(page 350)
9.1 Polar Coordinates
CHAPTER 9
9.1
POLAR COORDINATES AND
COMPLEX NUMBERS
Polar Coordinates
(page 350)
Circles around the origin are so important that they have their own coordinate system - polar coordinates.
The center at the origin is someti
MODEL ANSWERS TO HWK
1
1.1.20. (a) ~ Suppose that ~ = (a1 , a2 , a3 ). Then
0.
a
0 ~ = (0 a1 , 0 a2 , 0 a3 ) = (0, 0, 0) = ~
a
0.
The case of a vector in 2 is similar (and easier).
(b) ~ . Suppose that ~ = (a1 , a2 , a3 ). Then
a
a
a
1 ~ = (1 a1 , 1 a2 ,
Problems: Chain Rule Practice
One application of the chain rule is to problems in which you are given a function of x
and y with inputs in polar coordinates. For example, let w = (x2 + y 2 )xy, x = r cos and
y = r sin .
@w
1. Use the chain rule to nd
.
@r
MODEL ANSWERS TO HWK #2
(18.022 FALL 2010)
(1) (1.3.27)
(a) First let us x a labeling for the points: W1 , W2 and W3 are the centers of what we will
call the circles 1, 2 and 3, respectively, O is the point where all three circles intersect,
and A, B and
MODEL ANSWERS TO HWK
5
1. Let f : 2 ! 2 be the function given by f (x, y) = (x2 + y 2
1, y 2 x2 (x + 1). Then we are looking for solutions to the equation
f (x, y) = (0, 0).
We compute the derivative of f ,
Df (x, y) =
2x
2y
.
2x 3x2 2y
The determinant is
MODEL ANSWERS TO HWK #4
(18.022 FALL 2010)
(1) (i) f is nowhere continuous. Let = 1 . If x is rational, since irrational numbers are dense in
2
reals, for any > 0 there is irrational y such that |y x| < , and |f (y) f (x)| = 1 > .
Therefore f is not conti
Problems: Lagrange Multipliers
1. Find the maximum and minimum values of f (x, y) = x2 + x + 2y 2 on the unit circle.
Answer: The objective function is f (x, y). The constraint is g(x, y) = x2 + y 2 = 1.
Lagrange equations: fx = gx , 2x + 1 = 2x
fy = gy ,
MODEL ANSWERS TO HWK
(18.022 FALL 2010)
3
(1) In cartesian coordinates D is the region
(x a)2 + (y
1z3
a)2 a2
This translates to the cylindrical coordinates
r2 2ar(cos + sin ) a2
1z3
(2) The region D is
2a
a sin cos a
(3) (i) True. For any c 2 C there ex
Solution for PSet 2
1. (a) First, note T (1, 0) = (cos , sin ) and T (0, 1) = (cos( + /2), sin( +
/2) = ( sin , cos ) so the matrix is
cos sin
sin cos
(b) There are two ways to determine this problem, but perhaps the easiest is
to nd T . In that case T
1. (20pts) Let f: R3 > 1R be the function given by f(:i:,y,z) =
$33,; + yaz + Z31: 233112;.
(1) Find the gradient of f at P = (2, 1, 1).
Emma: EEK-M6? is" 2%)? AX: agarixz-degsmxgt
Valium: ('9 i 95: *(é+%~®&+(~1+6+M
: h + +
(ii) Find the directional deri
Solutions for PSet
1. (1.10:22)
(a) Let S = cfw_x1 , . . . , xk V . As L(S) = span(S), we can write:
L(S) = cfw_y : y 2 V where y =
k
X
i=1
ci xi ci is scalar
P
For cj = 1, ci = 0, i = j, we have y = i ci xi = xj 2 L(S). Thus xj 2 S
6
implies that xj 2
1. (20pts) Suppose that the four vectors t: 11, 17 and U7 lie in the
same plane H. Show that
(Exaxwxz)
3C A \l\ l3 OYMCSNQU to
So \C xe [Cs MM STo \hq
Sekkwhk 78 3&1)? \3 Odlmomx \"D
50 \3 Mwx lo M WM T
m at T ,3? m T {If m
H _N .4,
So (ltxlk)(v x?=?
(i
1. (20pts) Find the shortest distance between the plane If given by the
equation 23: y + 32 = 3 and the point of intersection of the two lines
ll and £2 given parametrically by
(m,y,z)=(2t-3,t,lt) and (3,352) =(1,1t,t).
a Pomk DJLwLE {We LNA L'HIEXL wet:
18.02 Problem Set 4, Part II Solutions
1. (a) The graphs of x ! F2 (x, t) = cos2 (x 2t) for t = 1, 0, 1 all have
the same sinusoidal shape f (u) = cos2 (u) shifted along the x-axis.
(b) This would represent the string displaced into the shape f and then
t
Solutions for PSet 3
1. First we note that T (P ) is again a parallelepiped. To see this notice that if
x T (P ) then x = T ( n=1 ci vi ) where 0 ci 1. By linearity x = i ci T (vi )
i
for 0 ci 1. That is, T (P ) = cfw_x Rn |x = i ci T (vi ), 0 ci 1 and th
Problems: Non-independent Variables
1. Find the total dierential for w = zxey + xez + yez .
Answer:
dw = zey dx + zxey dy + xey dz + ez dx + xez dz + ez dy + yez dz
= (zey + ez )dx + (zxey + ez )dy + (xey + xez + yez )dz.
2. With w as above, suppose we ha