Homework week 8
Problems.
1. Find the extreme values of f x y x3 y2 on the circle x2 y2 4. Use Lagrange Multipliers technics
(refer to Homework 7).Solution
The auxiliary function is
g x y x2 y2 4
and the condition that matches the normal vectors of the surface f and g is
∇f
which is
λ∇g
3x2 i 2yj 2λxi 2λyj
comparing component we obtain
3x
1
2λ
λ
that is λ 1 and that x 23 . Now from the auxiliary function we are only interested in case when
g x y 0, i.e. the original constraint. substituting the value of x into the constraint we obtain
g
2
y
3
that means that
y 4
9
y2 4 0
4
2
3
23 43 2 and at P2 23 2 4
2 3
4
f
2 2
3 3
3
3
2 4
2 3
3
f
2
2
3 3
3
2
so the extreme is at the point P1
obtain the values are
2. Indicate which integral corresponds to
. Evaluating on the function we
4
3 2
2
2
104
27
104
27
xx 12 yy π7 f x y dydx, is it
a)
12 b)
f x y dxdy. Give a short explanation. Solution:
The answer is (b), acording to the rules we integrate from inside to outside therefore the integration
limits 1 to 2 refere to the variable x.
7
π
7 2
π 1
f x y dxdy or
3. Sketch the region of integration and evaluate the integral
1
03 a)
0
2
x2 y 2xy dydx. Solution:
Using Fubini’s method, we integrate from inner integrals to outer integrals. first I take x constant
3
0
2
0
x2 y 2xy dydx 3
0
1 2 2
x y xy2
2
y 0
2x2 4x dx
x 3
2 3
x 2x2
3
x 0
18 18 0
y
2
3
dx
0
03 b)
2
0
4 y2 dydx. Solution:
3
0
2
0
4 y2 dydx 0
2π π
π
0
2
dx
0
16
dx
0 3
16 3 48
x 3 0
3
1 3
y
3
3
c)
4y 3
sin x cos y dydx. Solution
2π
π
π
0
sin x cos y dydx
y sin x sin y π0 dx
2π
π
2π
π
π sin xdx
π cos x 2π π 1 π 1 0
π
4. Find the volume of the figure whose base is in the given region R and whose top lies in the surface given
by f x y a) f x y 5 x2 y in the region R : The triangle with vertex 0 0 to 0 1 and 1 0 . Solution
From the sketch of the triangle it is easily seen that the lower boundary is the straight line y 0
x
(0,1)
y=−x+1
(0,0)
(1,0)
y
Figure 0.1: fig1
and the upper line is the straight line y x 1. The y limits of integration are then the functions:
upper: g1 x x 1 and lower g2 x 0.To find the x limits of integration, we observe that all
2
the vertical lines that touch the region of interest are in the interval x 0 to x 1. So the integral
is
x 1
x 0
y
y 0
x 1
5 x2 y dydx 1
0
0
5y 1
0
1
1
0
5
0
1
5 x2 y dydx
1 2 y x 1
x2 y dx
y
2 y 0
1 x 1 x2 x 1 x 1 2 dx
x 1
2
5x 5 x3 x2 1 x2 x 1 dx
2
x3 3 2
9
dx
x 4x 2
2
0
x 1
1 4 1 3
2 9
x
x 2x
x
4
2
2 x 0
1 1
9 1 2 8 18
2
4 2
2
4
2
9
4
b) f x y 5 x y in the region R: Lower boundary x y 1 and upper boundary x 2 y2 1.
Solution:
the y-integration limits are ylower x 1 and yupper 1 x2 the point of intersection of the
x 2 +y 2 =1
x
(0,1)
(1,0)
(0,0)
y
y=−x+1
Figure 0.2: fig2
two curves occure when ylower
yupper that is at
x 1 2 x2 2x 1 1 x2
1 x2
or
2x2 2x
0
x x 2
0
which is satified when
x
x
3
1
0
so the integral is
1
0
1 x2
1 x
5 x y dydx
5y xy 1
0
0
0
1
1 x2 5
1 x2 x
1 x2 5
1 x2 x
1 x2 5
1 x2 x
1 x2 5x 5 dx
1
dx
y 1 x
0
1 x2
1 x2 x
1
5
1
y
1 2
y
2
1 2 1 1 x 2 d
1 x 5 1 x x 1 x 2
2
1 1 2
1 1 2
x 5 5x x x2 x dx
x
2 2
2
2
1 2 2 1 2
1 1
5 dx
x x
x 5x x x 2
2
2 2
0
1 x2 dx 1
5
1
x
0
1 x2 dx 0
5 2
x 5x
2
1
0
the first integral can be solved by substituting
x
sin u
dx
(0.1)
cos u du
substituting these terms and making use of the trigonometric relations cos u dw
1 1
1
2
2 cos 2u , w 2u 2 du
1
1 x2 dx
1 sin cos u du
cos u du
1 1
2 du 2 cos 2u du
1 1
cos w dw
u
2 4 1
1
arc sin x 4 sin w 2
1 1
arc sin x 4 sin 2arc sin x
2
1
1
arc sin x 4 2x 1 x 2
1 1
arc sin x x 1 x 2
2
1
1
1
1 sin2 , cos2 u 2
0
2
1
0
1
0
1
2
1
0
1
0
0
1
2
0
2
π 0
2
0 0 2
π
where we use the trigonometric relation sin 2arc sin x 2x 1 x2 , this relation was really not
necesary since we could directly evaluate the integral from sin 2arc sin x , (see trigonometric notes
below)
The second integral can be solved by substituting the same change of variable as in (0.1)
1
x
0
1 x2 dx 1
4
sin u cos u du
2
now notice that
d cos u sin u du
so the integral reduces (using v cos u in the second step) to
1
0
x 1 x2 dx cos u v dv
1
v 3 1 cos u 3
1
2
d cos u 2
3
3
3
cos3 arc sin x 1 1 x2 3 2
3
3
1 0 1 3
1
0
0
1 x2 1 1 x2 1
1
3
1
0
where we used the trigonometric relation cos arc sin x 1 x2 . Again this last relation was
3
not absolutely necesary since we could evaluate cos arc sin x directly (see trigonometric note
below).
Putting all the results together
1
0
1 x2
1 x
5 x y dydx
5
arc sin x 2
5 1
π 0
4
3
5 17
π
4
6
5
1 2
x 1 x2 1 x 2
3
5
5
2
3 2
5 x2 5x
2
1
0
Notes on Trigonometry
The basic trigonometric relations we always need to remember or have at hand are
cos u v cos u cos v sin u sin v
cos u v cos u cos v sin u sin v
(0.2)
(0.3)
and
sin u v sin u v sin u cos v sin v cos u
sin u cos v sin v cos u
(0.4)
(0.5)
From these equations we can deduce three very commonly used trigonometric relations simply by considering
u v
5
Let us first consider equation (0.2) and (0.3)
cos2 u sin2 u
cos u sin2 u 2
cos 2u (0.6)
1
(0.7)
Mixing these last two equations we can find a relation between the cosine and the square of the cosine
cos 2u 2 cos 2 u 1
(0.8)
and also a relation between the square of the sine and the cosine
cos 2u 1 2 sin2 u
(0.9)
From equation (0.4) we can also find a relation a formula similar to (0.8) or (0.9). again consider u v then
equation (0.4) becomes
(0.10)
sin 2u 2 sin u cos u
Now the trick to find the formulas that relate the arc sin x and arc cos x . Consider the triangles in figure
(0.3) both triangles satisfy the Pithagoras formula c 2 a2 b2 , in both cases c 1 and both have the same
1
1
u
sin u
1−x 2
u
cos u
x
equivalent
descriptions
Figure 0.3: fig3
angle u therefore they are the same. this means that we can identify
in other words from (0.11)
cos u
sin u
x
1 x2
u arc cos x
(0.11)
(0.12)
(0.13)
and substituting this into (0.12) we finally obtain that
sin arc cos x 1 x2
(0.14)
similarly if we substitute (0.13) in the left hand side of (0.8) and (0.11) on the right hand side of (0.8) we
obtain that
cos 2arc cos x 2x2 1
(0.15)
Again the same process than before but now using formula (0.10) that give us
sin 2arc cos x 2x
6
1 x2
(0.16)
To obtain the remainig combinations, we draw another picture, see fig (0.4):
1
1
sin u
x
u
u
cos u
1−x 2
Figure 0.4: fig4
again the descriptions are equivalent but in this case the relations are
1 x2
cos u
sin u
and u is now defined as
x
(0.17)
(0.18)
u arc sin x
(0.19)
substituting this equation back into equation (0.17) we obtain
cos arc sin x 1 x2
Warning: you can NOT mix formulas (0.17) to (0.19) with equations (0.11) to (0.13), the reason is that our
calculations depend on the drawing.
We can substitute equations (0.19) and (0.17) into the left and right side of equation (0.8) respectively
cos 2arc sin x 1 x2
and similarly into equation (0.10) to obtain
2x 1 x2
sin 2arc sin x 7