EE2007 Revision
Dot, Cross and Scalar Triple Products
These are elementary exercises designed as a revision or for independent study. It can be used by
students as a self-test to determine if they have the necessary foundation for te vector calculus
module. Not all exercises need to be covered in the tutorial class, especially considering that
most tutors will be conducting CA in Weeks 8 or 9, and Week 10 is the e-Learning week. If a
student finds that there is a knowledge gap, then he/she should do something to help him/herself,
e.g. read relevant textbooks, ask friends, post questions to the discussion forum in edveNTUre
or consult your tutors.
Exercise 1.
(a) Find the components of the vecotr v with given initial point P : (2, 3, 0) and terminal point
Q : (1, 7, 3). Find |v|. Sketch v.
√
[ v = [−1, 4, 3]; |v| = 26 ]
(b) If v = [0, 2, −3] with initial point P : (0, −2, 4). Find the terminal point Q of v and the
lenght (norm) of v.
√
[ Q : (0, 0, 1);|v = 13 ]
(c) Let a = [3, −2, 1], b = [0, 3, 0], c = [4, 1, −1]. Find
|a + b|,
|a| + |b|,
(1/|a|)a,
5a − 4b + 3c
[
√
11;3 +
√
√
14;(1/ 14)[3, −2, 1]; [27, −19, 2] ]
(d) If the vectors p = [0, 1, 2], q = [3, −1, 4], u = [2, 0, 2], v = [3, −3, 0] represent forces, find the
resultant force and its magnitude.
√
[ [8, −3, 8]; 137 ]
(e) Geometrical applications can help you develop skill with vector methods. Try Problem Set
8.1/Q34 in text, pp.407.
KVL/Aug06
Exercise 2. [Dot Products]
(a) Let a = [1, 3, 2], b = [2, 0, −5], c = [4, −2, 1]. Find
(a) a · b, b · a
(b) a · (2b + 3c), 2a · b + 3a · c
(c) 2b · 5c, 10b · c
(d) (a − b) · b, a · b − b · b
(e) What properties of dot product do the above illustrate?
[ −8; −16; 30; −37 ]
(b) [Component or projection] Find the component of a in the direction of b.
(a) a = [3, 5, 1], b = [1, 0, 0]
(b) a = [3, −4, 7], b = [2, 5, 2]
[ 3; 0 ]
(c) [Angle between two straight lines] A straight line L1 : a1 x + a2 y = c can be written as
a · r = c with a = [a1 , a2 ] 6= 0 and r = [x, y]. The vector a is perpendicular to L1 . Find the
angle between x − y = 1 and x − 2y = −1.
[ 18.4o ]
Exercise 3. [Dot Product and Orthogonality] Orthogonality is an important concept in
vector algebra. Test yourself wit the following short questions.
h For what a1 are a = [a1 , 4, 3] and b = [5, −2, 1] orthogonal?
h Find all unit vectors a = [a1 , a2 ] orthogonal to [5, −2].
h Find all vectors orthogonal to a = [2, 1, 0].
h If the diagonals of a rectangle are orthogonal, what obvious conclusion can you draw and
how can you prove it by vectors?
[ 1; 5a1 = 2a2 with a21 + a22 = 1; b such that 2b1 + b2 = 0, b3 arbitrary; |a| = |b| ]
Exercise 4. [Cross and Scalar Triple Products]
h Find the area of a parallelogram if the vertices are (1, 1), (4, −2), (9, 3), (12, 0).
h Find the volume of a parallelpiped if the edge vectors are [1, −1, 0], [−2, 0, 2], [−2, 0, −3].
Make a sketch.
[ 30; 10 ]
KVL/Aug06
EE2007 Tutorial 10
Vector Calculus
(Derivative of Vector Function, Curves, Tangent, Surface and Surface Normal)
Exercise 1. [Derivative of Vector Function]
F , ∂ F , dF of the following vector functions F(x, y, z).
(a) Find the first partial derivatives, ∂∂x
∂y dz
(i) [y 2 , z 2 , x2 ]
(ii) [xy, yz, zx]
(iii) [ex cos y, ex sin y, 0]
[ ([0, 0, 2x], [2y, 0, 0], [0, 2z, 0]);
([y, 0, z], [x, z, 0], [0, y, x]);
x
x
x
([e cos y, e sin y, 0], [−e sin y, ex cos y, 0], [0, 0, 0]) ]
(b) A particle moves along a curve whose parametric equations are
x = e−t ,
y = 2 cos 3t,
z = 2 sin 3t
where t is the time.
(i) Determine its velocity and acceleration
(ii) What are the magnitudes of its velocity and acceleration at t = 0 and at t = 1?
[ v = r0 = [−e−t , −6 sin 3t, 6 cos 3t]; a = v0√= [e−t
√, −18 cos 3t, −18 sin 3t];
37, 325, 6.0113, 18.0038 ]
Exercise 2. [Parametric representation of curves] Represent the following curves parametrically and sketch them
(a) y 2 + (z − 3)2 = 9, x = 0
(b) x2 + y 2 = 1, y = z
(c) x2 + y 2 = 9, z = 5 arctan(y/x)
[ [0, 3 cos t, 3 + 3 sin t]; [cos t, sin t, sin t]; [3 cos t, 3 sin t, 5t], 0 ≤ t ≤ 2π ]
KVL/Aug06
Exercise 3. [Tangent Vectors and Tangents]
(a) Given a curve C : r(t), find (a) r0 , a tangent vector, (b) r0 at the given point P , and (c)
q, the tangent at P . Sketch the curve and tangent.
(i) r(t) = [t, t3 , 0], P : (1, 1, 0)
√
(ii) r(t) = [cos t, 2 sin t, 0], P : ( 21 , 3, 0)
[ The tangent at P is given by q = p + wr0 where r0 is the tangent vector at point P .
Once the value of the parameter t at point P is determined, the tangent vector at point
P can be found. Hence
2
([1, 3t , 0], t = 1, q(w)
+ w[1, 3, 0]);
√ = [1, 1, 0] √
([− sin t, 2 cos t, 0], t = π/3, q(w) = [1/2, 3, 0] + w[− 3/2, 1, 0]) ]
(b) Find the unit tangent vector at any point on the curve
x = t2 + 1,
y = 4t − 3,
z = 2t2 − 6t
p
[ [2t, 4, (4t − 6)]/ (2t)2 + (4)2 + (4t − 6)2 ]
Exercise 4. [Gradients]
(a) The force in an electrostatic field f (x, y) = xy has the direction of the gradient ∇f . Find
its value at point P : (1, 1).
(b) The flow of heat in a temperature field f (x, y, z) = sin(x + z) takes place in the direction of
maximum decrease of temperature. Find this direction at the point P : ( π8 , 1, π8 ).
[ ∇(xy) =
√ [y, x] = [1,
√ 1];
−∇(sin(x + z)) = −[cos(x + z), 0, cos(x + z)] = [−1/ 2, 0, −1/ 2] ]
Exercise 5. [Curve and Surface Normal] Find unit normal vectors for the given curves and
surfaces at the given points
(a) y = 43 x − 23 , P : (2, 2)
(b) y = 1 − x2 , P : (1, 0)
p
(c) z = x2 + y 2 , P : (6, 8, 10)
[ There are two possible normals ±∇f . The unit normal is√given by ±∇f /|∇f
√ |.
Hence the answers are: ±[−4, 3]/5; ±[2, 1]/ 5; ±[3, 4, −5]/ 50 ]
KVL/Aug06
Exercise 6. [Potential Field] We can handle potential fields f more easily than vector fields
v = ∇f . Find f for a given v or state that v has no potential.
(a) [2x, 4y, 8z]
(b) [xy, 2xy, 0]
[ x2 + 2y 2 + 4z 2 ; no potential ]
Exercise 7. [Parametric representation of surfaces] Find a parametric representation of
the follow surfaces and a normal vector to the surface.
(a) Plane x = z
(b) Plane 3x + 4y + 6z = 24
(c) Elliptic cylinder 9x2 + 4y 2 = 36
(d) Sphere x2 + (y − 1)2 + (z + 2)2 = 4
[ In general, the parametric representation of a surface is r(u, v) and a surface normal is
N = ru × rv . The answer gives one such representation and there are many others.
([u, v, u], [−1, 0, 1]);
([u, v, 4 − 21 u − 23 v], [1/2, 2/3, 1]);
([2 cos u, 3 sin u, v], [3 cos u, 2 sin u, 0]);
([2 cos v cos u, 1 + 2 cos v sin u, −2 + 2 sin v], [4 cos2 v cos u, 4 cos2 v sin u, 4 sin v cos v] ]
KVL/Aug06
EE2007 Tutorial 11
Vector Calculus
(Line and Surface Integrals)
R
Exercise 1. [Line Integral, work done by a force] Calculate C F · dr for the following
data. (If F is a force, this gives the work done in the displacement along C.)
(a) F = [y 2 , −x2 ], C is the straight line segment from (0, 0) to (1, 4).
(b) F = [y 2 , −x2 ], C : y = 4x2 from (0, 0) to (1, 4).
(c) [EE2007, April/May 2005, Q3(b)] Calculate the work done by the F = e2y i + xe2y j along
the following paths: (i) C1 : x = t, y = 1, z = t20 , 0 ≤ t ≤ 1, (ii) C2 : the straight line
joining (1, 0, 0) and (1, 1, 1).
[ 4, 6/5, e2 , (e2 − 1)/2 ]
Exercise 2. [Conservative Force Field]
(a) Show that F = [(2xy+z 3 ), x2 , 3xz 2 ] is a conservative force field and find its scalar potential.
Hence find the work done in moving an object in this field from (1, −2, 1) to (3, 1, 4).
(b) [EE2007, April/May 2005, Q3(c)] Show that the force F = e2y i + xe2y j can be made
conservative by modifying its component along the i direction.
[ 202; F = [e2y /2, xe2y ] ]
Exercise 3. [Surface Integral]
RR
(a) Evaluate the flux integral S F · ndA for the following data: F = [3x2 , y 2 , 0], S : r =
[u, v, 2u + 3v], 0 ≤ u ≤ 2, −1 ≤ v ≤ 1.
(b) [EE2007, April/May 2005, Q4b] Given that the vector field F = xi + yj + zk represents
electric flux density, calculate the amount of charge enclosed by a closed surface S bounded
by the surface z = x2 + y 2 and the plane z = 1.
[ −36, 3π/2 ]
KVL/Aug06
Solutions to Exercises
Solutions to Exercises
Exercise 1.
(a) The straight line segment is y = 4x, x : [0, 1]. So dy = 4dx.
Z
Z
F · dr =
2
2
Z
4
[16x2 , −x2 ] · [1, 4]dx = 4
[y , −x ] · [dx, dy] =
C
C
x=0
(b) y = 4x2 ⇒ dy = 8xdx.
Z
Z
F · dr =
C
2
2
Z
4
[16x4 , −x2 ] · [1, 8x]dx =
[y , −x ] · [dx, dy] =
C
x=0
6
5
(c) On path C1 : x = t, y = 1, z = t20 , 0 ≤ t ≤ 1.
Z
Z
F · dr =
C
[e2y , xe2y , 0] · [dx, dy, dz] =
C
Z
1
[e2 , te2 , 0] · [1, 0, 20t19 ]dt =
t=0
Z
1
e2 dt = e2
t=0
On path C2 : straight line joining (1, 0, 0) and (1, 1, 1). In other words
r = [1, 0, 0] + t[0, 1, 1] = [1, t, t], 0 ≤ t ≤ 1
Hence
Z
Z
Z
F·dr = [e2y , xe2y , 0]·[dx, dy, dz] =
C
C
1
t=0
[e2t , e2t , 0]·[0, 1, 1]dt =
Z
1
e2t dt = (e2 −1)/2.
t=0
Exercise 1
KVL/Aug06
Solutions to Exercises
Exercise 2.
(a) If ∇ × F = 0, then F is conservative. This can be easily verified:
i
j
k ∂
∂
∂
∇ × F = ∂x
∂y
∂z = 0
(2xy + z 3 ) x2 3xz 2 The scalar potential function of F can be obtained as follows:
fx = 2xy + z 3 ⇒ f = x2 y + xz 3 + g1 (y, z) . . .
fy = x2
⇒ f = x2 y + g2 (x, z)
2
fz = 3xz
⇒ f = xz 3 + g3 (x, y)
Thus the scalar potential function is
f = x2 y + xz 3 + c, where c is a constant.
Although work done can be evaluated by line integration, for conservative field, we can
evaluate work done easily by the scalar potential function at the start and end points.
Work done = f (3, 1, 4) − f (1, −2, 1) = 202.
(b) We need ∇ × F = 0, which gives
i
j
k
∂
∂
∂ 2y
2y
∇ × F = ∂x
∂y
∂z = [0, 0, e − 2e ]
e2y xe2y 0 Thus, choosing F = [e2y /2, xe2y ] would make F conservative.
Exercise 2
KVL/Aug06
Solutions to Exercises
Exercise 3.
(a) A vector normal to S is given by
i j k N = ru × rv = 1 0 2 = [−2, −3, 1]
0 1 3 ZZ
ZZ
Z
F · n dA =
1
Z
2
F · N dudv =
S
R
v=−1
[3u2 , v 2 , 0] · [−2, −3, 1] dudv = −36
u=0
(b) Let S1 : z = x2 + y 2 and S2 : z = 1. The charge enclosed is
ZZ
ZZ
Q=
F · n dA +
F · n dA
S1
S2
Parameterized S1 by
r = [v cos u, v sin u, v 2 ] with 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1
So
i
j
k
N1 = ru × rv = −v sin u v cos u 0
cos u
sin u 2v
= [2v 2 cos u, 2v 2 sin u, −v]
Similarly, S2 is parameterized by
r = [v cos u, v sin u, 1] with 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1.
So
i
j
k
sin u 0 = [0, 0, v]
N2 = rv × ru = cos u
−v sin u v cos u 0 Then
ZZ
ZZ
F · n dA +
Q =
Z
S1
1 Z 2π
=
F · n dA
S2
[v cos u, v sin u, v 2 ] · [2v 2 cos u, 2v 2 sin u, −v] dudv
v=0 u=0
Z 1 Z 2π
[v cos u, v sin u, 1] · [0, 0, 1] dudv
+
v=0
=
u=0
π
3
+π = π
2
2
Exercise 3
KVL/Aug06
EE2007 Tutorial 12
Vector Calculus
(Stokes’s Theorem and Divergence Theorem)
Exercise 1. [Stokes’s Theorem]
(a) Stokes’s theorem can be used for calculating the work done
over a closed path. If the vector field is F(x, y, z) = x2 i +
4xy 3 j + y 2 xk, find the work performed in bringing a particle
along the path C in the plane z = y as shown in the figure.
(b) If
RR the vector field is F(x, y, z) = [2z, 3x, 5y], calculate
S (∇ × F) · n dA across the surface S which is the portion
of the paraboloid z = 4 − x2 − y 2 for which z ≥ 0 with upward
orientation and C : x2 + y 2 = 4 forms the boundary of S in
the xy-plane.
[ −90; 12π ]
KVL/Aug06
Exercise 2. [Divergence Theorem] Divergence theorem may be used o help evaluate an
otherwise tedious surface integral
(a) Find the flux of the vector field F(x, y, z) = [0, 0, z] across the
outward-oriented sphere x2 + y 2 + z 2 = a2
(b) Find the outward flux of the vector field F(x, y, z)[x3 , y 3 , z 3 ]
across
p the surface of the region enclosed by the hemisphere
z = a2 − x2 − y 2 and the plane z = 0.
(c) Find the outward flux of the vector field F(x, y, z)[2x, 3y, z 2 ]
across the surface of the unit cube.
[ 4πa3 /3; 6πa5 /5; 6 ]
KVL/Aug06
Solutions to Exercises
Solutions to Exercises
Exercise 1. Stokes’s theorem relates surface and line integrals:
ZZ
Z
(∇ × F) · n dA =
F · dr
S
C
(a) Instead of computing the work done directly via the line integral, we use Stokes’s theorem
to compute the work done via the surface integral where the surface is bounded by path
C.
The surface bounded by path C can be parameterized by
r(u, v) = [u, v, v], 0 ≤ u ≤ 1, 0 ≤ v ≤ 3
Since the surface is oriented
i j
N = rv × ru = 0 1
1 0
downwards, the surface normal is
k 1 = [0, 1, −1]
0 Given that F(x, y, z) = x2 i + 4xy 3 j + y 2 xk, we have
i
j
k ∂
∂
∂ 2
3
∇ × F = ∂x
∂y
∂z = [2xy, −y , 4y ]
x2 4xy 3 y 2 x Thus
ZZ
ZZ
(∇ × F) · n dA =
S
F · N dudv
Z 1
=
[2uv, −v 2 , 4v 3 ] · [0, 1, −1] dudv
v=0 u=0
Z 3 Z 1
=
−v 2 − 4v 3 dudv = −90
Z
R
3
v=0
u=0
Confirm that the work done computed using the line integral
answer. Which is easier, line or surface integrals?
R
C
F · dr gives the same
(b) The circle can be parameterize as
r = [2 cos t, 2 sin t, 0], 0 ≤ t ≤ 2π
Hence
ZZ
Z
(∇ × F) · n dA =
S
F · dr
C
Z 2π
[0, 3(2 cos t), 5(2 sin t)] · [−2 sin t, 2 cos t, 0] dt
=
t=0
Z 2π
=
12 cos2 t dt = 12π
t=0
Verify your answer by evaluating the surface integral directly. Which is easier, line or
surface integral?
Exercise 1
KVL/Aug06
Solutions to Exercises
Exercise 2. Divergence theorem relates surface and volume integrals
ZZ
ZZZ
F · n dA =
∇ · F dV
S
V
(a) Instead of computing the flux directly using surface integral, we use the divergence theorem.
Since ∇ · F = 1, the flux across the outward-oriented sphere is
ZZZ
ZZZ
4
∇ · F dV =
dV = πa3 , (volume of sphere with radius a)
3
V
V
(b) Again, we use divergence theorem.
Given that F(x, y, z) = [x3 , y 3 , z 3 ], we have ∇ · F = 3x2 + 3y 2 + 3z 2 .
Using the spherical coordinates, dV = ρ2 sin φ dρ dφ dθ where x2 + y 2 + z 2 = ρ2 . Thus
ZZZ
2π
Z
∇ · F dV
π/2 Z a
Z
= 3
V
θ=0
φ=0
(ρ)2 ρ2 sin φ dρ dφ dθ
ρ=0
6 5
πa
5
=
You may wish to verify the divergence theorem for this case, by evaluating the surface
integral (which is fairly tedious):
ZZ
ZZ
ZZ
F · n dA =
F · n dA +
F · n dA
S
S1
S2
where S1 is the surface of the hemisphere of radius a, and S2 is the circular base of the
hemisphere.
(c) Again, instead of evaluating the surface integral directly, we use the divergence theorem.
Given that F(x, y, z) = [2x, 3y, z 2 ], we have ∇ · F = 5 + 2z. Thus,
ZZZ
Z
1
Z
1
Z
1
∇ · F dV =
V
(5 + 2z) dx dy dz = 6
y=0
x=0
z=0
You may wish to verify the divergence theorem for this case, by evaluating the surface
integral (which is fairly tedious):
ZZ
F · n dA =
S
6 ZZ
X
i=1
F · n dA
Si
where Si , i = 1, 2, . . . , 6 are six surfaces of the cube.
Exercise 2
KVL/Aug06