Download MATH 2720 Winter 2012 Assignment 4 Questions 1, 2, 6 and 8 were

MATH 2720
Winter 2012
Assignment 4
Questions 1, 2, 6 and 8 were marked.
√
1. Show that the curve c(t) = (t2 , 2t−1, t), t > 0, is a flow line of the vector field F (x, y, z) = (y+1, 2, 1/2z).
Solution. The curve is a flow line if
c0 (t) = F (c(t)).
We have
c0 (t) = (2t, 2, t−1/2 /2)
and
√
F (c(t)) = F (t2 , 2t − 1, t)
1
= (2t − 1) + 1, 2, √
2( t)
1
= 2t, 2, √
2 t
= c0 (t),
so c is a flow line of F .
2. Let c(t) be a flow line of a gradient field F = −∇V . Prove that V (c(t)) is a decreasing function of t.
Solution. We obviously have to consider a function V : Rn → R, since c : R → Rn and we do not know
what it means for a function to be decreasing except when it is scalar valued. So V (c(t)) is decreasing iff
d
V (c(t)) < 0.
dt
From the Chain Rule,
d
V (c(t)) = DV (c(t))Dc(t)
dt
= ∇V (c(t)) • c0 (t),
because of the dimensionality of the functions involved. By assumption, ∇V = −F , so
d
V (c(t)) = −F (c(t)) • c0 (t).
dt
Since c is a flow line of F , there holds that c0 (t) = F (c(t)), so the previous equation takes the form
d
V (c(t)) = −c0 (t) • c0 (t).
dt
Recall that for any vector u, u • u = kuk2 . Thus,
d
V (c(t)) = −kc0 (t)k.
dt
For any u, kuk ≥ 0 with kuk = 0 only when u = 0. It follows that V is decreasing except where c is not
smooth.
Remarks.
1. This property is true if c is piecewise-smooth. Indeed, in more advanced calculus/analysis, it is shown
that the definition of a decreasing function can be made more general; f : R → R is decreasing if
f 0 (x) < 0 except maybe at a denumerable set of points.
2. This property plays a fundamental role in the study of some differential equations. A system of
differential equations such that x0 (t) = F (x(t)) with F : Rn → Rn a vector field satisfying the
property that ∃V such that ∇V = −F is called a gradient system. Gradient systems of differential
equations enjoy a number of interesting properties, two of the most striking of which being that
nontrivial solution curves x(t) are orthogonal to the level curves of V and that such systems do not
admit nontrivial periodic solutions.
3. Let f : R3 → R. Prove that curl ∇f = 0.
Solution.
4. Let F (x, y, z) = 2xyez~i + ez x2~j + (x2 yez + z 2 )~k. Here is a method that will allow us to find a function
f : R3 → R such that F = ∇f .
4.a. Write that F = ∇f , that is,
∂f ∂f ∂f
2 z
2 ~
z 2~
z~
2xye i + e x j + (x ye + z )k =
,
,
.
(1)
∂x ∂y ∂z
Choose one term, say, ∂f /∂x. Compute the antiderivative of ∂f /∂x = 2xyez . You should obtain something
of the form f (x, y, z) = φ(x, y, z) + K, with φ to be determined.
4.b. Explain why, in the expression you just found, K should in fact be a function of y and z, i.e., the
antiderivative that you just computed should be f (x, y, z) = φ(x, y, z) + K(y, z), with K(y, z) unknown.
[Hint: differentiate the expression with respect to x.]
4.c. Now differentiate the function f you found in 4.a with respect to y. Using (1) and the same type of
reasoning as in 4.a and 4.b, show that this means that you can find an antiderivative of ∂K(y, z)/∂y of the
form K(y, z) = ψ(x, y)+L(z), with L(z) an unknown function of z only. Substitute this K into the expression
f (x, y, z) = φ(x, y, z)+K(y, z) found previously, giving you a function f (x, y, z) = φ(x, y, z)+K(y, z)+L(z).
4.d. Now differentiate the function f you found in 4.c with respect to z and obtain an expression for L0 (z).
Taking the antiderivative of this and substituting into the function found in 4.c, you obtain finally the f
such that F = ∇f . Check that this is indeed the case.
Solution.
5. A sphere of mass m, radius a and uniform density has potential u and gravitational force F at a distance
r from the centre (0, 0, 0) given by
u=
3m mr2
−
,
2a
2a3
m
u= ,
r
m
F = − 3r
(r ≤ a)
a
m
F = − 3r
(r > a),
r
where r = krk and r = x~i + y~j + z~k.
5.a. Check that F = ∇u on the inside and outside of the sphere.
5.b. Check that u satisfies Poisson’s equation uxx + uyy + uzz = K, with K a constant, inside the sphere.
5.c. Check that u satisfies Laplace’s equation uxx + uyy + uzz = 0 outside the sphere.
Solution.
6. Evaluate the double integral
ZZ
|y| cos
R
πx
dA
4
over the region R = [0, 2] × [−1, 0].
Solution. From Fubini’s theorem,
ZZ
πx
dA =
|y| cos
4
R
2Z 0
Z
|y| cos
0
−1
πx
dy dx.
4
Since the integrand takes the form f (x, y) = g(x)h(y) on a rectangular domain, we have
ZZ
πx
|y| cos
dA =
4
R
2
Z
0
Z
πx
cos
dx
4
2
Z
0
|y| dy
−1
Z 0
πx
dx
(−y) dy
4
0
−1
0
4
πx 2 −y 2
=
sin
π
4 0 2 −1
4 −1
=
π 2
2
=−
π
=
cos
Remark. To check your answer, you could have used the following Maple command:
int(int(abs(y)*cos(Pi*x/4),y=-1..0),x=0..2)
7. Let f be continuous on R = [a, b] × [c, d]. For x ∈ [a, b] and y ∈ [c, d], define
Z xZ y
F (x, y) =
f (u, v) dv du.
a
c
Using the fundamental theorem of calculus and Fubini’s theorem, show that
∂2F
∂2F
=
= f (x, y).
∂x∂y
∂y∂x
Solution.
8. Let D be the region bounded by the positive x and y axes and the line 3x + 4y = 10. Compute
ZZ
x2 + y 2 dA.
D
Solution. The region is as shown in Figure 1. Let us describe it as a y-simple region. The equation of the
Figure 1: The region D in Exercise 8.
line as a function of x is y = 5/2 − 3x/4. It intercepts the x-axis at x = 10/3 and thus
5 3
10
2
D = (x, y) ∈ R ; 0 ≤ x ≤ , 0 ≤ y ≤ − x .
3
2 4
As a consequence,
ZZ
2
2
Z
x + y dA =
D
10
3
0
Z
5
− 34 x
2
x2 + y 2 dy dx
0
5−3x
y3 2 4
2
=
x y+
dx
3 0
0
Z 10 3
3
1 5 3 3
2 5
=
x
− x +
− x
dx
2 4
3 2 4
0
"
# 10
3 4 1 5 3 4 3
5 3
x − x +
− x
=
6
16
9 2 4
0
4
3 10 4 1 5 3 10
5 10 3
−
+
−
.
=
6 3
16 3
9 2 4 3
Z
10
3
Remarks.
1. There was no need to evaluate this result any further. But just in case you want to check, the result
was 15625/1296.
2. In Maple, int(int(x^2+y^2,y =0..5/2-(3/4)*x),x = 0..10/3). If you want the antiderivative
obtained after integrating with respect to x, do not specify the bounds for x: int(int(x^2+y^2,
y=0..5/2-(3/4)*x),x)