Homework 8 solutions
Bowei Zhao
April 21, 2017
6xy − 6x
0
=
3x2 + 3y 2 − 6y
0
when 6xy − 6x = 0 and 3x2 + 3y 2 − 6y = 0
6xy − 6x = 0 implies x(y − 1) = 0, so x = 0 or y = 1.
If x = 0, 3y 2 − 6y = 0. so y(y − 2) = 0, y = 0 or y = 2. [2pt]
if y = 1, then 3x2 + 3 − 6 = 0, ie. 3x2 − 3 = 0, x = 1 or x = −1.
So the critical points are (0, 0), (0, 2), (1, 1) and (−1, 1). [1pt]
We compute the second derivatives:
fxx = 6y − 6, fxy = fyx = 6x, fyy = 6y − 6
Now at each of the critical points:
At (0, 0), D = fxx (0, 0)fyy (0, 0)−(fxy (0, 0))2 = (−6)(−6)−0 = 36 > 0, and fxx (0, 0) = −6 < 0,
hence a local maximum.
At (0, 2),D = fxx (0, 2)fyy (0, 2) − (fxy (0, 2))2 = 6 · 6 − 0 = 36 > 0, and fxx (0, 2) = 6 > 0, hence
a local minimum. [1pt]
At (1, 1),D = fxx (1, 1)fyy (1, 1) − (fxy (1, 1))2 = 0 · 0 − 62 = −36 < 0, hence a saddle point.
At (−1, 1), D = fxx (−1, 1)fyy (−1, 1) − (fxy (−1, 1))2 = 0 · 0 − (−6)2 = −36 < 0, hence a saddle
point. [1pt]
1. ∇f (x, y) =
4. First write
the system in matrix form:
x
2 −1 →
→
−
−
−
Let x = 1 , then →
x0 =
x [1pt]
x2
0 3
To find eigenvalues and eigenvectors for the above matrix A, let det(A − λI) = 0. Then
(2 − λ)(3 − λ) = 0, λ = 2 or 3. [1pt]
0 −1
When λ = 2, A − λI =
0 1
1
Since there is one free variable, eigenvector could be taken to be
. [1pt]
0
−1 −1
When λ = 3, A − λI =
0
0
1
An eigenvector could be taken as
. [1pt]
−1
1 2t
1
−
So general solution is →
x = c1
e + c2
e3t ,
0
−1
ie. x1 = c1 e2t + c2 e3t ,
x2 = −c2 e3t . [1pt]
6. First write
the system in matrix form:
x
−6 4 →
→
−
−
−
Let x = 1 , then →
x0 =
x
x2
3 −5
To find eigenvalues and eigenvectors for the above matrix A, let det(A − λI) = 0. Then
(−6 − λ)(−5 − λ) − 12 = 0, λ = −2 or−9. [1pt]
−4 4
When λ = −2, A − λI =
3 −3
1
Since there is one free variable, eigenvector could be taken to be
. [1pt]
1
3 4
When λ = −9, A − λI =
3 4
1
4
An eigenvector could be taken as
. [1pt]
−3
1 −2t
4
→
−
So general solution is x = c1
e
+ c2
e−9t ,
1
−3
ie. x1 = c1 e−2t + 4c2 e−9t ,
x2 = c1 e−2t − 3c2 e−9t .
Now to find c1 and c2 we plug in the initial condition.
c1 + 4c2 = 1
c1 − 3c2 = 0
So c1 = 37 , c2 = 17 .
So x1 = 37 e−2t + 47 e−9t ,
x2 = 73 e−2t − 73 e−9t . [2pt]
1
1
9. In 4, the eigenvectors are
and
, with both corresponding eigenvalues positive.
0
−1
Therefore it is a source, as both parts of the fundamental solutions will go to infinity when t goes
to infinity. [2pt] we first sketch the eigenvectors, with directions positive. Then the other solutions
follow these directions.
[3pt]
11. For 6, both eigenvalues are negative. Therefore solutions will be going towards 0 when t
goes to infinity. That means the origin is a sink. [2pt] The graph is as follows.
2
[3pt]
3