ch.5 - Chapter 5 Vector Functions for 0 r < 0 and 0 < 2...

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5.1. Vector-valued functions In this chapter we study vector-valued functions of n variables : f : R n -→ R m . Such a function assigns to each x = ( x 1 ,...,x n ) T R n a unique vector y = ( y 1 ,...,y m ) T R m . A vector-valued function can be studied by looking at the m real-valued functions f 1 ,f 2 ,...,f m where ( y 1 ,...,y m ) T = f ( x 1 ,...,x n ) = ( f 1 ( x 1 ,...,x n ) ,f 2 ( x 1 ,...,x n ) ,...,f m ( x 1 ,...,x n )) T . The m real-valued functions f 1 ,f 2 ,...,f m are called the component functions of f . Example 5.1 Let f : R -→ R 2 be defined by f ( t ) = (cos t, sin t ) T . Thus f assigns a real number t to a vector (cos t, sin t ) T R 2 . The component functions are f 1 ( t ) = cos t and f 2 ( t ) = sin t . In this case the graph of f is a unit circle. 2 Example 5.2 Let g : R -→ R 3 be defined by g ( t ) = (2 cos t, 2 sin t,t ) T . The component functions are g 1 ( t ) = 2 cos t , g 2 ( t ) = 2 sin t and g 3 ( t ) = t . Here is a sketch of this curve: Example 5.3 The equations f 1 ( r,θ,φ ) = r sin θ cos φ f 2 ( r,θ,φ ) = r sin θ sin φ f 3 ( r,θ,φ ) = r cos θ for 0 r < , 0 θ π and 0 φ < 2 π , define a vector-valued function f ( r,θ,φ ) whose component functions are f 1 , f 2 and f 3 . Observe that the function f relates the spherical polar coordinates and the Cartesian coordinates. 2 Example 5.4 Let A = ( a ij ) M m × n ( R ) . Then the function f : R n -→ R m defined by f ( x 1 ,...,x n ) = f ( x ) = A x where x = ( x 1 ,...,x n ) T R n is a vector-valued function with component functions f i ( x 1 ,...,x n ) = n X j =1 a ij x j for i = 1 ,...,m . 2 Definition 5.5 Let f : R n -→ R m be a vector-valued functions of n variables. The m × n matrix J f = ± ∂f i ∂x j ² = ∂f 1 ∂x 1 ∂f 1 ∂x 2 ··· ∂f 1 ∂x n ∂f 2 ∂x 1 ∂f 2 ∂x 2 ··· ∂f 2 ∂x n . . . . . . . . . . . . ∂f m ∂x 1 ∂f m ∂x 2 ··· ∂f m ∂x n is called the Jacobian matrix of the function f . The ij -th entry of J f is the partial derivative of the component function f i with respect to x j . Note that the i -th row of the Jacobian matrix J f is the transpose of the gradient vector f i of the i -th component function f i . In the special case when f is real-valued, i.e. m = 1 , we have J f = ( f ) T . Example 5.6 Let f : R 2 R 3 be defined by f ( x,y ) = ( x 2 y, 2 xy, 3 y ) T . Compute J f (2 , 1) . Solution . By definition, J f ( x,y ) = ∂f 1 ∂x ∂f 1 ∂y ∂f 2 ∂x ∂f 2 ∂y ∂f 3 ∂x ∂f 3 ∂y = 2 xy x 2 2 y 2 x 0 3 . At the point
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This note was uploaded on 12/20/2011 for the course MATH 1813 taught by Professor Wu during the Fall '11 term at HKU.

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ch.5 - Chapter 5 Vector Functions for 0 r < 0 and 0 < 2...

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