
scheme.
)
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10.4
DERIVING
difference
formula
for
ANY
order
derivative
to
ANY
error
order
7
pts
Use
Taylor
series
expansions
to
derive
a
forward
difference
scheme,
over
equally-spaced
points,
using
any
or
all
of
f;,
fi.1,
fiu2,
fir3
and
fis
that
approximates
d?f/dx?
(2"
derivative
at
x;)
to
order
(Ax3)
discretization
error.
Hint:
DON’T
write
a
Taylor
Series
for
f;.
f;is
justitself
(or
just
written
as
f(x;)).
There’s
nothing
else
you
can
do
with
it!
Only
ever
create
Taylor
Series
expansions
for
points
other
than
f;.
Show
all
your
work!
Write
all
appropriate
Taylor
Series
out,
and
describe
your
goals
—
what
terms
do
you
need
to
keep,
what
terms
do
you
need
to
eliminate?
If
you
follow
the
methodology
from
class,
you
should
end
up
with
4
equations
for
4
unknowns
(i.e.
coefficients
a,
b,
¢,
d
that
you're
using
to
weight
each
Taylor
Series).
It’s
fine
if
you
then
use
MATLAB
to
solve
for
them
—
you
should
get
nice
“round”
fractions
in
your
final
scheme.
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10.5
Fundamentals:
Characterizing
ODEs
(Chapter
10)
6
pts
For
each
of
the
following
three
ODEs
...
i.
characterize
its
order
(e.g.
1%*-order,
2"%-order,
etc.),
ii.
characterize
it
as
an
IVP
(Initial
Value
Problem)
or
BVP
(Boundary
Value
Problem),
iii.
write
the
ODE
in
standard
form.
(A)
qer=r%+\/z’—7q,
q(2)=0,
q'(0)=1
(B)
y(j—;)=sin(x)+dfiz3),
x(—4)=1
r
+dt2
dr’
dt
copda,
|40
0V
o)
g(0)=1
Ay
e
=rte
[0
)0
g
(o)
I\\
SQLW\.A
order|
J
7\\I‘BV\,
L}
{_j;_‘
term
H\
J
Ottereit
tives
—s
qy=o
,
9@
-
i)
ot
g
-
#
(e
i)
bv*
dr
lewike
as:
-
s
>
Take
Aff‘\“"H\"
(fé}
}’(—)=sin(x)+
Sj),
x(—4)=1
£ivet
<
A(K?)
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3)
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T,ivia
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value
mn
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Shindod
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=
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Cencide
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*
HW10
(10.1
-
10.6)
due
Monday
April
17
10.6
Application:
(Single-Step,
Explicit)
Euler
Method
to
Solve
an
IVP
.
-
dT
'
43
1t
.
5
pts
You
want
to
solve
the
following
1*-order
Initial
Value
Problem:
—
=—-—4
————
withthe
initial
condition
T(t
=
0)
=
T,
=
30.
dt
18
45
300
-
Why??
Because
your
manufacturing
floor
is
at
an
excruciating
T,
=
30°C
(about
86
F),
when
you
.
turn
on
the
air-conditioning.
What
you
want
to
determine
is
how
long
it
will
take
to
get
the
room
"
temperature
down
to
18°C.
The
room
has
a
volume
V
=
(10m)
x
(20m)
x
(6m)
=
1800
m?,
so
-
knowing
the
density
of
air
=
1.2
kg/m?
there’s
m
=
2160
kg
of
air
you
need
to
cool.
-
Your
air-conditioner
can
only
Q
=
14400
—
120t
provide
120
kg/min
of
cold
air
watts
of
solar
heat
\\
.
(at
Tae
=
10°C,
about
50F).
That
m
=
2160
kg
of
air,
initially
at
T,
=
30°C
will
mix
around
with
the
air
in
::}
Room
temp
.
the
room,
so
an
equal
amount
.
)
_
'
L
m
=
120
kg/min
of
=T(t)
.
of
air
will
leave
the
room
at
)
:
cold
air
at
T,
=
10°C
-
:
whatever
.the
ter:nperature
of
m
=
120
kg/min
of
-
the
roomis.
Let’s
call
that
T{t).
room
air
leaving
at
T{(t)
-
That
all
wouldn’t
be
too bad,
except
the
sun
is
still
coming
in
the
windows,
putting
in
a
lot
of
heat!
-
It’s
near
the
end
of
the
day,
so
the
amount
of
solar
heat
is
dropping
off
with
time,
according
to
the
.
formula
Q
=
14400
-
120t,
where
t
is
in
minutes,
and
Q
in
Watts
(so
it
will
be
at
0
after
2
hours).
.
So
what
does
all
this
mean?
You'll
learn
in
a
Thermodynamics
class
that
the
temperature
of
the
'
room
T{(t)
will
go
up
and
down
as
the
thermal
energy
in
the
room
goes
up
and
down.
The
solar
-
heat
raises
the
thermal
energy.
The
difference
between
the
cold
air
in
and
room-temp
air
out
.
lowers
the
thermal
energy.
The
“energy
balance”
is
given
by
the
1*
order
ODE
dT
m
(
dT
.
.
:
.
.
me—=mcT,
.
—mcT
+Q
,
which
can
be
rewrittenas
—
=
—
TC.—T)+Q
dt
dt
m
4
mc
-
The
“c”
in
the
equation
represents
the
“thermal
capacity”
of
the
air
(how
much
heat
it
takes
to
.
raise
the
temperature
of
1
kg
of
air
by
1°C).
That’s
known
to
be
1000
J/kg°C.
When
|
plug
all
the
*
values
for
the
problem
in
the
equation,
and
convert
units
so
I’'m
working
with
time
(t)
in
minutes,
|
-
get
the
equation
|
started
with
at
the
top
of
the
page!
So
let’s
get
back
to
that
(since,
as
you
-
probably
figured
out
by
now,
you
didn’t
really
have
to
understand
anything
in
this
“Why??”
box
to
.
solve
the
problem).
................................................................................................
dT
T
43
¢
(a)
Use
the
(single-step,
explicit)
Euler
method
to
solve
the
1*-order
IVP
——
=
-
—
4
——
where
T
is
the
temperature
of
the
room
(in
°C),
and
t
is
time
(in
dt
18
45
300
minutes).
Solve
the
problem
over
the
range
t
=
[0,
40]
minutes,
7(0)=30
using
step
size
At
=
10
min
(i.e.
calculate
T,
at
t,=
0,
t;
=
10,
t,
=
20,
t;
=
30,
and
t;
=
40
minutes).
Show
all
your
work
in
table
form,
like
we
did
in
class,
showing
how
you
start
from
each
(t;,
T;)
to
get
the
next
t;,;,
“slope/”,
and
T,;.
(b)
Make
a
sketch
of
T;(t,)
(i.e.
draw
and
connect
the
5
dots),
and
comment
about
what
time
you
think
(if
ever!)
the
temperature
of
the
room
will
get
down
to
18°C.
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