Download Problem 1. Consider the function f(x, y)=3y2 - 2y3

Problem 1.
Consider the function
f (x, y) = 3y 2 − 2y 3 − 3x2 + 6xy.
Find the critical points of the function and determine their nature.
We calculate
fx = −6x + 6y = 0 =⇒ x = y
fy = 6y − 6y 2 + 6x = 0 =⇒ x + y = y 2 =⇒ 2y = y 2 =⇒ y = 0 or y = 2.
We find the second derivatives
fxx = −6, fxy = 6, fyy = 6 − 12y.
We apply the second derivative test. When x = y = 0, the Hessian is
−6 6
Hf =
6 6
which is indefinite since the determinant is negative. The point (0, 0) is a saddle. When y = 2, the Hessian
equals
−6
Hf =
6
6
−18
which is negative definite since the trace is negative while the determinant is positive. We found (2, 2) is a
local maximum.
Problem 2.
Assume that u(x, y) and v(x, y) are harmonic conjugates i.e. they satisfy the Cauchy-Riemann equations
ux = vy , uy = −vx .
Show that
f (x, y) = u(x2 − y 2 , 2xy), g(x, y) = v(x2 − y 2 , 2xy)
are also harmonic conjugates.
Write z = x2 − y 2 and w = 2xy. Then
zx = 2x, zy = −2y,
wx = 2y, wy = 2x.
Using the chain rule, we calculate
fx = uz · zx + uw · wx = 2xuz + 2yuw
and
gy = vz · zy + vw · wy = −2yvz + 2xvw .
Since
uz = vw , uw = −vz
1
we deduce fx = gy . The equation
fy = −gx
is similar.
Problem 3.
Calculate the total derivative of the function
f : Matn×n → Matn×n
given by
f (A) = AAT .
We find
f (A + H) − f (A) = (A + H)(A + H)T − AAT = (AAT + AH T + HAT + HH T ) − AAT = AH T + HAT + HH T .
We claim that
Df (A)(H) = AH T + HAT .
To this end, it suffices to check that
f (A + H) − f (A)(H) − (AH T + HAT )
= 0,
H→0
||H||
lim
or equivalently
HH T
= 0.
H→0 ||H||
lim
Note however that
as H → 0. This shows
HH T ||HH T ||
||H||||H T ||
= ||H T || → 0
||H|| = ||H|| ≤
||H||
HH T
=0
H→0 ||H||
lim
as claimed.
Problem 4.
Two paraboloids
z = (x − 2)2 + (y − 2)2
and
z = 20 − x2 − y 2
intersect along a curve C. Find the point of C which is closest to the point (1, 1, 0).
We minimize the function
f (x, y, z) = (x − 1)2 + (y − 1)2 + z 2
subject to the constraints
g1 (x, y, z) = (x − 2)2 + (y − 2)2 − z = 0, g2 (x, y, z) = 20 − x2 − y 2 − z = 0.
We find
∇f = 2(x − 1, y − 1, z)
∇g1 = (2(x − 2), 2(y − 2), −1), ∇g2 = (−2x, −2y, −1).
We note that if ∇g1 and ∇g2 are dependent then they must be equal since the last coordinates are. But then
the first coordinates do not match. Therefore, we must have that
∇f = λ∇g1 + µ∇g2
which gives
2(x − 1) = 2(x − 2)λ − 2xµ
2(y − 1) = 2(y − 2)λ − 2yµ
2z = −λ − µ.
The first equation gives
x(λ − µ − 1) = 2λ − 1
while the second gives
y(λ − µ − 1) = 2λ − 1.
If
λ−µ−1=0
we must have λ =
1
2
and this µ = − 12 . This yields via the third equation z = 0 hence x = y = 2 because
g1 = 0. This set of numbers does not satisfy the second constraint g2 = 0. Thus
λ − µ − 1 6= 0 =⇒ x = y =
2λ − 1
.
λ−µ−1
The two constraints g1 and g2 become
z = 2(x − 2)2 = 20 − 2x2 =⇒ x2 + (x − 2)2 = 10 =⇒ x2 − 2x = 3 =⇒ (x − 1)2 = 4 =⇒ x = −1 or 3.
When x = y = −1 we obtain z = 18 which gives f (−1, −1, 18) = 4 + 4 + 182 . When x = y = 3 we obtain
z = 2 which gives f (3, 3, 2) = 4 + 4 + 4 = 12. The point we are searching is (3, 3, 2).
Extra Credit. [5 points]
Consider a differentiable function f : Rn → R and let
S = {(x1 , . . . , xn ) : f (x1 , . . . , xn ) = c}
be a level set. Consider P a point of S. Show that the gradient ∇f (P ) is ”orthogonal” to the level set S in
the following sense:
Let γ be a differentiable curve passing through P and lying entirely on S i.e.
γ : R → Rn
has the property that γ(t) is contained in S for all t and γ(0) = P . Using the chain rule, show that the
gradient ∇f (P ) is perpendicular to the tangent vector γ 0 (0) of the curve.
Since γ lies on S we must have
f (γ(t)) = c.
Differentiating we obtain
d
(f ◦ γ)(t) = 0 =⇒ ∇f (γ(t)) · γ 0 (t) = 0.
dt
Making t = 0 and recalling γ(0) = P we obtain
∇f (P ) · γ 0 (0) = 0
which shows that the gradinent ∇f (P ) is perpendicular to the tangent vector.