1 Homework week 2
Find the function’s a) Domain, b) Range and c) Level Curves of the next functions
1. f x y xy. Solution:
a) first and third cuadrant of the xy plane
b) all positive reals
c) lines give by y 2. f x y a) set
1
x y.
c2
x
Solution:
x y x y 0 that is all points but the line y x
b) The range is the union of the open intervals
c) c 3. f x y 1
x y
so y x ex ey
ex ey .
1
c
∞ 0 and 0 ∞ define the level lines
Solution:
a) Entire plane
b) range is the open interval of the real numbers
1 1
c) c curves can be ploted in the xy plane with the formula y ln ex
so the level
substitute eC 11 cc so y ln ex C and y x ln 11 cc ex ey
ex ey
1
11 cc 2 Tutorial Work
In this tutorial we want the student to be able to draw a surface in the three dimensional space. We
use the cartesian coordinate system and we use the level curves method as a first aproximation.
We start with the function
1
z
1 x 2 y2
First information to notice is the domain of the function
D
x y x2 y2 1
That means that the points on the circle of radius
x2 y2
1
are not a possible set of values for the function.
The second information we learn how to extract was the range of the function, in this case notice
that any value between 0 1 will not be valid. Consider the constant values of z a then
a
1
1
x 2 y2
a 1
a
x2 y2 but then we can see that for a 0 1 the value of a a 1 is negative and that cannot be since
x2 y2 0. Also notice that for a 0 then a a 1 is not well defined. Putting all this results
together, the range is
R z z 0 1 We start then with the level curves. Level curves are defined by taking a CONSTANT value of
the function.
ci
f x y
1
1 x 2 y2
where i is labeling the different constant values. This constant term can be seen as the intersection of the surface z f x y with the flat plane at c conts (see next fig1) The level curve is
then obtain since we can write
c
f x y in our case we find
y2 x2
which are circles of radius c 1
c
y g x
c 1
c
so let us call C y C2 x2
2
c 1
c
2 Tutorial Work
Once we are at this stage we can draw the circles of radius C, for example for C See fig2.( we didn’t use C=1 because that is not in the domain for this function)
5 3 1 1 1 1
2 2 2 3 4 5.
The next step is to find out wich oune of this level curves has the highest value. Using our
C radius
z c hight
5 2
4 21 0 19
1 3 0 33
2
3 2
4 5 0 8
c 1
1
1
48
relation C c or c C21 1 :
09
52
1 2
4 3 1 33
1 3
9 8 1 25
1 4
16 15 1 07
1 5
25 24 1 04
Which give us a profile of the curve: see fig3 revolving this lines we finally obtain fig4
z
z
z=f(x,y)
f(x,y)=1−x2 −y2
c=1−x2 −y2
y= (1−c)−x2
x
y
x
y
Figure 2.1: fig1
1
0.8
0.6
0.4
0.2
0
0.2
0.4
x
0.6
Figure 2.2: fig2
3
0.8
1
2 Tutorial Work
Now compare this results with the ones using Maple V: fig5 and fig6, it is not that easily find
what the shape of the plot really is. The main problem in this case is that the singularities blown
up in the z direction. As I understand in more resent versions, of Mathematica and Maple, this
can be controled.
z
5
1
x
−3
−2
−1
−1
1
−4
Figure 2.3: fig3
Table 2.1: fig4
4
2
3
2 Tutorial Work
1
0.8
y 0.6
0.4
0.2
–1 –0.8 –0.6 –0.4 –0.2 0
–0.2
–0.4
–0.6
–0.8
–1
Figure 2.4: fig5
5
0.2 0.4 0.6 0.8
x
1
2 Tutorial Work
10000
8000
6000
4000
2000
0
–2
–2
–1
–1
0x
y0
1
1
2
2
Figure 2.5: fig6
6