Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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please answer parts iv), v) and vi) of this question, i have attached the first the given answers to the first three parts of the question for assistance
![Step 1: Introduction
"Since you have posted a question with multiple sub parts, we will provide the solution only to the first three
sub parts as per our Q&A guidelines. Please repost the remaining sub parts separately."
The given function of several
variables is
f(x,y) = sin(xy.
we
and
have
range
to
-x-3y
+3)
determine the domain
determ
of the function f.
Step 2: Determine the domain and range of f
For the domain, we
where the
Since the
need to consider..
function f is defined.
sine function is defined for
all real numbers,
so, the domain of f(x,y) is all pains
real numbers, i... IR2.
of
As the
function is [-1]).
So the
[-]
range
of the
sine
range
of f(x,y) will
be
Step 4: Find the relationship for the x cross-section
To find the cross-sections by
setting x =k, where k is a real
number.
f(x)=sin(x-vy-D}.
let's tabulate the relationships for
<= 0, 1,..., 6.
0
K
Relationship
sin(+3(y-1))
-ser (-2 (y-1))
Sirr(y-1)
0
sin (y-1)
sin (2(y-1))
sin(3(y-1).
- sense, VER
2
3
4
5
6
[Note:
sir (-x) =
].
The
is
-
ху
x(y-1)-3(y-1)
-x-
argument
-3y+3
of the sine function
for K=3,
f(x,y) = 0,
cross-section is a
720, regardless
flat line at
of
y.
Step 3: Factor the argument of the sin function
180
the
For K= 4, 5, 6, the function
=
(x-3)(y-1).
sin((x-3)(y-1)).
will.
oscillates as
varies. The
function oscillates
y
-ent of the
So, the factored form of the
sine function is
argum
with different frequencies and
amplitudes, but it's
still centered
(x-3)(y-1).
around
y= 1.
The factor form indicates the function
might have
some kind of symmetry or
reqularity with respect to the
factors
x-3 and
y-1.
For
It's
K = 0, 1, 2, the function oscillates
with same frequency
amplitude
for
still centered
and
opposite
K = 4, 5, 6,
around
y = 1
The cross-sections for k=4,5,6
The cross-sections for k=0,1,2
‣ Solution
Therefore, the domain of the function f is R2
and the range of the function f is -1,1]](https://content.bartleby.com/qna-images/question/c1750264-b177-480f-8500-2593618fe8d1/44f634b7-5746-4b81-aa76-eb0edd051630/e2hn3g5_processed.png)
Transcribed Image Text:Step 1: Introduction
"Since you have posted a question with multiple sub parts, we will provide the solution only to the first three
sub parts as per our Q&A guidelines. Please repost the remaining sub parts separately."
The given function of several
variables is
f(x,y) = sin(xy.
we
and
have
range
to
-x-3y
+3)
determine the domain
determ
of the function f.
Step 2: Determine the domain and range of f
For the domain, we
where the
Since the
need to consider..
function f is defined.
sine function is defined for
all real numbers,
so, the domain of f(x,y) is all pains
real numbers, i... IR2.
of
As the
function is [-1]).
So the
[-]
range
of the
sine
range
of f(x,y) will
be
Step 4: Find the relationship for the x cross-section
To find the cross-sections by
setting x =k, where k is a real
number.
f(x)=sin(x-vy-D}.
let's tabulate the relationships for
<= 0, 1,..., 6.
0
K
Relationship
sin(+3(y-1))
-ser (-2 (y-1))
Sirr(y-1)
0
sin (y-1)
sin (2(y-1))
sin(3(y-1).
- sense, VER
2
3
4
5
6
[Note:
sir (-x) =
].
The
is
-
ху
x(y-1)-3(y-1)
-x-
argument
-3y+3
of the sine function
for K=3,
f(x,y) = 0,
cross-section is a
720, regardless
flat line at
of
y.
Step 3: Factor the argument of the sin function
180
the
For K= 4, 5, 6, the function
=
(x-3)(y-1).
sin((x-3)(y-1)).
will.
oscillates as
varies. The
function oscillates
y
-ent of the
So, the factored form of the
sine function is
argum
with different frequencies and
amplitudes, but it's
still centered
(x-3)(y-1).
around
y= 1.
The factor form indicates the function
might have
some kind of symmetry or
reqularity with respect to the
factors
x-3 and
y-1.
For
It's
K = 0, 1, 2, the function oscillates
with same frequency
amplitude
for
still centered
and
opposite
K = 4, 5, 6,
around
y = 1
The cross-sections for k=4,5,6
The cross-sections for k=0,1,2
‣ Solution
Therefore, the domain of the function f is R2
and the range of the function f is -1,1]
![Let f(x, y) = sin(xy-x-3y+3).
(i) Determine the domain and range of f.
(ii) Factor the argument of the sin function. Does this factored form give any hints about the appear-
ance of the surface z = = f(x, y)?
(iii) Find a relationship for the x cross-sections by letting x = k, k = R. Then,
⚫ tabulate the relationships for k = 0, 1, ..., 6;
-
• sketch the relationships for k = 3, 4, 5, 6 on the interval y € [1 − π, 1 + π] ; and
⚫ with respect to the cross-sections for k = 4, 5, 6, describe the cross-sections for k = 0, 1, 2.
(iv) Find a relationship for the level curves / contours by letting z = m, m Є R. You must recognise
the periodic nature of f and ensure the relationship accounts for it.
(v) Sketch the contours m = 0 on the interval x € [−1,7], y ≥ [−3, 5] .
Ꮖ
(vi) Which of the surfaces below is a graph of this function?](https://content.bartleby.com/qna-images/question/c1750264-b177-480f-8500-2593618fe8d1/44f634b7-5746-4b81-aa76-eb0edd051630/mfu32k_processed.png)
Transcribed Image Text:Let f(x, y) = sin(xy-x-3y+3).
(i) Determine the domain and range of f.
(ii) Factor the argument of the sin function. Does this factored form give any hints about the appear-
ance of the surface z = = f(x, y)?
(iii) Find a relationship for the x cross-sections by letting x = k, k = R. Then,
⚫ tabulate the relationships for k = 0, 1, ..., 6;
-
• sketch the relationships for k = 3, 4, 5, 6 on the interval y € [1 − π, 1 + π] ; and
⚫ with respect to the cross-sections for k = 4, 5, 6, describe the cross-sections for k = 0, 1, 2.
(iv) Find a relationship for the level curves / contours by letting z = m, m Є R. You must recognise
the periodic nature of f and ensure the relationship accounts for it.
(v) Sketch the contours m = 0 on the interval x € [−1,7], y ≥ [−3, 5] .
Ꮖ
(vi) Which of the surfaces below is a graph of this function?
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