Assignment 11
Assignment 10
for F(x) = 2y i - 3z j + x k, with S being the part of the sphere x2+y2+ z2 = 4 in the first octant. The normal to S points away from the origin, and C is the positively oriented curve that serves as the boundary of S.
for F(x) = 2y i + 3x j - z3 k, with S being the surface of the closed cylinder (top, bottom, and curved side) 0 <= z <= 2, x2 + y2 = 25, with outward drawn normal. (This is the same cylinder as in Problem 4(b), Assignment 9.)
Assignment 9
Assignment 8
Assignment 7
A | = | 1 | -1 |
1 | -1.001 |
B | = | 1 | 4 | 7 |
2 | 5 | 8 | ||
3 | 6 | 10 |
Assignment 6
3 4 2 1 -1 -1 -1 2 4 -1 1 -1 4 -1 1 -1 4
-2 1 5 1 2 3 0 4 9 -1 -6 4 1 -3 7 0 1 -8 5 1 -1 1 5 3
I 0 0 Ris invertible and has inverse
I 0 0 R-1.
U V 0 W,where U is an r×r upper triangular matrix, V is an r×m matrix, W is an m×m, and 0 is the m×r zero matrix, then S-1A S has the same structure.
z1 * 0 A1where A1 is (n-1)×(n-1).
z1 * * 0 z2 * 0 0 A2.
5 1 4 0 2 0 -1 1 1
Assignment 5
x x x x x x x x 0 0 x x 0 0 x xwhere the x's indicate possible nonzero entries.
1 1 1 0 1 1 0 0 1
Assignment 4
10 -5 0 0 -5 10 -5 0 0 -5 10 -5 0 0 -5 10
Assignment 3
p0(t)=(1/2)½
p1(t)=(3/2)½t
p2(t)= (5/2)½(½)(3t2 -
1)
1 -1 2 1 2 -1 0 1 1 2 1 1
Due Thursday, Sept. 21.
Assignment 2
0 0 1 1/3 1/3 -1 -1/3 2/3 2 1/3;Hint: Take E=B.
Assignment 1
1 | -2 | 3 | 3 |
2 | -5 | 7 | 3 |
-1 | 3 | -4 | 3 |