Math 5120
Homework 1 Solutions
∂f
∂x
1. (a) For f = x2 + y 2 :
= 2x;
∂2f
∂x∂y
= 0;
∂2f
∂y∂x
= 0; ∇f = (2x, 2y).
fxx fxy
There is a critical point at (0, 0), satisfying ∇f = (0, 0). Recall that if M = det
,
fyx fyy
and both M >0 and fxx < 0 are true, then the critical point is a local maximum. Here,
2 0
M = det
= 4 with fxx > 0, so there is no local maximum here.
0 2
mesh
meshc
50
50
40
40
30
30
20
20
10
10
0
5
0
5
5
0
y
5
0
0
−5
−5
y
x
meshz
0
−5
−5
x
surface
5
50
50
40
40
30
y
30
20
0
20
10
0
5
10
5
0
y
(b) For f = xy:
∂f
∂x
= y;
0
−5
−5
∂2f
∂x∂y
= 1;
−5
−5
x
∂2f
∂y∂x
0
x
5
0
= 1; ∇f = (y, x).
0 1
There is a critical point at (0, 0), satisfying ∇f = (0, 0). Here, M = det
= −1, so
1 0
there is no local maximum here.
1
mesh
meshc
100
100
50
50
0
0
−50
−50
−100
10
−100
10
10
y
10
5
0
−10
−5
−10
5
0
0
y
x
meshz
0
−10
−5
−10
x
surface
10
100
5
50
0
0
100
50
y
0
−50
−100
10
−5
−50
10
5
0
y
0
−10
−10
−10
−5
−10
x
−5
0
x
5
10
−100
2. (a) F = (x + y, x − y): This is a gradient field provided that if F = (M, N ), then My = Mx .
Since My = Nx = 1, we know that there exists a φ such that F = ∇φ. To find φ, we
integrate:
Z
1
φ(x, y) = (x + y) dx = x2 + xy + c1 (y)
2
Z
1
φ(x, y) = (x − y) dy = xy − y 2 + c2 (x)
2
1 2
1 2
⇒ c1 (y) = − y + C; c2 (x) = x + C
2
2
1 2
1 2
⇒ φ(x, y) = x + xy − y + C
2
2
(b) F = (x2 y, y 2 x): Assuming that F = (M, N ), notice that My 6= Nx , and so F cannot be a
gradient field.
3. (a) The generalized 1d conservation equation is given by
∂
∂
[c(x, t) · A(x, t)] = − [J(x, t) · A(x, t)] ± σ(x, t) · A(x, t),
∂t
∂x
where J is the flux and σ is the creation/destruction of digested material. Substituting the
given expression for A(x, t) into the above equation gives
o
o
a
a
∂ n
∂ n
a
c(x, t) · [2 + cos(x − vt)] = −
J(x, t) · [2 + cos(x − vt)] ±σ(x, t)· [2+cos(x−vt)].
∂t
2
∂x
2
2
2
To solve for
∂c
∂t ,
expand the left-hand side:
∂c a
a
· [2 + cos(x − vt)] + c · [− sin(x − vt) · −v]
∂t 2 2
a ∂J
a
=−
· [2 + cos(x − vt)] + J · [− sin(x − vt)] ± σ · [2 + cos(x − vt)]
2 ∂x
2
⇒
∂c
∂t
− ∂J
∂x · [2 + cos(x − vt)] + J · sin(x − vt) ± σ · [2 + cos(x − vt)] − v · c · sin(x − vt)
2 + cos(x − vt)
(J − vc) sin(x − vt)
∂J
±σ+
.
=−
∂x
2 + cos(x − vt)
=
(b) If material is absorbed from the gut into the bloodstream at a rate proportional to its
concentration, then σ(x, t) = k · c(x, t) is subtracted from the conservation equation, where
k is a constant. This, along with the given flux of J(x, t) = 1, yields the following balance
equation:
∂c
(1 − vc) sin(x − vt)
= −kc +
.
∂t
2 + cos(x − vt)
(c) Now, if J(x, t) = 0 and σ(x, t) = 0, then
∂c
vc sin(x − vt)
=−
.
∂t
2 + cos(x − vt)
Since this is nonzero, the concentration will continue to change.
3