2
1.5
1
0.5
1. Find the Domain, level curves and graph of the function f x y this function is
D x y x y the level curves are given by c
2 x
2 y
2 x 2 y 2 . Solution: the domain of which are concentric circles around the origin. The bigger value of c, the bigger the circles, all this confirming that f is the equation of kind of cone.
For c 1 we will have circles of radius 1, for c 2 circles of radius 2 and so on, therefore the plots are shown in fig1 and fig2.
y
0.5
1 1.5
2 c=1.5
c=1 c=2 c=0.5
x z=2 z=1.5
z=1 z=0.5
y x
2
Figure 0.1: fig 1 and fig2
2. The radius of the Earth is about R 6 300 km, How accurate would it be to say that the Dublin area A given by A
∆ x
∆
y with sides
∆ x 10km and
∆ y 10km is a flat surface.
Hints: a) approximate the Earth by the surface of a sphere (we only need half sphere) call that surface f x y b) Linearise that function at the point, lets say 0 0 .
c) Consider Dublin to be at the north pole of this sphere.
d) Calculate the second derivatives to find out the error function E x y
1
2
∆ x then the porcentual error. from there find out how accurate the approximation is.
∆ y
2
M and
1

Solution: the equation of a sphere of radius R is
R
2 x
2 y
2 z
2 the surface of the Earth can be written as z f x y in the form f x y R 2 x 2 y 2 first of all we calculate all the derivatives of this function up to second order: x f x
R 2 x 2 y y 2
1
2 f y f xx f yy
R 2
R
R
2
2 x 2
1 x x
2
1
2 xy y 2 y 2 y 2
1
2
1
2
1
2
R 2
R 2 x x
2
2 y
2 x 2 y 2 y 2
3
2
3
2 f xy
R 2 x 2 y 2
3
2
Now we assume that Dublin is at the north pole of this sphere. We want to simulate Dublin area with a plane. The plane that passes through that point 0 0 R is the horizontal plane, and its equation is simply z R
The standard linear approximation says then that f x y z R in a region A around the point 0 0 R . The whole point of the exercise is to check to what size of A this assumption is a good approximation so we calculate the error
E
1
2 where M is the highest number among f xx
∆ x
∆ y
10
10 f yy f f xx yy
0
0
0
0
1
R
1
R f yy
0 0 0
∆ x
∆ y
2
M f xy
1
R
1
R
. The values we need for the Dublin area are
6
6
1
300
1
300
0
0
00016
00016
Therefore the error is
E
1
∆ x
∆ y
2
1
10 10
2
0 032
2
2
0 00016
0 00016
(0.1)
2

As a percentage of the radius of the earth f 0 0 R, the error is no greater than
E
f 0 0
100
0
6
032
300
100 0 000508%
3. Now find the same information for the area span by St Stephen’s Green park
∆ x
∆ y 200m and
Trinity College Dublin
∆ x 800m
∆ y 200m.
The only information I need to change is
∆ x
∆ y 0 2km for St Stephen’s Green park and
∆ x
0 8km
∆ y 0 2km for Trinity College and substitute it into equation (0.1):
E
St Stephen s Green
1
0 2 0 2
2
0 000013
2
0 00016
E
Trinity College
1
0 8 0 2
2
0 000080
2
0 00016 therefore the porcentual error compared with the radius of the Earth is
E
St Stephen s Green
R
100
0 000013
6 300
100 0 21 10
6
E
Trinity College
R
100
0 000080
6 300
100 0 13 10
5
4. Proceed as in problem 2. but now consider the radius that Christopher Columbus used in his calculations
R 3250 miles. Consider the Dublin area and Trinity College Dublin as your test areas.Solution:
That radius in kilometres is (1mile 1 6093km)
R 5230km so the error for Dublin would it be
1
R
E
1
5230
0 00019 and using the previous formulas
1
10 10
2
0 038
2
0 00019
The percentage error with respect to the radius of the earth is
E
f 0 0
100
0 038
5230
100 0 00073%
Experiment: Looking through a window with a view to the sea, we can compare the curvature of the horizon against a straight line. If the view spans about 10km of the horizon, the previous results tell us that we will need some precision instruments to detect the real error and therefore the radius of the earth.
5. Find the total derivative of the function
To solve the following problems we use the chain rule so d f dt
∂ f
∂ x dx dt
∂ f
∂ y dy dt
3

a) f x y x
2 b) f x y e xy y
3
x such that x t
2 and y t 1 at t 2. Solution: d f dt d f dt t 2
4t
3 t
4
2x y
3
2t
2 dx dt t 1
3
2
4
3t
2t t
3
3t
3 2
3
1
2
3 t
3y
2
2t
3t
3 2
2 x
2 dy dt
3 t 1 t 1
2
2 54
2 t
2
1 such that x t and y 1 t at t 3. Solution:
1 d f dt ye xy
1 xe xy
0
1 t t
1 e
1 t
1
2 c) f x y x ln y x such that x t and y 1 t 1 at t 4.Solution
first of all we can rewrite the function as f x y x 1 ln y d f dt
1 ln y 1
1 t 1
2 y
1
1 ln t ln t
1
1
1 t t 1 t 1
1
1
2 d f dt t 4
0 8094 d) f x y xy such that x cos t and y sin t at t 0.Solution: d f dt d f dt t 0 y sin t x cos t sin
2 t cos
2 t sin
2
0 cos e) f x y y x such that x sin t and y cos t at t 1.
2
0 1 d f dt d f dt t 1 y
1 x 2 cos t
1 x sin t x 2
1 sin
2
1 t
y cos t x sin t cos
2 t sin
2 t sin
2 t
1
0 70873 sin
2
1
4

6. Prove Euler’s Theorem (in the web notes the second part of the theorem is missing). Solution in Blue font :
If f :
2 is a class C
2 function well defined in an open region R of the domain, then
∂
∂ x
2
∂ f y x
0 y
0
∂ 2 f
∂ y
∂ x x
0 y
0
Proof:
We are interested to see that this is correct for any arbitrary point x
0
R defined by the points x
0
, x
0
∆
x, y
0 and y
0
∆
y.
Consider the expression y
0
. We start by analysing the region f x
0
∆ x y
0
∆ y f x
0
∆ x y
0 f x
0 y
0
∆ y f x
0 y
0
(0.2) and the function g x f x y
0
∆ y f x y
0
(0.3)
It is easy to show that we can reproduce equation (0 2) by using equation (0 3), let see how this is done, substitute x x
0
∆
x and x x
0 in equation (0 3) we then have g x
0
∆ x g x
0 f x
0 f x
0 y
∆ x y
0
0
∆ y
∆ y f x
0 f x y
0
0
∆ x y
0
So we conclude that another way to write is g x
0
∆ x g x
0
(0.4)
Now, from the mean value theorem we know that the right hand side of equation (0 4) is where x is some point in the interval x
0 g x
0 x
0
∆ x g x
0
∂ g x
∂ x
∆ x
∆ x , therefore it follows that we can also write equation (0 2) as
∂ g
∂ x a
∆ x
But let us now substitute back the definition of g, equation (0 3), into this expression. We obtain
∂ f
∂ x a y
0
∆ y
∂
∂ x f a y
0
∆ x but we notice that again the quantity between square brackets is according to the mean value theorem (notice that this time is applied to the y):
∂ f
∂ x a y
0
∆ y
∂
∂ f x a y
0 ∂
∂ y
2
∂ f x a b
∆ y
This means that finally we can right equation (0 2) as where b is in the interval y
0 y
0
∂ 2 f
∂ y
∂ x a b
∆ y
∆ x
∆ y . Notice that the point a b R
5

So after all this tricks now we know that the second derivative of the function with respect to x and with respect to y, at the point a b is given by
∂ 2 f
∂ y
∂ x a b
∆ y
∆ x
Here is where the continuity is important, the point a b is somewhere in the region R the question is
Where?... we don’t know. What we know is that if we make
∆
x and
∆
y reduce to nearly cero, the region R reduces to nearly cero as well and so the point a b will coincide with the point x
0 y
0
; which is well defined and is the point we are interested in. For this process to be possible we need that the function is continuous and therefore that the limit exists:
∂ 2 f
∂ y
∂ x x
0 y
0 lim
∆ x
∆ y 0 0
∆ y
∆ x
(0.5)
This is the end of the road, we have a final relation for the application of the derivatives in the following order: first
∂
∂ x and
∂
∂ y second.
It will be let to the student to check that now we can arrive exactly to the same result by defining the function h y f x
0
∆ x y f x
0 y (0.6)
So we start from here, we check that substituting y y
0 and y y
0 h y
0 h y
0
∆ y f x
0
∆ x y
0 f x
0
∆ x y
0 f x
0 y
0
∆ y f x
0
∆
y into equation (0.6) then we obtain y
0
∆ y and therefore h y
0
∆ y h y
0 but applying the mean value theorem the right hand side of the equation (wich is only a function of y) can be rewritten as
∂ h
∂ y n
∆ y substituting back the value of h in terms of f , equation (0.6),we obtain
∂
∂ y f x
∂ f x
0
0
∆ x
∂ y
∆ x n n
∂ f x f x
∂
0
0 y n
∆ y n
∆ y
Again the mean value theorem (but now on the x variable) can be applied to the function beween square brackets so:
∂
∂ x
∂ f m n
∂ y
∆ x
∆ y or
∂ 2 f m n
∂ x
∂ y
∆ x
∆ y or
∂ 2 f
∂ y
∂ x m n
∆ y
∆ x
6

where the point m n R but is different than the point a b . So it will follow that by continuity arguments we obtain
∂ 2 f
∂ x
∂ y x
0 y
0 lim
∆ x
∆ y 0 0
∆ y
∆ x
(0.7) so that the theorem is complete compare (0 5) and 0 7
∂ 2 f
∂ y
∂ x x
0 y
0
∂ 2 f
∂ x
∂ y x
0 y
0
7