Musa beyah
10/1/16
53103353
(AP Statistics MA14)
289B02SP
Pages 360361 Exercises 7.1, 7.5, 7.7
7.1
One broken egg
a. Discrete b. Continuous c. Discrete d. Discrete e. Continuous
7.5
y is continuous variable. Y is measured in feet and the value could vary
Chain Rule
1. The temperature on a hot surface is given by
T = 100 e
(x2 +y 2 )
.
A bug follows the trajectory r(t) = ht cos(2t), t sin(2t)i.
a) What is the rate that temperature is changing as the bug moves?
b) Draw the level curves of T and sketch the b
Problems for 18.112 Final Examination
(Open Book)
Dec. 20, 2006
1. (20) Let a, b, c be complex numbers satisfying
ba
ac
=
.
ca
bc
Considering the triangle with vertices a, b, c. Prove
b a = c a = b c.
2. (15) Find where the series
n=1
zn
1 + z 2n
co
Solution for 18.112 Final Examination
Problem 1.
Method 1 (Geometric way): Let A, B, C be points on the complex plane corresponding to complex numbers a, b, c. Then
AB = b a, BC = c b, CA = a c
and
A = arg
By the condition
we get
ba
cb
ac
, B = arg
, C =
Derivatives  vector functions.
of
Recall that if x is a point of
a scalar function of
x,

R
"
and if
then the derivative of
f
f ( 5 ) is
(if it
exists) is the vector
For some purposes, it will be convenient to denote the derivative
of
f

by a row matri
Notes  double integrals.
on
(Read 11.111.5 of Apostol.)
Just as for the case of a single integral, we have the
following condition for the existence of a double integral:
 defined is
on
f  integrable  Q   only is
on
if and
if
 functions s  t with
Solution for 18.112 ps 1
1(Prob1 on P11).
Solution:
a < 1, b < 1 = (1 aa)(1 b < 1
b)
= 1 aa b + a < 1
b
abb
ab ab > a + b ab ab
= 1 + a b
ab
a
b
= (1 a ab) > (a b)(
b)(1
a b)
ab
=
< 1.
1 ab
2(Prob4 on P11).
Solution:
If there is a solution, th
Solution for 18.112 ps 3
1(Prob 2 on P83).
Solution: First we will prove the following two lemmas.
Lemma 1. Reection carries circles to circles.
Proof: Reection is the composition of two maps: conjugation
zz
and linear transformation
z R2 /(z a) + a,
both
® Let C be the ellipse x2/4 + y2/9 = 1 traversed once in
the counterclockwise direction, and define
:2 r + 2
G(z) =  d{ for any 2 inside C
C I 2
Then find G(1) , G' (i) and G"(i)
00 x HINT:
G) Evaluate f dx ,
_°° sinh x '
where sinh x = (ex e'x)/2
Rough Solutions for PSet 12
1. (12.10:7) We rst parameterize the sphere of radius a by r(, ) = (a cos sin , a sin sin , a cos )
where 0 2 and 0 . (We write rst so that the normal points
outward.) Then the integrals become
S
xy dy dz =
2
a2 sin2 cos sin (a
Solution for 18.112 ps 2
1(Prob 7 on P58).
Solution: Let A be a discrete set in separable metric space (X, d), then for any
a A, there is ra > 0 such that
A Nra (a) = cfw_a.
It follows that for any a = b in A,
Nra /2 (a) Nrb /2 (b) = .
Let E be a countabl
Solutions for PSet 11
1. (11.28:14) We substitute u = x y and v = x + y. the Jacobian of the
transformation (x, y) (u, v) is
1 1
1 1
This has determinant 2 and the vertices of the parallelogram S correspond to
< u < , < v < 3. Thus:
3
S
(x y)2 sin2 (x +