Lista Aula 1- Física Matematica

Prévia do material em texto

Lista exercícios - Aula 1 
 
1. 
2. 
 
3. 
 
 Q até P. (d) Encontre o divergente e o rotacional de G no ponto Q.
	 (e) Obtenha o gradiente do divergente de G no ponto P. 
	 (f) Obtenha o Laplaciano de G no ponto P. 
4. Mostre que 
	 	 	 Dica: siga a resolução do livro texto, cap. 6, M. Boas, ou desenvolva os dois 
	 	 	 lados e compare. 
5. 	 Use a identidade acima e verifique se (AXB) X C = A X (BXC).
Lista 1 - Eletromagnetismo
Lista 1 - Eletromagnetismo
Lista 1 - Veja Cap. 6, MATHEMATICAL METHODS IN THE PHYSICAL SCIENCES, Mary 
Boas, 3a. Edição
I - Operações com vetores:
II - Calcule a divergência e o rotacional dos seguintes campos:
III - Outros exercícios: 
IV - Responda: qual o significado físico e geométrico do gradiente de um campo 
escalar?
294 Vector Analysis Chapter 6
the component of ∇f in any direction u is the directional derivative df/ds in that
direction. (We are changing the function from φ to f since φ is used as an angle in
spherical, and sometimes in cylindrical and polar, coordinates.) The element of arc
length ds in the r direction is dr so the directional derivative in the r direction is
df/dr (θ and z constant) which we write as ∂f/∂r. In the θ direction, the element
of arc length is r dθ (Chapter 5, Section 4) so the directional derivative in the θ
direction is df/(r dθ) (with r and z constant) which we write as (1/r)∂f/∂θ. Thus
we have in cylindrical coordinates (or polar without the z term)
(6.7) ∇f = er ∂f
∂r
+ eθ
1
r
∂f
∂θ
+ ez
∂f
∂z
in cylindrical coordinates.
In a similar way we can show (Problem 21) that
(6.8) ∇f = er ∂f
∂r
+ eθ
1
r
∂f
∂θ
+ eφ
1
r sinφ
∂f
∂φ
in spherical coordinates.
PROBLEMS, SECTION 6
1. Find the gradient of w = x2y3z at (1, 2,−1).
2. Starting from the point (1, 1), in what direction does the function φ = x2−y2+2xy
decrease most rapidly?
3. Find the derivative of xy2 + yz at (1, 1, 2) in the direction of the vector 2i− j+ 2k.
4. Find the derivative of zex cos y at (1, 0,π/3) in the direction of the vector i+ 2j.
5. Find the gradient of φ = z sin y − xz at the point (2,π/2,−1). Starting at this
point, in what direction is φ decreasing most rapidly? Find the derivative of φ in
the direction 2i+ 3j.
6. Find a vector normal to the surface x2 + y2− z = 0 at the point (3, 4, 25). Find the
equations of the tangent plane and normal line to the surface at that point.
7. Find the direction of the line normal to the surface x2y + y2z + z2x+ 1 = 0 at the
point (1, 2,−1). Write the equations of the tangent plane and normal line at this
point.
8. (a) Find the directional derivative of φ = x2+sin y−xz in the direction i+2j−2k
at the point (1, π/2,−3).
(b) Find the equation of the tangent plane and the equations of the normal line
to φ = 5 at the point (1, π/2,−3).
9. (a) Given φ = x2 − y2z, find ∇φ at (1, 1, 1).
(b) Find the directional derivative of φ at (1, 1, 1) in the direction i− 2j + k.
(c) Find the equations of the normal line to the surface x2 − y2z = 0 at (1, 1, 1).
For Problems 10 to 14, use a computer as needed to make plots of the given surfaces and
the isothermal or equipotential curves. Try both 3D graphs and contour plots.
284 Vector Analysis Chapter 6
PROBLEMS, SECTION 3
1. If A = 2i− j−k, B = 2i− 3j+k, C = j+k, find (A ·B)C, A(B ·C), (A×B) ·C,
A · (B× C), (A× B)× C, A× (B× C).
For Problems 2 to 6, given A = i+ j − 2k, B = 2i− j+ 3k, C = j − 5k:
2. Find the work done by the force B acting on an object which undergoes the dis-
placement C.
3. Find the total work done by forces A and B if the object undergoes the displacement
C. Hint: Can you add the two forces first?
4. Let O be the tail of B and let A be a force acting at the head of B. Find the torque
of A about O; about a line through O perpendicular to the plane of A and B; about
a line through O parallel to C.
5. Let A and C be drawn from a common origin and let C rotate about A with an
angular velocity of 2 rad/sec. Find the velocity of the head of C.
6. In Problem 5, draw B with its tail at the head of A. If the figure is rotating as in
Problem 5, find the velocity of the head of B. With the same diagram, let B be a
force; find the torque of B about the head of C, and about the line C.
7. A force F = 2i− 3j+ k acts at the point (1, 5, 2). Find the torque due to F
(a) about the origin;
(b) about the y axis;
(c) about the line x/2 = y/1 = z/(−2).
8. A vector force with components (1, 2, 3) acts at the point (3, 2, 1). Find the vector
torque about the origin due to this force and find the torque about each of the
coordinate axes.
9. The force F = 2i − j − 5k acts at the point (−5, 2, 1). Find the torque due to F
about the origin and about the line 2x = −4 y = −z.
10. In Figure 3.5, let r′ be another vector from O to the line of F . Show that r′× F =
r× F. Hint: r − r′ is a vector along the line of F and so is a scalar multiple of F.
(The scalar has physical units of distance divided by force, but this fact is irrelevant
for the vector proof.) Show also that moving the tail of r along n does not change
n · r× F. Hint: The triple scalar product is not changed by interchanging the dot
and the cross.
11. Write out the twelve triple scalar products involving A, B, and C and verify the
facts stated just above (3.3).
12. (a) Simplify (A · B)2 − [(A× B)× B] ·A by using (3.9).
(b) Prove Lagrange’s identity : (A×B) ·(C×D) = (A ·C)(B ·D)−(A ·D)(B ·C).
13. Prove that the triple scalar product of (A×B), (B×C), and (C×A), is equal to
the square of the triple scalar product of A, B, and C. Hint: First let (B×C) = D,
and evaluate (A× B)×D. [See Am. J. Phys. 66, 739 (1998).]
14. Prove the Jacobi identity : A× (B×C) +B× (C×A) +C× (A×B) = 0. Hint:
Expand each triple product as in equations (3.8) and (3.9).
284 Vector Analysis Chapter 6
PROBLEMS, SECTION 3
1. If A = 2i− j−k, B = 2i− 3j+k, C = j+k, find (A ·B)C, A(B ·C), (A×B) ·C,
A · (B× C), (A× B)× C, A× (B× C).
For Problems 2 to 6, given A = i+ j − 2k, B = 2i− j+ 3k, C = j − 5k:
2. Find the work done by the force B acting on an object which undergoes the dis-
placement C.
3. Find the total work done by forces A and B if the object undergoes the displacement
C. Hint: Can you add the two forces first?
4. Let O be the tail of B and let A be a force acting at the head of B. Find the torque
of A about O; about a line through O perpendicular to the plane of A and B; about
a line through O parallel to C.
5. Let A and C be drawn from a common origin and let C rotate about A with an
angular velocity of 2 rad/sec. Find the velocity of the head of C.
6. In Problem 5, draw B with its tail at the head of A. If the figure is rotating as in
Problem 5, find the velocity of the head of B. With the same diagram, let B be a
force; find the torque of B about the head of C, and about the line C.
7. A force F = 2i− 3j+ k acts at the point (1, 5, 2). Find the torque due to F
(a) about the origin;
(b) about the y axis;
(c) about the line x/2 = y/1 = z/(−2).
8. A vector force with components (1, 2, 3) acts at the point (3, 2, 1). Find the vector
torque about the origin due to this force and find the torque about each of the
coordinate axes.
9. The force F = 2i − j − 5k acts at the point (−5, 2, 1). Find the torque due to F
about the origin and about the line 2x = −4 y = −z.
10. In Figure 3.5, let r′ be another vector from O to the line of F . Show that r′× F =
r× F. Hint: r − r′ is a vector along the line of F and so is a scalar multiple of F.
(The scalar has physical units of distance divided by force, but this fact is irrelevant
for the vector proof.) Show also that moving the tail of r along n does not change
n · r× F. Hint: The triple scalar product is not changed by interchanging the dot
and the cross.
11. Write out the twelve triple scalarproducts involving A, B, and C and verify the
facts stated just above (3.3).
12. (a) Simplify (A · B)2 − [(A× B)× B] ·A by using (3.9).
(b) Prove Lagrange’s identity : (A×B) ·(C×D) = (A ·C)(B ·D)−(A ·D)(B ·C).
13. Prove that the triple scalar product of (A×B), (B×C), and (C×A), is equal to
the square of the triple scalar product of A, B, and C. Hint: First let (B×C) = D,
and evaluate (A× B)×D. [See Am. J. Phys. 66, 739 (1998).]
14. Prove the Jacobi identity : A× (B×C) +B× (C×A) +C× (A×B) = 0. Hint:
Expand each triple product as in equations (3.8) and (3.9).
Section 4 Differentiation of Vectors 285
15. In the figure u1 is a unit vector in the direction of
an incident ray of light, and u3 and u2 are unit
vectors in the directions of the reflected and re-
fracted rays. If u is a unit vector normal to the
surface AB, the laws of optics say that θ1 = θ3 and
n1 sin θ1 = n2 sin θ2, where n1 and n2 are constants
(indices of refraction). Write these laws in vector
form (using dot or cross products).
16. In the discussion of Figure 3.8, we found for the angular momentum, the formula
L = mr× (ω× r). Use (3.9) to expand this triple product. If r is perpendicular to
ω, show that you obtain the elementary formula, angular momentum = mvr.
17. Expand the triple product for a = ω × (ω × r) given in the discussion of Figure
3.8. If r is perpendicular to ω (Problem 16), show that a = −ω2r, and so find
the elementary result that the acceleration is toward the center of the circle and of
magnitude v2/r.
18. Two moving charged particles exert forces on each other because each one creates
a magnetic field in which the other moves (see Problem 4.6). These two forces are
proportional to v1× [v2× r] and v2× [v1× (−r)] where r is the vector joining the
particles. By using (3.9), show that these forces are equal and opposite (Newton’s
third “law”) if and only if r× (v1 × v2) = 0. Compare Problem 14.
19. The force F = i+ 3j + 2k acts at the point (1, 1, 1).
(a) Find the torque of the force about the point (2 ,−1, 5). Careful! The vector r
goes from (2 ,−1, 5) to (1, 1, 1).
(b) Find the torque of the force about the line r = 2 i− j+5k+(i− j+2k)t. Note
that the line goes through the point (2 ,−1, 5).
20. The force F = 2 i− 5k acts at the point (3,−1, 0). Find the torque of F about each
of the following lines.
(a) r = (2 i− k) + (3j− 4k)t.
(b) r = i+ 4j + 2k+ (2 i+ j− 2k)t.
4. DIFFERENTIATION OF VECTORS
If A = iAx + jAy + kAz , where i, j, k are fixed unit vectors and Ax, Ay , Az are
functions of t, then we define the derivative dA/dt by the equation
(4.1)
dA
dt
= i
dAx
dt
+ j
dAy
dt
+ k
dAz
dt
.
Thus the derivative of a vector A means a vector whose components are the deriva-
tives of the components of A.
298 Vector Analysis Chapter 6
where the subscripts on ∇ indicate which function is to be differentiated. Since φ
is a scalar, it can be moved past the dot. Then
∇φ · (φV) = (∇φφ) ·V = V · (∇φ),
where we have removed the subscript in the last step since V no longer appears
after ∇. Actually you may see in books (∇φ) ·V meaning that only the φ is to be
differentiated, but it is clearer to write it as V · (∇φ). [Be careful with (∇φ)×V,
however; assuming that this means that only φ is to be differentiated, the clear way
to write it is −V× (∇φ); note the minus sign.] In the second term of (7.5), φ is a
scalar and is not differentiated; thus it is just like a constant and we can write this
term as φ(∇ ·V). Collecting our results, we have
(7.6) ∇ · (φV) = V ·∇φ+ φ(∇ ·V).
In Chapter 10, Section 9, we will derive the formulas for divV = ∇ · V and
∇2f in cylindrical and spherical coordinates. However, it is useful to have the
results for reference, so we state them here. Actually, these can be done as partial
differentiation problems (see Chapter 4, Section 11), but the algebra is messy.
In cylindrical coordinates (or polar by omitting the z term):
∇ ·V = 1
r
∂
∂r
(rVr) +
1
r
∂
∂θ
Vθ +
∂
∂z
Vz(7.7)
∇2f = 1
r
∂
∂r
(
r
∂f
∂r
)
+
1
r2
∂2f
∂θ2
+
∂2f
∂z2
.(7.8)
In spherical coordinates:
∇ ·V = 1
r2
∂
∂r
(
r2Vr
)
+
1
r sin θ
∂
∂θ
(Vθ sin θ) +
1
r sin θ
∂Vφ
∂φ
(7.9)
∇2f = 1
r2
∂
∂r
(
r2
∂f
∂r
)
+
1
r2 sin θ
∂
∂θ
(
sin θ
∂f
∂θ
)
+
1
r2 sin2 θ
∂2f
∂φ2
.(7.10)
PROBLEMS, SECTION 7
The purpose in doing the following simple problems is to become familiar with the formulas
we have discussed. So a good study method is to do them by hand and then check your
results by computer.
Compute the divergence and the curl of each of the following vector fields.
1. r = xi+ y j+ zk 2. r = xi+ y j
3. V = zi+ y j+ xk 4. V = yi+ zj+ xk
5. V = x2i+ y2j+ z2k 6. V = x2yi+ y2xj+ xyzk
7. V = x sin y i+ cos y j+ xyk 8. V = sinh z i+ 2y j+ x cosh z k