1 Homework week 4
1. Consider the function
f x y x2 y2 x
find:
a) the Domain. Solution:
D x y x
x b) the level curves.Solution: the level curves can be find by setting
c x 2 y2 x
so that
y x x2 c
it is clear that the domain of m y will allways be such c x x2 . Tables:
c 1
x
y 1 x x2 c 2
x
y 2 x x2 0 01
0 61
1
0
0 5
1 12
0 5
0 5
0
0
1
1 41
0 5
1 12
0 5
1 50
1
1
1
1 4
1 61
0 01
1 5
1 12
0
2
c 3
x
y 3 x x2 c 4
x
y 4 x x2 0 1
0 08
1 3
1 56
1
0 5
1
1 5
0 5
1
1 5
1 41
0
1 73
0
2
0 5
1 8
0 5
2 06
1
1 7
1
2
2 3
0 1
2 56
0 08
and so on...this is all we need to find the level curves. See the level curves c 1, c 2
and c 3 in fig1
There is an interesting alternative way to find the shape of this level curves, notice that
the level curves can be written as
x2 y2
1
R x
1 Homework week 4
where
R x
c x
we immediatly notice that for x 0 then R 0 c, also that for values x 0 then the
“radius-square” is R x 0 c finaly we notice that when x 0 then the radius-square
R x 0 c.
As a reference we draw the circles corresponding to the equation x 2 y2 R c, notice
the change on the radius-square R x with respect to the radius-square c.
y
2
0.5
1
1.5 2
c=3
c=2
c=1
1.5
1
0.5
x
Figure 1.1: fig1
c) using the level curves, plot the surface associated with this function. Solution:
z=x2 + y 2 + x
y
x
Figure 1.2: fig2
2
1 Homework week 4
2. Calculate the partial derivatives, f x and fy , of the following functions1 :
a) f x y 2x2 3y 4. Solution: f x
f) f x y 3
x2 1 y 2 . Solution f x 2x y 2 and f y xy 1 2 . Solution: f x 2y xy 1 and f y 2x
x2 y2 . Solution: fx x x2 y2 1 2 and fy 1
1 2 and f y 1 2
x y . Solution: f x x y
x y
1
1
and fy ln x y . Solution: f x g) f x y exy ln
b) f x y c) f x y d) f x y e) f x y 4x and fy
x y
y. Solution: f x
x2 1 xy 1 x x2 y2 1 2
x y
yexy ln y
xexy ln y and fy
1 xy
ye
h) f x y xy . Solution: a bit more tricky, notice that: x y exp ln xy exp y ln x then it follows that f x yx ey ln x yx xy yxy 1 and fy ey ln x ln x xy ln x.
i) f x y j) f x y tan
fy
sin
sin
3. Show that f xy
2
y x . Solution: f x 1
1
xy e x xy cos xy
y 1
y
x2
. Solution: f x
1 x
2 xy
y x sec 2 y x and fy 2
tan
sin
2
xy cos xy
e
x y 1
1
x
tan
1 y
2 xy
ey ln x
y x sec 2 y x 2
e x
y 1
and
fyx for the following functions.
a) f x y x y xy. Solution: f x
fxy 1 and fyx fxy .
1 y so fyx
1 on the other hand f y
1 x therefore
b) f x y x2 y cos y y sin x. Solution: f x 2xy y cos x so f yx 2x cos x on the other
hand fy x2 sin y sin x and f xy 2x cos x again the condition f xy fyx is satified.
c) f x y xey y 1. Solution: f x ey 1 so fyx
again the condition f xy fyx is satified.
d) f x y 1
x y
2
ln x y . Solution: f x therfore it’s satidfied that f xy
1
x y
and also f y
fyx .
ey ; now fy
1
x y
e) f x y ex x ln y y ln x. Solution, first derivatives: f x
Second derivatives: f yx 1y 1x fxy .
4. The Laplace equation in three dimensions, is defined as
∂2 f
∂x2
∂2 f
∂y2
∂2 f
∂z2
0
Show if the following functions satisfy the Laplace equation
1 Notation:
fx ∂f
∂x
fxy ∂2 f
∂x∂y
∂
∂x
fxx ∂2 f
∂x∂x
∂
∂x
3
∂f
∂y ∂f
∂x so fyx
xey 1 and fxy
x y
ex ln y y
x
and fy
2 and
ey so
x
y
fxy
ln x.
1 Homework week 4
a) f x y z x2 y2 z2 . Solution: f xx
doesn’t satify the Laplace equation.
2, fyy
2 and fzz
2 therfore 2 2 2 0 it
b) f x y z e 2y cos 2x. Solution: trivialy f zz 0 now fx 2e 2y sin 2x and f xx 4e 2y cos 2x
while fy 2e 2y cos 2x so f yy 4e 2y cos 2x. If we add the results: f xx fyy fzz 4e 2y cos 2x 4e 2y cos 2x 0 0 so it does satisfy the Laplace equation.
c) f x y z x
x2 y2
2
x2 y2
ln x2 y2 . Solution: f x y z 1
2
ln x2 y2 so fzz
2
so fxx x2 y2 22x 2 2 similarly for f yy
x y 2 x2 y2 2 x 2 y2 2 x 2 y2 2 2 2
2 2 2
0. It satifies
x y x y 1
1
x2 y2
2y2
2 2
x y 2
then
y2
2y n x2 y2 z2 2z n x2 y2 z2 fz
2x n x2 y2 z2 fx
fy
x2
fxx fyy fzz
the Laplace equation.
d) f x y z e3x 4y cos 5z. Solution: f xx 9 f and fyy 16 f while f zz fzz 9 16 25 f 0
5. for which values of n the function f x y z equation is satisfied?
0 trivialy and f x
z2 n
25 f so fxx fyy the tree dimensional Laplace
n 1
n 1
n 1
So
2n x2 y2 z2 fxx
fyy
fzz
2n x2 y2 z2 2n x2 y2 z2 The Laplace form would be f xx fxx fyy fyy 4x2 n n 1 x2 y2 z2 n 1
n 1
n 1
4y2 n n 1 x2 y2 z2 4z2 n n 1 x2 y2 z2 3
1
n 2
n 2
fzz
6n x2 y2 z2 fzz
n 2
n 1
4n n 1 x2 y2 z2 2 n 1 2n x2 y2 z2 2n 2n x2 y2 z2 So according to me only for n 0 and n and n 12 )
1
2
4
n 1
n 1
n 1
(Thomas and Finney say that the result is n 0