Math 113 - Worksheet 1 - 9/4/2015
1. (a) True. (b) False. (c) True.
2. ~a = h 11
, 22 i and ~b = h 45 , −2
i.
5 5
5
3. x = 1.
4. (2/3, 4/3, 4/3).
5. (a) Just factor out the A.
(b) We have
B
C 2
2
z = A x + xy + y
A
A
!
2
2
B
By
C
B
= A x2 + xy +
+ y2 −
y2
A
2A
A
4A2
!
2
B
4AC 2
B2 2
=A
x+
y +
y −
y
2A
4A2
4A2
2
B 2 − 4AC 2
B
=A x+
y −
y .
2A
4A
2
B
(c) If B 2 − 4AC < 0, then −(B 2 − 4AC) > 0. This means that the coefficients of x + 2A
y
and y 2 have the same sign (that sign is positive if A > 0 and negative if A < 0). In this case,
the surface is an elliptic paraboloid, because the origin is a local extremum (if A > 0 the elliptic
paraboloid opens up and the origin is a local min - if A < 0 the elliptic paraboloid opens down
and the origin is a local max).
2
If B 2 − 4AC > 0, then the signs of the coefficients A and − B −4AC
are opposite. This means
4A
that in one direction the surface has a parabola that opens up, and in another direction it has
a parabola which opens down. This means that the surface is a hyperbolic paraboloid, and the
origin is a saddle point.
√
6. A = 13.
1