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 =
2hx, yi. rf
Two possibilities: x = 0 ! = 1 ! y = 0; y 6= 0 ! = 1 ! x = 0. So
rf = r
18.02 Lecture 21.
Tue, Oct 30, 2007
Test for gradient elds.
Observe: if F = M + N is a gradient eld then Nx = My . Indeed, if F = rf then M = fx ,
N = fy , so Nx = fyx = fxy = My .
Claim: Conversely, if F is dened and dierentiable at every point of
18.02 Lecture 33.
Thu, Dec 6, 2007
Handouts: PS12 solutions; exam 4 solutions; review sheet and practice nal.
Applications of div and curl to physics.
Recall: curl of velocity eld = 2 angular velocity vector (of the rotation component of motion).
18.02 Lecture 11.
Tue, Oct 2, 2007
Recall in single variable calculus: y =p (x) ) dy = f 0 (x) dx. Example: y = sin 1 (x) ) x = sin y,
dx = cos y dy, so dy/dx = 1/ cos y = 1/ 1 x2 .
Total dierential: f = f (x, y, z) ) df = fx dx + fy dy +
1. We give a worked example here. A fuller explanation will be given in the next session.
w = x3 y 2 + x2 y 3 + y
and assume x and y satisfy the relation
x2 + y 2 = 1.
We consider x to be the independent variable, then, becau
18.02 Lecture 8. Tue Sept 25 2007
Functions of several variables.
Recall: for a function of 1 variable, we can plot its graph, and the derivative is the slope of the
tangent line to the graph.
Plotting graphs of functions of 2 variables: examples z = y, z
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
Vectors. A vector (notation: A) has a direction, and a length (|A|). I
18.02 Lecture 16.
Thu Oct 18 2007
Handouts: PS6 solutions, PS7.
Recall integral in 1-variable calculus: a f (x) dx = area below graph y = f (x) over [a, b].
Now: double integral R f (x, y) dA = volume below graph z = f (x, y) over
Solutions for PSet 5
(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].
E.g., cycloid: ~(t) = ht sin t, 1 cos ti.
Velocity vector: ~ (t) =
= h , i. E.g., c
18.02 Lecture 3. Tue Sept 11 2007
B = B , = 0.
Application of cross product: equation of plane through P1 , P2 , P3 : P = (x, y, z) is in the
plane i det(P1 P , P1 P2 , P1 P3 ) = 0, or equivalently, P1 P N = 0, where N is the normal vector
18.02 Lecture 26.
Tue, Nov 13, 2007
Spherical coordinates (, , ).
= rho = distance to origin. = ' = phi = angle down from z-axis. = same as in cylindrical
coordinates. Diagram drawn in space, and picture of 2D slice by vertical plane with z, r coordinat
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,
dx dy =
ab du dv = ab
du dv = ab.
(x/a)2 +(y/b)2 <1
u2 +v 2 <1
u2 +v 2 <1
(substitution works here as in 1-var
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: @u/@t
Solutions for PSet 4
(t, t cos ) for 0 < t 1,
(0, 0) for t = 0.
f (t) =
f : R R2 , thus it is continuous if both f1 (t) = t and
f2 (t) =
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.
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
(a) dR = fr dr + fw dw = k(w( 4r 5 )dr + r 4 dw).
4 dr +
is more sensitive to dr = relative change in r. Opposite signs in
(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)
Solutions for PSet 6
(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
= 2 cos(y + 2x)
= 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
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:
18.02 Lecture 29.
Tue, Nov 20, 2007
Recall statement of divergence theorem:
F dS =
div F dV .
Del operator. r = h@/@x, @/@y, @/@zi (symbolic notation!)
rf = h@f /@x, @f /@y, @f /@zi = gradient.
r F = h@/@x, @/@y, @/@zi hP, Q, Ri = Px + Qy +