ENGI 2422
1.1
Fundamentals – Lines and Planes
Page 1-01
Equation of a Plane
A non-zero vector n in 3 will fix the orientation of a plane, to be at right angles to n .
Let P be a point that is known to be on the plane. Let the Cartesian coordinates of P be
(xo, yo, zo). Its position vector, a  OP  xo , yo , zo , allows us to pick out exactly one
plane from the infinite set of parallel planes that share the same orientation defined by the
plane normal vector n .
The two vectors, together, allow us to define a single plane completely.
Let r  OR  x, y, z
be the position vector of a general point R, with Cartesian
coordinates (x, y, z), in 3.
But OR  OP  PR

PR  OR  OP  r  a
Note that the normal vector n is at right angles to any vector lying in the plane .
R on 
 PR  n
("perpendicular to")

PR n  0
 r  a n  0
OR
rn  an
which is the vector equation of a plane.
Fundamentals – Lines and Planes
ENGI 2422
Page 1-02
Let n  A, B, C , (so that A, B, C are the Cartesian components of the normal vector),
then
a n  Axo  B yo  Czo   D
a constant 
r n  Ax  B y  C z
The vector equation of the plane, r n  a n , then leads to the
Cartesian equation of the plane:
Ax  B y  C z  D  0
Example 1.1.1
Find an equation for the plane through the point (3, 2, 4), which is normal to the vector
2ˆi  3ˆj  kˆ .
n  A, B, C  2, 3, 1
a  xo , yo , zo  3,  2, 4
rn  an
 2 x  3 y  1z  2  3  3   2   1  4
Therefore the Cartesian equation of the required plane is
2x  3y  z  4
Fundamentals – Lines and Planes
ENGI 2422
Page 1-03
Example 1.1.2
Find a unit vector orthogonal to the plane
x  2y + 2z = 7
From the Cartesian equation of a plane, one may read off the Cartesian coordinates of a
vector that is at right angles to that plane.
n  1,  2, 2
 n  n 
12   2   22 
2
9  3
Therefore a unit normal is
nˆ 
1
3
1,  2, 2
[Note that there is one other acceptable final answer, namely the unit vector that points in
the opposite direction, nˆ  13 1, 2,  2 .]
Fundamentals – Lines and Planes
ENGI 2422
Page 1-04
Equations of a Line
A line L is determined uniquely by two vectors, a line direction vector, v, which orients
the line in 3, and the position vector, a, of a point P known to be on that line.
Let R be a general point, with Cartesian coordinates (x, y, z), that is constrained to lie on
the line L.
OR  OP  PR
 r  a  PR
But the line L is parallel to the direction vector v.
 PR  t v
for some value of t.
Therefore the parametric vector equation of the line is
r  a  tv
t  
[Note that there is one free parameter, t, which can be any real number. The number of
free parameters (one) matches the dimension of the geometric object (the line is a onedimensional object).]
The Cartesian equations of the line can be found from the vector parametric form.
Fundamentals – Lines and Planes
ENGI 2422
Page 1-05
Example 1.1.3
Find the Cartesian equations of the line that passes through the point (3, 1, 2) and that is
parallel to the vector ˆi  2ˆj  3kˆ .
a  3, 1, 2 ,
v  1,  2, 3
The vector parametric form is
r  a  tv
or
x, y , z 
 3  t  , 1  2t  ,  2  3t 
Making t the subject of all three simultaneous equations
x = 3+t,
y = 1  2t
and
z = 2 + 3t ,
we obtain the Cartesian symmetric form for the equations of the line:
t 
x3
y 1
z2


1
2
3
General case:
A line r  a  t v which passes through the point (xo, yo, zo) and is parallel to the
vector v  v1, v2 , v3 , (where v1, v2, v3 are all non-zero) has a Cartesian symmetric
form
x  xo
y  yo
z  zo


v1
v2
v3
If v1 = 0, then separate out the equation x = xo.
If v2 = 0, then separate out the equation y = yo.
If v3 = 0, then separate out the equation z = zo.
Fundamentals – Lines and Planes
ENGI 2422
Page 1-06
Example 1.1.4
Find the Cartesian equations of the line through (1, 1, 1) that is perpendicular to the
plane 2x + 3z = 1.
The normal vector n to the plane is at right angles to the plane  .
The line L is also at right angles to the plane  .
Therefore the line’s direction vector v must be parallel to n.
v  n  2, 0, 3
a  1, 1,  1
The vector parametric form is
r  a  tv
or
x, y , z 
1  2t  ,
1 ,  1  3t 
Making t the subject of the equations where possible (for x and z ), the Cartesian
symmetric form then follows:
x 1
z 1
y 1,

2
3
Fundamentals – Lines and Planes
ENGI 2422
Page 1-07
More Equations of Lines and Planes
Three non-collinear points, A, B, C, define a plane.
The three vectors joining the three points to each other lie in that plane. Any two of
them may be used as the basis for a coordinate grid that can be laid out on the entire
plane.
The vector parametric equation of the plane then follows:
r  a  su  t v
s , t
where
a  any one of OA, OB, OC;
u, v  any two of AB, BC , CA;
and u  v  n .
The other vector equation for the plane can be recovered from this form:
Vectors AB and AC are both in the plane
 n  AB  AC is normal to the plane.
Let R be a general point in space, with Cartesian coordinates (x, y, z).
If and only if point R lies in the plane, then
AR AB  AC  0
 r  a n  0


Fundamentals – Lines and Planes
ENGI 2422
Page 1-08
Example 1.1.5
Find the Cartesian equation of the plane that passes through the points A(1, 0, 0),
B(2, 3, 4) and C(1, 2, 1).
AB 
 2  1 ,  3  0  ,  4  0 
 1, 3, 4
AC  2, 2, 1
ˆi ˆj kˆ
u  v  AB  AC  1 3 4
2 2 1
 n  5,  9,  8
Let R be a general point in 3, with Cartesian coordinates (x, y, z). Then
r  a  AR   x  1 ,  y  0  ,  z  0   x  1, y , z
For R to be on the plane,
r  a n


or
 0



AR AB  AC  0
(x  1)(5) + y(9) + z(8) = 0
5x  9y + 8z = 5
5x  9 y  8z  5
Alternative solution (next page):
Fundamentals – Lines and Planes
ENGI 2422
Page 1-09
Alternative Solution to Example 1.1.5:
The plane  is Ax + By + Cz + D = 0
This equation must be true at all three points on the plane.
An under-determined linear system of three equations for the four unknown coefficients
then follows:
A B C D
(1, 0, 0) on :
[ 1 0 0 1 | 0 ]
(2, 3, 4) on :
[ 2 3 4 1 | 0 ]
(1, 2, 1) on :
[ 1 2 1 1 | 0 ]
Row reduction leads to the reduced row echelon form
[ 1
[ 0
[ 0

 A, B, C , D 
0
1
0
0
0
1
1 | 0 ]
9/5 | 0 ]
8/5 | 0 ]
   D,  95 D, 85 D, D  , D 
Select a convenient non-zero value for D that leaves all four coefficients as integers and
makes at least two of A, B and C non-negative. Therefore select D =  5:
5x  9 y  8z  5  0
Fundamentals – Lines and Planes
ENGI 2422
Example 1.1.6
y2
z3

3
2
Find the angle between the line L : x  1 ,
and the plane  : x  2 y  z  4 .
sin 

nv
nv

1, 2,1 0,3,2
1 4 1 0  9  4
Therefore
Find the intersection of the planes
1 :
x  2y + z = 1 and
 2 : 3x  5y + 2z = 4.
Solve the under-determined linear system
x y z
[ 1 2 1 | 1 ]
[ 3 5 2 | 4 ]
R2 ← R2  3 R1 :
[ 1 2 1 | 1 ]
[ 0 1 1 | 1 ]
R1 ← R1 + 2 R2 :
[ 1
[ 0

x = t + 3,
y = t + 1,
z = t,
4
6  13
  26.9
Example 1.1.7
1
2

t
0 1 | 3 ]
1 1 | 1 ]
 .4529
Page 1-10
Fundamentals – Lines and Planes
ENGI 2422
Example 1.1.7 (continued)
 t 
x3
y 1
z 0


1
1
1
The intersection is a LINE.
In vector parametric form, it is
r = a + t v , with a = 3, 1, 0 and v = 1, 1, 1 .
Summary for Lines and Planes:
Vector forms
r  a  tv
Line
Plane
a = point on line,
v = direction vector
r  a  su  t v,
a = point on plane,
u, v = vectors in plane
OR
r n  a n , where
n  uv
Cartesian form
x  xo
y  yo
z  zo


v1
v2
v3
where a  xo , yo , zo
and
v  v1 , v2 , v3
Ax + By + Cz + D = 0 ,
n  A, B, C
Page 1-11
Fundamentals – Areas
ENGI 2422
1.3
Area, Arc Length, Tangents and Normals, Curvature
b
A 
 f  x  dx
a
With parameterization:
A 
tb
dx
 y dt dt
ta
where x(ta) = a , x(tb) = b , a < b and f (x) > 0 on [a, b].
Example 1.3.1
Find the area enclosed in the first quadrant by the circle x2 + y2 = 4.
x = 2 cos  , y = 2 sin  .
x  0    2
x2   0
x  2 cos 

dx
  2sin 
d
Page 1-12
Fundamentals – Areas
ENGI 2422
Example 1.3.1 (continued)
 /2
0

A 
  2sin  2sin  d

 4
/2

0
sin 2  d
0
 /2
 /2
 4

1
2
1
1  cos 2  d  2   sin 2 
2

0
 


 2   0    0  0  

 2

Therefore
A
Check: The area is the interior of a quarter-circle.
1
1
 r 2    2 2    

4
4
Review of the Tangent:
dx dy dz
, ,
dt dt dt
The unit tangent is
T 

dr
dt
T 
dr dr

dt
dt
Page 1-13
Fundamentals – Arc Length, Tangent
ENGI 2422
Page 1-14
Arc Length
In 2:
 s 
2
  x    y 
2
2
In 3:
 s 
2
  x    y    z 
2
2

The vector
ds

dt
2
2

2

dr
dt
dr
points in the direction of the tangent T to the curve defined parametrically
dt
by r = r(t).
dr
dr
d r ds
d r dt





dt
dt
dt dt
dt ds
dr
T 
ds
 T 
2
 dx 
 dy 
 dz 
       
 dt 
 dt 
 dt 
Fundamentals – Arc Length, Tangent
ENGI 2422
Page 1-15
Example 1.3.2
(a)
(b)
Find the arc length along the curve defined by
r  14  2t  sin 2t  , 14 cos 2t , sin t , from the point where t = 0 to the point
where t = 4π.
Find the unit tangent T .
dx
1
2
1
  2  2 cos 2t   1  cos 2t    2sin 2 t   sin 2 t
dt
4
4
2
(a)
dy
1
2
   2sin 2t     2sin t cos t    sin t cos t
dt
4
4
dz
 cos t
dt

dr
 sin 2 t ,  sin t cos t , cos t
dt

ds
dr


dt
dt

ds
1
dt
sin 4 t  sin 2 t cos 2 t  cos 2 t 
sin 2 t  sin 2 t  cos 2 t   cos 2 t
1 sin 2 t  cos 2 t  1
4
 s 
 1 dt
 t  0
4
0
Therefore
s  4
(b)
T 
But
dr dr

dt
dt
dr
 1 t
dt
"for all t" .
Therefore
T  sin 2 t ,  sin t cos t , cos t
[Note that, in this example, | t | itself is the arc length from the point (0, 1/4, 0), (where
t = 0). There are very few curves for which the parameterization is this convenient!]
Fundamentals – Arc Length, Tangent
ENGI 2422
Page 1-16
Review of Normal and Binormal:
T  1  TT  1

dT
dT
dT
T  T
 0  2T
 0
ds
ds
ds

dT
 0 or
ds
dT
 T
ds
(Note: a unit vector can never be zero).
Select the principal normal N to be the direction in which the unit tangent is,
instantaneously, changing:
dT
d T ds
N 


ds
dt dt
The unit principal normal then follows:
N 
dT dT

dt
dt
The binormal is at right angles to both tangent and principal normal:
B  T N
Fundamentals – Curvature
ENGI 2422
Page 1-17
Curvature:
The derivative of the unit tangent with respect to distance travelled along a curve is the
dT
principal normal vector, N 
. Its direction is the unit principal normal N . Its
ds
magnitude is a measure of how rapidly the curve is turning and is defined to be the
curvature:
  N 
dT
dT
dr


ds
dt
dt
 N  N
The radius of curvature is  
1
.

The radius of curvature at a point on the curve is the radius of the circle which best fits
the curve at that point.
Example 1.3.2 (continued)
(c)
Find the curvature κ(s) at any point for which s > 0, for the curve
r  s   14  2s  sin 2s  , 14 cos 2 s, sin s
where s is the arc length from the point (0, 1/4, 0).
From part (b):
T  sin 2 s,  sin s cos s, cos s

dT
 2sin s cos s,  cos 2 s  sin 2 s,  sin s  sin 2 s,  cos 2 s,  sin s
ds
  s 
  
1


1
1  sin 2 s
dT

ds
1  sin 2 s
Fundamentals – Curvature
ENGI 2422
Another formula for curvature:
Let r 
dr
dt
and s 
ds
dr

 r , then
dt
dt
r  sT
 r  sT  s
But
dT
dt
 
dT
d T ds


 N s
dt
ds dt
 

 r r  s T  s T  s 2 N  0  s 3B

r  r   s 3 , but
s  r .
Therefore
 
rr
r
3
Page 1-18
Fundamentals – Curvature
ENGI 2422
Page 1-19
Example 1.3.3
Find the curvature and the radius of curvature for the curve, given in parametric form by
x = cos t , y = sin t , z = t . Assume SI units.
Let c = cos t ,
s = sin t .
r  c, s , t
 v 
a 
r   s , c, 1
r   c,  s , 0
ˆi
ˆj kˆ
r  r  s c 1
c  s 0

r 
s2  c2  1 
s2  c2  1 
  

rr
r
3

 s ,  c, s 2  c 2

 s ,  c, 1
2
2
2
1

2
2 2
Therefore
 
1 m1
2
 constant 
and
 
1

 2m
ENGI 2422
1.4
Fundamentals – Conic Sections
Page 1-20
Conic Sections
All members of the family of curves known as conic sections can be generated, (as the
name implies), from the intersections of a plane and a double cone. The Cartesian
equation of any conic section is a second order polynomial in x and y. The only cases
that we shall consider in this course are such that any axis of symmetry is lined up along
a coordinate axis. For all such cases, the Cartesian equation is of the form
A x2 + C y2 + D x + E y + F = 0
where A, C, D, E and F are constants. There is no “xy” term, so B = 0.
The slope of the intersecting plane is related to the eccentricity, e of the conic section.
x2 + y2 = r2
A parametric form is
(x, y) = (r cos , r sin  ), (0 <  < 2).
x2
y2

 1
a2
b2
where b2  a 2 1  e 2 
The circle is clearly a special case of the ellipse, with e = 0 and b = a = r.
The longest diameter is the major axis (2a). The shortest diameter is the minor axis (2b).
If a mirror is made in the shape of an ellipse, then all rays emerging from one focus will,
after reflection, converge on the other focus.
A parameterization for the ellipse is r    a cos ˆi  b sin  ˆj ,
 0    2  .
ENGI 2422
Fundamentals – Conic Sections
Page 1-21
y2 = 4ax
One vertex is at the origin.
The centre and the other vertex and focus are at infinity.
If a mirror is made in the shape of a parabola, then all rays emerging from the focus will,
after reflection, travel in parallel straight lines to infinity (where the other focus is). The
primary mirrors of most telescopes follow a paraboloid shape.
x2
y2

 1
a2
b2
where b2  a 2  e 2  1
The hyperbola has two separate branches.
As the curve retreats towards infinity, the curve approaches the asymptotes
x2
y2
bx 


 0 ,  y    .
2
2
a
b
a 

The distance between the two vertices is the major axis (2a).
If a mirror is made in the shape of an hyperbola, then all rays emerging from one focus
will, after reflection, appear to be diverging from the other focus.
Circles and ellipses are closed curves. Parabolas and hyperbolas are open curves.
Fundamentals – Conic Sections
ENGI 2422
Page 1-22
A special case of the hyperbola occurs when the eccentricity
is e  2 and it is rotated 45 from the standard
orientation. The asymptotes line up with the coordinate
axes, the graph lies entirely in the first and third quadrants
and the Cartesian equation is xy = k.
This is the rectangular hyperbola.
Degenerate conic sections arise when the intersecting plane passes through the apex of
the cone. Two cases are:
0  e  1:
x2
y2

 0
a2
b2
point at the origin.
e  1:
x2
y2

 0
a2
b2
line pair through the origin.
Another degenerate case is
x2
y2

 1
a2
b2
nothing !
Example 1.4.1
Classify the conic section whose Cartesian equation is 3y2 = x2 + 3 .
Rearranging into standard form,
y2
x2
3 y2  x2  3


 1
1
3
Compare with
X2
Y2

 1: X  y , Y  x , a  1 , b  3
a2
b2
Therefore this is an hyperbola,
rotated through 90°,
with vertices at (0, ±1) and
3
x.
asymptotes x   3 y or y  
3
[The asymptotes make angles of 30° with the x axis.]
ENGI 2422
Fundamentals – Conic Sections
Page 1-23
Example 1.4.2
Classify the conic section whose Cartesian equation is 21x2 + 28y2 = 168x + 168y .
Rearranging into standard form, first complete the square.
21x2 + 28y2  168x  168y = 0

21(x2  8x) + 28(y2  6y) = 0

21((x  4)2  16) + 28((y  3)2  9) = 0

21(x  4)2 + 28(y  3)2 = 588
2
2
x  4
y  3




 1
28
21
Compare this with the standard form
X2
Y2

 1:  X  x  4 , Y  y  3 , a  28 , b  21
a2
b2
The conic section is therefore an ellipse,
semi major axis 28 , semi minor axis 21 ,
(from which the eccentricity can be found to be exactly e = 1/2),
centre (4, 3).
It happens to pass exactly through the origin.
Moving up to three dimensions, we have the family of quadric surfaces.
Fundamentals – Quadric Surfaces
ENGI 2422
1.5
Page 1-24
Classification of Quadric Surfaces
Again, we shall consider only the simplest cases, where any planes of symmetry are
located on the Cartesian coordinate planes. In nearly all cases, this eliminates “crossproduct terms”, such as xy, from the Cartesian equation of a surface. Except for the
paraboloids, the Cartesian equations involve only x2, y2, z2 and constants.
The five main types of quadric surface are:
The ellipsoid (axis lengths a, b, c)
x2
a2

y2
b2

z2
c2
 1
The axis intercepts are at
(±a, 0, 0), (0, ±b, 0) and (0, 0, ±c).
All three coordinate planes are
planes of symmetry.
The cross-sections in the three
coordinate planes are all ellipses.
Special cases:
a = b > c : oblate spheroid (a “squashed sphere”)
a = b < c : prolate spheroid (a “stretched sphere” or cigar shape)
a = b = c : sphere
Hyperboloid of One Sheet (Ellipse axis lengths a , b ; aligned along the z axis)
x2
a2

y2
b2

z2
c2
 1
For hyperboloids, the central axis is
associated with the “odd sign out”.
In the case illustrated, the hyperboloid is
aligned along the z axis.
The axis intercepts are at
(±a, 0, 0) and (0, ±b, 0).
The vertical cross sections in the x-z and
y-z planes are hyperbolae.
All horizontal cross sections are ellipses.
Fundamentals – Quadric Surfaces
ENGI 2422
Page 1-25
Hyperboloid of Two Sheets (Ellipse axis lengths b , c ; aligned along the x axis)
x2
a2

y2
b2

z2
c2
 1
For hyperboloids, the
central axis is associated
with the “odd sign out”.
In the case illustrated, the
hyperboloid is aligned
along the x axis.
The axis intercepts are at
(±a, 0, 0) only.
Vertical cross sections
parallel to the y-z plane are either ellipses or null.
All cross sections containing the x axis are hyperbolae.
Elliptic Paraboloid
(Ellipse axis lengths a , b ;
aligned along the z axis)
z x2 y 2


c a 2 b2
For paraboloids, the central axis is
associated with the “odd exponent out”.
In the case illustrated, the paraboloid is
aligned along the z axis.
The only axis intercept is at the origin.
The vertical cross sections in the x-z and y-z
planes are parabolae.
All horizontal cross sections are ellipses (for
z > 0).
Fundamentals – Quadric Surfaces
ENGI 2422
Hyperbolic Paraboloid
Page 1-26
(Hyperbola axis length a ; aligned along the z axis)
z x2 y 2


c a 2 b2
For paraboloids, the central
axis is associated with the
“odd exponent out”.
In the case illustrated, the
paraboloid is aligned along
the z axis.
The only axis intercept is at
the origin.
The vertical cross section in the x-z plane is an upward-opening parabola.
The vertical cross section in the y-z plane is a downward-opening parabola.
All horizontal cross sections are hyperbolae, (except for a point at z = 0).
The plots of the five standard quadric surfaces shown here were generated in the software
package Maple. The Maple worksheet is available from a link at
"http://www.engr.mun.ca/~ggeorge/2422/programs/index.html".
Degenerate Cases:
x2 y 2 z 2


0
a 2 b2 c2
:
A single POINT at the origin.
x2 y 2 z 2


0
a 2 b2 c2
:
Elliptic CONE, aligned along the z axis;
[asymptote to both types of hyperboloid].
x2 y 2 z 2


 1
a 2 b2 c2
:
NOTHING
x2 y2

1
a 2 b2
:
ELLIPTIC CYLINDER, aligned along the z axis.
x2 y 2

1
a 2 b2
:
HYPERBOLIC CYLINDER, aligned along the z axis.
y x2

b a2
:
PARABOLIC CYLINDER, vertex line on the z axis.
Fundamentals – Quadric Surfaces
ENGI 2422
x2 y 2

0
a 2 b2
:
LINE (the z axis)
x2 y 2

 1
a 2 b2
:
NOTHING
x2 y 2

0
a 2 b2
:
PLANE PAIR (intersecting along the z axis)
x2
1
a2
:
Parallel PLANE PAIR
x2
0
a2
:
Single PLANE (the y-z coordinate plane)
x2
 1
a2
:
NOTHING
Example 1.5.1
Classify the quadric surface, whose Cartesian equation is 2x = 3y2 + 4z2 .
Rearranging into standard form,
x
y2
z2


6
4
3
Compare to the standard form
Z
X2
Y2
 2  2 : Z  x, X  y, Y  z, a 2, b 3, c6
c
a
b
The quadric surface is therefore an elliptic
paraboloid, aligned along the x axis, with its vertex at
the origin.
Page 1-27
ENGI 2422
Fundamentals – Quadric Surfaces
Page 1-28
Example 1.5.2
Classify the quadric surface, whose Cartesian equation is z2 = 1 + x2 .
Rearranging into standard form,
z2  x2 = 1
Compare to the standard form
X2
Y2
 2  1: X  z , Y  x , a  b  1
a2
b
The quadric surface is therefore one of the
degenerate cases.
It is a hyperbolic cylinder, centre at the origin,
aligned along the y axis.
Example 1.5.3
Classify the quadric surface, whose Cartesian equation is x2  y2 + z2 + 1 = 0 .
Rearranging into standard form,
y2  x2  z2 = 1
Compare to the standard form
X2
Y2
Z2


 1: X  y , Y  x , Z  z , a  b  c  1
a2
b2
c2
The quadric surface is therefore an
hyperboloid of two sheets,
centre at the origin,
aligned along the y axis.
ENGI 2422
1.6
Fundamentals – Surfaces of Revolution
Page 1-29
Surfaces of Revolution
Consider a curve in the x-y plane, defined by the equation y = f (x).
If it is swept once around the line y = c , then it will generate a surface of revolution.
At any particular value of x, a thin cross-section through that surface, parallel to the y-z
plane, will be a circular disc of radius r, where
r 
f  x  c
Let us now view the circular disc face-on, (so that the x axis and the axis of rotation are
both pointing directly out of the page and the page is parallel to the y-z plane).
Let (x, y, z) be a general point on the surface of
revolution.
From this diagram, one can see that
r2 = (y  c)2 + z2
Therefore, the equation of the surface generated,
when the curve y = f (x) is rotated once around
the axis y = c , is
 y  c
2
 z2 
 f  x  c
2
Special case: When the curve y = f (x) is rotated once around the x axis, the equation of
the surface of revolution is
y2  z2 
 f  x 
2
or
y2  z2  f  x 
Fundamentals – Surfaces of Revolution
ENGI 2422
Page 1-30
Example 1.6.1
Find the equation of the surface generated, when the parabola y 2 = 4ax is rotated once
around the x axis.
The solution is immediate:
y2 + z2 = 4ax ,
which is the equation of an elliptic paraboloid,
(actually a special case, a circular paraboloid).
The Curved Surface Area of a Surface of Revolution
For a rotation around the x axis,
the curved surface area swept out by the element of arc
length s is approximately the product of the
circumference of a circle of radius y with the length s.
A  2 y s
Integrating along a section of the curve y = f (x) from
x = a to x = b, the total curved surface area is
A  2 
x b
x a
f  x  ds
For a rotation of y = f (x) about the axis y = c, the curved surface area is
A  2 
x b
x a
f  x   c ds  2 
b
a
2
ds
 ds 
 dy 
f x  c
dx and    1   
dx
 dx 
 dx 
Therefore
A  2

b
a
f  x  c
1   f   x   dx
2
2
Fundamentals – Surfaces of Revolution
ENGI 2422
Page 1-31
Example 1.6.2
Find the curved surface area of the circular paraboloid generated by rotating the portion
of the parabola y 2 = 4cx (c > 0) from x = a ( > 0) to x = b about the x axis.
A  2 
x b
x a
y 2  4cx
 2 y y   4c
 dy 
1  
 dx 

ds

dx

A  2  2 cx
b
a
2


y 
4c 2
1 2 
y
y ds
2c
y
4c 2
1

4cx
xc
x
b
xc
1/ 2
dx  4 c   x  c  dx
a
x
b
  x  c 3/ 2 
 4 c 

3

 a
2
Therefore
A 
8 c
3
 b  c 
3/ 2
 a  c
3/ 2

ENGI 2422
1.7
Fundamentals – Hyperbolic Functions
Page 1-32
Hyperbolic Functions
When a uniform inelastic (unstretchable) perfectly flexible cable is suspended between
two fixed points, it will hang, under its own weight, in the shape of a catenary curve.
The equation of the standard catenary curve is most concisely expressed as the hyperbolic
cosine function, y = cosh x, where
e x  e x
2
The solutions to some differential equations can be expressed conveniently in terms of
hyperbolic functions.
cosh x 
The other five hyperbolic functions are
e x  e x
sinh x 
,
2
sinh x
e x  e x
tanh x 
 x
,
cosh x
e  e x
coth x 
1
e x  e x
 x
tanh x
e  e x
sech x 
1
2
 x
,
cosh x
e  e x
and csch x 
1
2
 x
.
sinh x
e  e x
Unlike the trigonometric functions, the hyperbolic functions are not periodic.
However, parity is preserved:
Of the six trigonometric function, only cos  and sec  are even functions.
Of the six hyperbolic functions, only cosh  and sech  are even functions.
The other functions are odd.
Fundamentals – Hyperbolic Functions
ENGI 2422
Page 1-33
There is a close relationship between the hyperbolic and trigonometric functions.
From the Euler form for e j , e j = cos  + j sin  ,

e –j = cos  – j sin 
e j  e  j
1
sin  
 sinh  j    j sinh  j 
2j
j
cos 
e j  e  j
 cosh  j 
2
 tan  
sinh  j 
sin 
1

 tanh  j    j tanh  j 
cos
j cosh  j 
j
Identities:
Let x = j :
sin  + cos 
2
2
≡ 1

 sinh 2 x 
2
 j 2   cosh x  1 


cosh2x  sinh2x ≡ 1
1 + tan2 ≡ sec2
 tanh 2 x 
 1 
1  
 
 

2
2
 cosh x 
 j

1  tanh2x ≡ sech2x

etc.
Fundamentals – Hyperbolic Functions
ENGI 2422
Page 1-34
Derivatives
d
 sinh x  
dx
d  e x  e x 
e x  e x

dx  2 
2
Therefore
d
 sinh x   cosh x
dx
d
d  e x  e x 
e x  e x

 cosh x  
dx
dx  2 
2
Therefore
d
 cosh x    sinh x
dx
[Note the different sign from the trigonometric version, (cos x) ' =  sin x .]
d
 cosh x  cosh x   sinh x sinh x   1
d  sinh x 

 tanh x  


dx
dx  cosh x 
cosh 2 x
cosh 2 x
Therefore
d
 tanh x   sech 2 x
dx
OR
Let x = j , then  = jx
tanh(j) = j tan 


tanh x = j tan(jx)
d
 tanh x   j  sec2   j x      j    sec2   j x   sec2 
dx
= sech2(j) = sech2x .
Therefore
d
 tanh x   sech 2 x
dx
Fundamentals – Hyperbolic Functions
ENGI 2422
d
 csch x  
dx
 

Page 1-35

d
1
2
 sinh x     sinh x   cosh x 
dx
1
cosh x

sinh x sinh x
Therefore
d
 csch x    csch x coth x
dx
etc.
A list of identities and derivatives for hyperbolic functions is presented in the suggestions
for a formula sheet, in Appendix A to these lecture notes.
Fundamentals – Integration by Parts
ENGI 2422
1.8
Page 1-36
Integration by Parts
Review:
Let u(x) and v(x) be functions of x. Then, by the product rule of differentiation,
d
du
dv
 uv    v  u 
dx
dx
dx
Integrating with respect to x :
uv 

 uv dx

 uv dx
This leads to the formula for integration by parts:
 uv  
 uv dx
 uv dx
Example 1.8.1
Find
I 
3 x
 x e dx .
2
[v' must be identified with a factor of the integrand that can be antidifferentiated easily.]
v   2 x e  x
u   12 x 2

v  e x
u   x
2
2
 I  uv    uv dx    12 x 2e x     x e x dx


2
 e x   12 x 2 
2
Therefore
1
2
2
C
I   12  x 2  1 e  x  C
2
Fundamentals – Integration by Parts
ENGI 2422
Example 1.8.2
Find

I 
x
[Repeated use of integration by parts]
2
cos x dx .
u = x2
v' = cos x
u' = 2x
v = sin x
 I  x 2 sin x 
 2ux  sinv x dx
Using integration by parts again, with u = 2x and v' = sin x ,
v =  cos x
u' = 2

 I  x 2 sin x  2 x cos x 
 2 cos x dx 
= x2sin x + 2x cos x  2 sin x + C
Therefore
I 
x
2
cos x dx   x 2  2  sin x  2 x cos x  C
Check:
I' = 2x sin x + (x2  2) cos x + 2 cos x  2x sin x + 0
= x2 cos x 
Page 1-37
Fundamentals – Integration by Parts
ENGI 2422
Page 1-38
Shortcut (a tabular form for repeated integrations by parts):
I 
x
2
cos x dx :
Reading off the diagonals,
I = x2sin x + 2x cos x  2 sin x + C
Note:
There are three ways the table can end:
Example 1.8.3
Find
I 
x e
4
x
1)
column 'D' reduces to 0 (as in the
examples on this page);
2)
the product across a row is easy
to integrate;
3)
the product across a row is a
constant multiple of the original
integrand. (ex. 1.8.4 next page)
dx .
Therefore
I = (x4  4x3 + 12x2  24x + 24) ex + C
Check:
I' = (x4  4x3 + 12x2  24x + 24
+ 4x3  12x2 + 24x  24 + 0) ex
= x4 ex

Fundamentals – Integration by Parts
ENGI 2422
Example 1.8.4
Find
I 
e
Page 1-39
(recursive use of integration by parts)
ax
sin bx dx .
Either
or
cos bx
sin bx  a 2

I   eax
 aeax 2   2 I
b
b  b


eax
eax  b2
I    sin bx
 b cos bx 2   2 I
a
a  a

 a2 
 1  2  I 
 b 
ea x
a sin bx  b cos bx 
b2
 a 2  b2  I  ea x a sin bx  b cos bx 
 b2 
 1  2  I 
 a 
ea x
a sin bx  b cos bx 
a2
 a 2  b2  I  ea x a sin bx  b cos bx 
Therefore
eax
 e sin bx dx  a 2  b2  a sin bx  b cos bx   C
ax
Check:
Let s = sin bx and c = cos bx , then s' = bc , c' = bs and
eax
I  2
 as  bc   C
a  b2
dI
eax

 2
a  as  bc    a  bc   b  bs  
dx
a  b2
eax
eax
2
2
 2
a
s

abc

abc

b
s

a 2  b2  s  eax sin bx

2 
2
2 
a b
a b



Fundamentals – Leibnitz Differentiation
ENGI 2422
1.9
Page 1-40
Leibnitz Differentiation of a Definite Integral

d
dx
y  g x
y  f  x

H  x, y  dy  H  x, g  x  
dg
df
 H  x, f  x  

dx
dx

y  g x
y  f x
H
dy
x
If the limits of integration are both constant, then just differentiate the integrand with
respect to x, treating all other terms as constants.
d
dx

b
a

f  x, t  dt 

b
a
f
dt
x
Example 1.9.1
dI
, where I  t  
dt
Evaluate
2t
t
zt dz .
Using Leibnitz differentiation:
dI

dt
  2t  t   dtd  2t 

  t  t   dtd  t 
z  2t
 z2 
 4t  t   
 2  z t
2
2
 3t 2 

 z 1 dz
2t
t
3t 2
9t 2

2
2
Directly:
I t  
2t

2t
t
 z2 
zt dz  t  
 2 t
 4t 2  t 2 
3t 3
 t


2
 2 
dI
d  3t 3 
9t 2



dt
dt  2 
2
See Problem Set 3 and Section 5.10 for more practical examples of Leibnitz
differentiation.
End of Chapter 1