HW 06 3.2 - 4. Answer: f(x, y)=1

HW 06
MATH 110. 202
3.2 - 4. Answer:
f (x, y) = 1 − 2x2 − 2y 2 + · · ·
Q: To compute the second order Taylor series expansion of f (x, y) = 1/(x2 +y 2 +1)
at (0, 0).
Need to find the partial derivatives:
f (0, 0) = 1
∂f
−2x
(0, 0) = 2
|(0,0)
∂x
(x + y 2 + 1)2
=0
∂f
(0, 0) = 0
∂y
(x2 + y 2 + 1)2 (−2) − (−2x)(some term)
∂2f
(0,
0)
=
|(0,0)
∂x2
(x2 + y 2 + 1)4
= −2
∂2f
(0, 0) = −2
∂y 2
∂2f
(−2x)(some term)
∂2f
(0, 0) =
(0, 0) =
|(0,0)
∂x∂y
∂y∂x
(x2 + y 2 + 1)4
=0
by symmetry
by symmetry
The Taylor series expansion is given by
f (x, y) = 1 − x2 − y 2 + · · ·
3.2 - 10. Answer:
f (x, y) = 1 + (1 + 2π)(x − 1) − π 2 /2.(y − 2)2 + π(x − 1)(y − 2) + · · ·
or
f (x + 1, y + 2) = 1 + (1 + 2π)x − π 2 /2.y 2 + πxy + · · ·
Q: To compute the second order Taylor series expansion of f (x, y) = x cos(πy)−
y sin(πx) at (1, 2).
1
2
MATH 110. 202
Need to find the partial derivatives:
f (1, 2) = 1
∂f
(1, 2) = cos(πy) − πy cos(πx)|(1,2)
∂x
= 1 + 2π
∂f
(1, 2) = −πx sin(πy) − sin(πx)|(1,2)
∂y
=0
∂2f
(1, 2) = π 2 y sin(πx)|(1,2)
∂x2
=0
∂2f
(1, 2) = −π 2 x cos(πy)|(1,2)
∂y 2
= −π 2
∂2f
∂2f
(1, 2) =
(1, 2) = −π sin(πy) − π cos(πx)|(1,2)
∂x∂y
∂y∂x
=π
The Taylor series is then any of the two equivalent forms:
f (x, y) = 1 + (1 + 2π)(x − 1) − π 2 /2.(y − 2)2 + π(x − 1)(y − 2) + · · ·
f (x + 1, y + 2) = 1 + (1 + 2π)x − π 2 /2.y 2 + πxy + · · ·
3.3 - 14. Answer:
(0, 0) - saddle point,
{(x, y) : xy = 2nπ + π/2 for any integer n} - maxima,
{(x, y) : xy = (2n − 1)π + π/2 for any integer n} - minima
Q: Find critical points of f (x, y) = log(2 + sin xy).
∂f
y cos xy
=
∂x
2 + sin xy
∂f
x cos xy
=
∂y
2 + sin xy
Setting these equal to 0 reduces the problem to solving the simultaneous system,
y cos xy = 0
x cos xy = 0
The solutions of this are (x, y) = (0, 0) and xy = nπ + π/2 for any integer n. The
critical points (x, y) = (0, 0) is an isolated critical point and here we can use the
Second Derivative test.
HW 06
3
To find the type of critical point compute the Hessian,
(y cos xy).(y cos xy) − (2 + sin xy)(−xy sin xy)
|(0,0)
fxx (0, 0) =
(2 + sin xy)2
=0
(y cos xy).(x cos xy) − (2 + sin xy)(cos xy − xy sin xy)
|(0,0)
(2 + sin xy)2
1
=−
2
(x cos xy).(x cos xy) − (2 + sin xy)(−xy sin xy)
fyy (0, 0, 0) =
|(0,0)
(2 + sin xy)2
=0
fyx (0, 0, 0) = fxy (0, 0, 0) =
And so,
Hess(f )(0,0) =
0
−1/2
−1/2
0
det Hess(f )(0,0) = −1/4
<0
And so (0, 0) is a saddle point.
One way to verify our answer is by observing that the function increases away
from 0 along the direction y = x and decreases along the direction y = −x.
The other set of critical points is not isolated and we need to check these by
hand. At xy = nπ + π/2 the function log(2 + sin xy) takes values either log(2 + 1)
or log(2 − 1). Because log is an increasing function we see that these values are in
fact the absolute maxima and minima respectively and hence also the local maxima
and minima respectively.
3.3 - 18. Answer: k ≤ 2.
a) Given f (x, y, z) = x2 + y 2 + z 2 − kyz, verify that (0, 0, 0) is the critical point
for f .
∇f (x, y, z) = (2x, 2y − kz, 2z − ky)
⇒ ∇f (0, 0, 0) = (0, 0, 0)
hence (0, 0, 0) is a critical point.
b) Local minima at (0, 0, 0)?
The Hessian of f is


fxx fxy fxz
Hess(f )(0, 0, 0) = fyx fyy fyz 
fzx fzy fzz (0,0,0)


2 0
0
= 0 2 −k 
0 −k 2
4
MATH 110. 202
The three determinants to check are:
2 =2
0
=4
2

0
−k  = 2(4 − k 2 )
2
det
2
det
0

2
det 0
0
0
2
−k
By the second derivative test, for local minima all three determinants should be
> 0 and so the condition is k < 2.
However when k = 2 the second derivative test is inconclusive as the determinant
is 0 and we need to check the function by hand. At k = 2,
f (x, y, z) = x2 + y 2 + z 2 − 2yz
= x2 + (y − z)2
And we see that (x, y, z) = (0, 0, 0) is a local minima, (and so are all the points of
the form (0, t, t)).
3.3 - 28. Answer: (40/9, −20/9, 40/9).
Q: Find the point on the plane 2x − y + 2z = 20 closest to the origin.
You can solve this in 3 different ways:
Method 1 : Converting
this into equations, this is asking us to find the point
p
2 + y 2 + z 2 subject to the constraint 2x − y + 2z = 20. Minix
which minimizes
p
mizing x2 + y 2 + z 2 is the same thing as minimizing x2 + y 2 + z 2 .
Use the constraint to get rid of one of the variables. It is easiest to get rid of y.
2x − y + 2z = 20
⇒ y = 2x + 2z − 20
So we need to minimize:
x2 + (2x + 2z − 20)2 + z 2
= 5x2 + 5z 2 + 8xz − 80x − 80z + 400
=: f (x, z)
HW 06
5
Finding critical points:
∂f
= 10x + 8z − 80
∂x
∂f
0=
= 10z + 8x − 80
∂z
40
⇒x=z=
9
40
40
⇒ y = 2. + 2. − 20
9
9
−20
=
9
Now one can argue either geometrically or by using the Hessian that this is a local
minima.
0=
Method 2: This is THE situation when one uses Lagrange multipliers.
f (x, y, z) := x2 + y 2 + z 2
g(x, y, z) := 2x − y + 2z − 20
⇒ ∇f (x, y, z) = (2x, 2y, 2z)
⇒ ∇g(x, y, z) = (2, −1, 2)
The Lagrange multipliers method requires us to find a λ (pronounced lambda) such
that
(2x, 2y, 2z) = λ(2, −1, 2)
⇒ x = z = λ,y = −λ/2
Plugging back in the constraint
2λ − (−λ/2) + 2λ = 20
⇒ 9λ/2 = 20
⇒ λ = 40/9
⇒x = z = 40/9, y = −20/9
Again one needs to argue that this is indeed a minima by comparing it to nearby
points.
Method 3: One can see geometrically that this is going to be the point where
the line through the origin in the direction normal to the plane intersects it.
The normal direction is n
ˆ = (2, −1, 2) and hence the equation of the line is
(x(t), y(t), z(t)) = (2, −1, 2).t
Finding the intersection of this line with the plane 2x − y + 2z = 20 gives us the
point x = z = 40/9, y = −20/9.