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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 ro and v
so,
So the relation,
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)
Describing the graph of vector-valued function
so,
r(t) = ro + tv...........................................(1)
equation (1) represents, r(t) passes through the point of position vector ro 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
sin t kSolution: we get x= sin t, y=2 cos t z= sin t
sin tx2+ y2+ z2=sin2t+4cos2t+3sin2t
=4(sin2t+cos2t)=4
So we get Z=
x as z= sin tThus 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)
