Slide_01 - MA1104 Multivariable Calculus Lecture 1 Dr KU...

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Unformatted text preview: MA1104 Multivariable Calculus Lecture 1 Dr KU Cheng Yeaw Monday Jan 12, 2008 Overview In this lecture, • we introduce coordinate systems for three-dimensional space. This provides the setting for our study of calculus of functions of two variables. • we introduce vectors geometrically and study their algebraic properties. We emphasize the power of algebraic manipulation of vectors. 3-D Coordinate Systems We can represent a point on a plane by an ordered pair ( a,b ) of real numbers, where a is x-coordinate and b is the y-coordinate. To locate a point in space R 3 , we need three numbers. First, we need to set up a coordinate system as follows: • A fixed point O (the origin) • Three directed lines through O that are perpendicular to each other. They are labeled as x-axis, y-axis and z-axis respectively. This is how we always draw the axes. The direction of the z-axis is determined by the right-hand rule Any two of the axes determine a plane. Example 1. Describe and sketch the surface in R 3 represented by the equation x + y = 2 . Solution. x + y = 2 represents a line on the xy-plane. However, in R 3 , it represents the plane containing all points whose x- and y-coordinate sum to 2 . This is a vertical plane. Distance Formula in 3-D The distance | P 1 P 2 | between the points P 1 ( x 1 ,y 1 ,z 1 ) and P 2 ( x 2 ,y 2 ,z 2 ) is | P 1 P 2 | = p ( x 2- x 1 ) 2 + ( y 2- y 1 ) 2 + ( z 2- z 1 ) 2 . Construct a rectangle box as follows: Note that | P 1 P 2 | 2 = | P 1 B | 2 + | BP 2 | 2 . Clearly, | BP 2 | = | z 2- z 1 | . By the Pythagorean Theorem, | P 1 B | 2 = | P 1 A | 2 + | AB | 2 , = | x 2- x 1 | 2 + | y 2- y 1 | 2 . Combining these equations, | P 1 P 2 | 2 = | x 2- x 1 | 2 + | y 2- y 1 | 2 + | z 2- z 1 | 2 , as desired. Consequently, we have the following equation of a sphere: Equation of a sphere An equation of a sphere with center C ( h,k,l ) and radius r is ( x- h ) 2 + ( y- k ) 2 + ( z- l ) 2 = r 2 . Vectors Imagine a particle moving in space. Knowing the position of this particle is unsatisfactory as this only gives a static picture of the particle. We need a way to tell us the direction of this moving particle. This can be done by using a vector . A vector is often represented by an arrow. • The length of the arrow represents the magnitude of the vector. • The arrow points in the direction of the vector. For instance, suppose a particle moves along a line segment from point A to point B . The vector v has initial point A (the tail) and terminal point B (the tip). We indicate this by writing v =--→ AB . Call this the displacement vector of a particle from A to B . We denote a vector by either: • Printing a letter in boldface v , or • Putting an arrow above the letter ~v ....
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Slide_01 - MA1104 Multivariable Calculus Lecture 1 Dr KU...

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