18.02 Problem Set 6, Part II Solutions
~
~
Problem 1 (a) f (x, y) = x2 y 2 , rf = 2hx, yi, g(x, y) = x2 + y 2 , rg =
~ = rg ) hx, yi = hx, yi, or x = x, y =
~
y.
2hx, yi. rf
Two possibilities: x = 0 ! = 1 ! y = 0; y 6= 0 ! = 1 ! x = 0. So
6
~
~
~
~
rf = r
Solutions for PSet 5
1. (8.14:10)
(a) By hypothesis, we know f (x) = 0 for all x B(a). By denition
f (x) = (D1 f (x), . . . , Dn f (x), thus D1 f (x) = = Dn f (x) = 0. By
the 1-dimensional theorem, this means that f (x) is constant on every line
x + te1 ,
18.02 Lecture 6. Tue, Sept 18, 2007
Handouts: Practice exams 1A and 1B.
Velocity and acceleration. Last time: position vector ~(t) = x(t) + y(t) [+z(t)k].
r
E.g., cycloid: ~(t) = ht sin t, 1 cos ti.
r
d~
r
dx dy
Velocity vector: ~ (t) =
v
= h , i. E.g., c
18.02 Lecture 16.
Thu Oct 18 2007
Handouts: PS6 solutions, PS7.
Double integrals.
Rb
Recall integral in 1-variable calculus: a f (x) dx = area below graph y = f (x) over [a, b].
RR
Now: double integral R f (x, y) dA = volume below graph z = f (x, y) over
18.02 Lecture 1. Thu Sept 6 2007
Handouts: syllabus; PS1; ashcards.
Goal of multivariable calculus: tools to handle problems with several parameters functions of
several variables.
~
~
Vectors. A vector (notation: A) has a direction, and a length (|A|). I
18.02 Lecture 30.
Tue, Nov 27, 2007
Handouts: practice exams 4A and 4B.
Clarication from end of last lecture: we derived the di usion equation from 2 inputs. u =
concentration, F = ow, satisfy:
1) from physics: F = kru,
2) from divergence theorem: @[email protected]
Solutions for PSet 4
1. (B63:3)
(t, t cos ) for 0 < t 1,
t
(0, 0) for t = 0.
f (t) =
f : R R2 , thus it is continuous if both f1 (t) = t and
f2 (t) =
t
t cos
for 0 < t 1,
0 for t = 0.
are continuous. f1 is clearly continuous at any point t, and so is f2 a
18.02 Lecture 14.
Thu Oct 11 2007
Handouts: PS5 solutions, PS6, practice exams 2A and 2B.
Non-independent variables.
Often we have to deal with non-independent variables, e.g. f (P, V, T ) where P V = nRT .
Question: if g(x, y, z) = c then can think of z
18.02 Problem Set 5, Part II Solutions
Problem 1 R = f (r, w) = kwr
4
(k constant)
(a) dR = fr dr + fw dw = k(w( 4r 5 )dr + r 4 dw).
(b)
dR
R
(c)
dR
R
dw
w
=
4 dr +
r
dw
.
w
is more sensitive to dr = relative change in r. Opposite signs in
r
(or in dr and
1. (ZOpts) Find a recursive formula for a sequence of points (£130,913),
(whyl), ., (93mm), whose limit (momyoo), if it exists, is a point of
intersection of the curves 2. (ZOpts) Suppose that F: R3 > R2 is differentiable at P = (3, 2, 1)
with derivative
Solutions for PSet 6
1. (8.22:14)
(a) f (x, y) = f1 (x, y)i + f2 (x, y)j, where f1 (x, y) = ex+2y , and f2 (x, y) =
sin(y + 2x). Computing all the partial derivatives
f1
x
f2
x
f1
y
f2
y
= ex+2y
= 2 cos(y + 2x)
= 2ex+2y
= cos(y + 2x)
So the matrix for the
1. (20pts) For what values of A does the function f: R3 > R,
f($1y,z)= A562 - My + 92 + A22,
have a nondegenerate local minimum at (O, 0, 0)? . (20pts) Let f: R3 > R be the function at, 3/, z) = $2 y2 + 22.
(1) Show that f has a global maximum on the elli
1. (20pts) Let 11' and *5 be two vectors. Show that the vectors
5 = uffi |17|7 and b = |'r1'|17~ are orthogonal.
WQ (Ma Mr 3317:0-
31. \3 : (mm + \angmw nvu
: WV MEL WW m
: O_
-1 H3
So 0\ WAR \0 CAPE
(ii) Show that. the vector (1' = }- bisects the angle
18.02 Lecture 24.
Tue Nov 6 2007
Simply connected regions. [slightly dierent from the actual notations used]
Recall Greens theorem: if C is a closed curve around R counterclockwise then line integrals can
be expressed as double integrals:
I
ZZ
I
ZZ
~ d~
18.02 Lecture 18.
Tue, Oct 23, 2007
Change of variables.
Example 1: area of ellipse with semiaxes a and b: setting u = x/a, v = y/b,
ZZ
ZZ
ZZ
dx dy =
ab du dv = ab
du dv = ab.
(x/a)2 +(y/b)2 <1
u2 +v 2 <1
u2 +v 2 <1
1
(substitution works here as in 1-var