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: 25%, 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.
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
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 ||
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