Vector-valued function, Lecture no 2, Calculus II

Date: 14.02.2011

Describing the graph of vector-valued function

Let r(t) be a vector valued fuction which  can be represented as the resultant vector of two vector called r_o and v
so,

r(t) = r_{o + tv...........................................(1)}

_{equation (1) represents, r(t) passes through the point of position vector} r_{o and as the direction or parallel to v.}

_{Example 1: a problem has taken from the book of  Haward Anton  page no: 868}

_{problem no: 13}

_{Describe the graph of the vector valued function given as, r(t)= (3-2t) i + 5j}

_{Solution: Given r(t)=}(3-2t) i + 5j

                                       _{= 3 i + 0. j + ( -2 i + 5 j ) t}

_{According to the above theory we can say that the graph of r is a straight line in two dimensional space passing through the point (3,0)  and parallel to the line -2 i + 5 j. we can show this in figure bellow-}

_{Example 2: Describe the graph of} 

_{r(t)= 3cos t i +2 sin t j - k}

_{Solution: we get x=3cos t,   y= 2sin t  and z= -1}

 _{the relation between x and y} 

 _{the graph of r(t) is an ellipse in the plane z=-1 at the center(0,0,-1)}

_{major axis length is 6, parallel to X axis and minor axis length is 4, parallel to Y axis.}

Example 3: Describe the graph of 

                r(t)= 2ti-3j+(1+3t)k

Solution:  Given,

                r(t)= 2ti-3j+(1+3t)k

                    = 2ti-3j+k+3tk

                    = 0.i-3j+k+(2i+0.j+3k)t

So the graph of r(t) is a straight line in three dimensional space passing through the point (0,-3,1) and as the direction or parallel to the line 2i+0.j+3k.

Example 4: Describe the graph of 

              r(t)= 2cos t i -3 sin t j + k

Solution: Given,

                        r(t)= 2cos t i -3 sin t j + k

we get, x=2cos t,  y= -3sin t  and z=1

the relation between x and y 

the graph of r(t) is an ellipse in the plane z=1, the center is (0,0,1) .

Graph Sketching 

Norm of a vector valued function : Norm of a  vector valued function r(t) is denoted by ІІr(t)ІІ and is defined by 

ІІr(t)ІІ= Sqrt ((x(t))²+ (y(t))²+ (z(t))²)

 Example 1: Show that the graph of  r(t) is a circle where 

            r(t)= sin t i + 2 cos t j +sin t k

Solution: we get  x= sin t,   y=2 cos t        z= sin t

So the  relation,

x²+ y²+ z²=sin²t+4cos²t+3sin²t

                =4(sin²t+cos²t)=4

So we get Z= x   as      z= sin t
Thus the graph of r(t) lies on the sphere  x²+ y²+ z²= 4 and on the plane  Z= x of which center is (0,0,0) and radius is  2. the figure as bellow

 

(rest of the part of lecture 2 will be published soon)