Math 304
Multivariable Calculus & Differential Equations
Wafik Lotfallah
Course Material
T1) Thomas' Calculus, 11th edition, by M.D. Weir, J. Hass,
and F.R. Giordano, Pearson Add, 2005
T2) Calculus, 5th edition, by J. Stewart, Brooks/Cole, 2003
T3) Advanced Engineering Mathematics, 8th edition, by E.
Kreyszig, J. Wiley & Sons, 1999
T4) Shaum’s Outline of Advanced Mathematics for
Engineers and Scientists, 1st edition, by M.R. Spiegel,
McGraw-Hill, 1971
T5) Supplementary Material on the inter/intranet.
Course website: http://math.guc.edu.eg/math304/
Course Assessment System
 Homework + Classwork: 10%
 Quizzes: 25%, both counted (NO make-ups)
 Midterm: (35%)
 Final: (45%, comprehensive)
Prerequisites: Math 103, Math 203.
However, apart from being able to differentiate
and integrate, we will review anything you
need to know for the course.
Lecture 1
Revision on Vectors
 Find and interpret
 the sum of two vectors
 the product of a vector by a scalar
 the dot product of two vectors
 the magnitude of a vector
 the unit vector in a specified direction
 the angle between two vectors
 the cross product of two vectors
 the work done by a constant force in moving objects
along a straight line segment
 Find and graph equations of
 spheres
 Planes
 Straight lines
Lecture Objectives
The 3 Dimensional Space
The 3 Coordinate Planes
Points in Space
Distances
Spheres
Vectors
Equality of
Vectors
Vectors in
Standard
Position
Notation:
v = v1, v2, v3
= x2 x1, y2 y1, z2 z1
Length of
Vectors
Subtraction:
An Alternative Representation:
Letting i = 1, 0, 0; j = 0, 1, 0; k = 0, 0, 1;
we can write:
v = x, y, z = x 1, 0, 0 + y 0, 1, 0 + z 0, 0, 1
= x i + y j + z k
Example: 1, 2, 3 = i + 2 j + 3 k
2, 1, 0 = 2 i  j
Properties of the Norm: Let u and v be vectors and c a
scalar. Then:
 || v || = 0 iff v = 0
 || cv || = |c| || v ||
Vector Direction:
Example: Find a vector of length 5 in the
same direction of the vector v = 2, 1, 2.
The Dot Product
Equivalent Definition:
uv = |u| |v| cos
Orthogonal (Perpendicular)
Vectors:
Letting  = /2 in uv = |u||v|cos():
Example: Show that 1, 2, 3 and 2, 1, 0
are orthogonal.
Work
The Cross Product
An Alternative Definition:
Also note:
6. u  u = 0
Equations of Planes
Example: Find the equation of the plane through the
origin (0, 0, 0) and perpendicular to 2, 0, 5.
Solution: 2(x  0) + 0(y  0) + (5)(z  0) = 0
2x  5z = 0
Example: Find the equation of the plane through
the points P(1, 0, 0); Q(0, 2, 0); R(0, 0, 3).
Solution 1: First find a normal vector by taking
the cross product:
Then use the plane formula:
6(x  1) + 3(y  0) + 2(z  0) = 0
or 6x + 3y + 2z = 6
k
j
i
k
j
i
2
3
6
3
0
1
0
2
1
3
,
0
,
1
0
,
2
,
1 









PR
PQ
Example: Find the equation of the plane through
the points P(1, 0, 0); Q(0, 2, 0); R(0, 0, 3).
Solution 2: Start with Ax + By + Cz = D, and
substitute the points
P(1, 0, 0); Q(0, 2, 0); R(0, 0, 3) to get:
A = D; 2B = D; 3C = D
Letting D = 1, we get A = 1, B = 1/2, C = 1/3, so
the equation becomes: x + (1/2)y + (1/3)z = 1,
or 6x + 3y + 2z = 6
Equations of Straight Lines
Symmetric Equations for a Line: Eliminate the parameter t
3
0
2
0
1
0
v
z
z
v
y
y
v
x
x
t






Thank you for listening.
Wafik

Math Lecture 11 For engineering students

  • 2.
    Math 304 Multivariable Calculus& Differential Equations Wafik Lotfallah
  • 3.
    Course Material T1) Thomas'Calculus, 11th edition, by M.D. Weir, J. Hass, and F.R. Giordano, Pearson Add, 2005 T2) Calculus, 5th edition, by J. Stewart, Brooks/Cole, 2003 T3) Advanced Engineering Mathematics, 8th edition, by E. Kreyszig, J. Wiley & Sons, 1999 T4) Shaum’s Outline of Advanced Mathematics for Engineers and Scientists, 1st edition, by M.R. Spiegel, McGraw-Hill, 1971 T5) Supplementary Material on the inter/intranet. Course website: http://math.guc.edu.eg/math304/
  • 4.
    Course Assessment System Homework + Classwork: 10%  Quizzes: 25%, both counted (NO make-ups)  Midterm: (35%)  Final: (45%, comprehensive)
  • 5.
    Prerequisites: Math 103,Math 203. However, apart from being able to differentiate and integrate, we will review anything you need to know for the course.
  • 6.
  • 7.
     Find andinterpret  the sum of two vectors  the product of a vector by a scalar  the dot product of two vectors  the magnitude of a vector  the unit vector in a specified direction  the angle between two vectors  the cross product of two vectors  the work done by a constant force in moving objects along a straight line segment  Find and graph equations of  spheres  Planes  Straight lines Lecture Objectives
  • 8.
  • 9.
  • 10.
  • 11.
  • 12.
  • 13.
  • 14.
  • 15.
    Vectors in Standard Position Notation: v =v1, v2, v3 = x2 x1, y2 y1, z2 z1
  • 16.
  • 19.
  • 21.
    An Alternative Representation: Lettingi = 1, 0, 0; j = 0, 1, 0; k = 0, 0, 1; we can write: v = x, y, z = x 1, 0, 0 + y 0, 1, 0 + z 0, 0, 1 = x i + y j + z k Example: 1, 2, 3 = i + 2 j + 3 k 2, 1, 0 = 2 i  j
  • 22.
    Properties of theNorm: Let u and v be vectors and c a scalar. Then:  || v || = 0 iff v = 0  || cv || = |c| || v ||
  • 23.
    Vector Direction: Example: Finda vector of length 5 in the same direction of the vector v = 2, 1, 2.
  • 24.
  • 25.
  • 26.
    Orthogonal (Perpendicular) Vectors: Letting = /2 in uv = |u||v|cos(): Example: Show that 1, 2, 3 and 2, 1, 0 are orthogonal.
  • 27.
  • 28.
  • 29.
  • 31.
    Also note: 6. u u = 0
  • 32.
  • 34.
    Example: Find theequation of the plane through the origin (0, 0, 0) and perpendicular to 2, 0, 5. Solution: 2(x  0) + 0(y  0) + (5)(z  0) = 0 2x  5z = 0
  • 35.
    Example: Find theequation of the plane through the points P(1, 0, 0); Q(0, 2, 0); R(0, 0, 3). Solution 1: First find a normal vector by taking the cross product: Then use the plane formula: 6(x  1) + 3(y  0) + 2(z  0) = 0 or 6x + 3y + 2z = 6 k j i k j i 2 3 6 3 0 1 0 2 1 3 , 0 , 1 0 , 2 , 1           PR PQ
  • 36.
    Example: Find theequation of the plane through the points P(1, 0, 0); Q(0, 2, 0); R(0, 0, 3). Solution 2: Start with Ax + By + Cz = D, and substitute the points P(1, 0, 0); Q(0, 2, 0); R(0, 0, 3) to get: A = D; 2B = D; 3C = D Letting D = 1, we get A = 1, B = 1/2, C = 1/3, so the equation becomes: x + (1/2)y + (1/3)z = 1, or 6x + 3y + 2z = 6
  • 37.
  • 39.
    Symmetric Equations fora Line: Eliminate the parameter t 3 0 2 0 1 0 v z z v y y v x x t      
  • 42.
    Thank you forlistening. Wafik