Transcript Green Circles - Himmadika UNS

Calculus III
Second Lecture Notes
By
Rubono Setiawan, S.Si.,M.Sc.
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Contents of this presentation
• 1.2. Equations of Lines and Curve
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Lines
• A line in the xy-plane is determined when a point on the line
and the direction of the line ( its slope or angle of inclination)
are given. The equation of the line can then be written using
the point-slope form.
• Likewise, a line L in three- dimensional space is determined
when we know a point Po(X0,Y0,Z0) on L and the direction of
L. In three dimensional space the direction of line is
conveniently described by a vector, so we let v be a vector
parallel to L. Let P(x,y,z) be an arbitrary point on L and let r0
and r be the position vectors of P0 and P
( that is, they have representations OP0 and OP )
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• If a is a vector with representation P0 P then the Triangle Law
for vector addition gives r = r0 + a. But, since a and v are
parallel vectors, there is a scalar t such that a = tv. Thus
r = r0 + tv
which is a vector equation of L .
• Each value of parameter t gives the position vector r gives
the position vector r of a point on L. In other words, as t varies,
the line is traced out by the tip of the vector r. As Following
• figure, positive values of t correspond to points on L that lie on
one side of P0 , whereas negative values of t correspond to
points on L that lie on the other side of P0
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• If the vector v that gives the direction of the line L is written in
component form as v = < a, b, c > , then of course, we have
tv=< ta, tb, tc > .We also can write r = <x,y,z> and r0 =< x0,y0,z0>
so the vector position of P (x, y, z) becomes
<x,y,z>= < x0+ta, y0+tb, z0+tc >
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• Two vectors are equal if and only if
corresponding components are equal.
• Therefore, we have the scalar equations
x = x0+ at y = y0 + bt z= z0+ct .... *)
where t  R . Equations *) are called
parametric equations of the line L
through the point Po(X0,Y0,Z0) and
parallel to the vector v = <a,b,c>. Each
value of parameter t gives point (x,y,z)
on L
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Example 1
1. Find a vector equation and parametric
equation for the line that passes
through the point ( 5,1,3) and its
parallel to the vector i + 4j – 2k . Then,
find two other points on the line.
2. Find a vector equation and parametric
equations for the line through the
point (-2,4,10) and parallel to the
vector < 3,1,-8>
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• In general , if a vector v = <a,b,c> is used
to describe the direction of a line L, then
the numbers a, b, and c are called
direction numbers of L. Since any vector
parallel to v could also be use, we see that
any three numbers proportional to a, b, c
could also be used as a set of direction
numbers of L
•Another way to describing a line L is to
eliminate the parameter t from Equation
*). If none of a, b, or c is 0, we can solve
each equation for t, equate the results,
and
obtain Templates
:
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x  x0 y  y 0 z  z 0


a
b
c
• These equations are called symmetric
equations.
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Example 2
1. a. Find the parametric equations and
symmetric equations of the line that
passes through the points A (2,4,-3)
and B ( 3,-1,1)
b. At what point does this line intersect
the xy – plane ?
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SKEW LINES
In general
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SKEW LINES
Two lines with certain parametric equations are
called by skew lines, if they do not intersect and
are not parallel ( and therefore do not lie in
same plane )
Example
Show that the lines L1 and L2, with parametric
equations
:
x  1  t , y  2  3t , z  4  t
x  2s, y  3  s, z  3  4s
are skew lines !
Solution
The lines are not parallel, because the
corresponding vectors < 1,3,-1> and <2,1,4> are
not parallel ( Their components are not
proportional
). Templates
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SKEW LINES
If L1 and L2 had a point of intersection, there
would be values of t and s, such that
1+t = 2s
-2+3t=3+s
4-t=-3+4s
But if we solve the first two equations, we get
t=11/5 and s=8/5, and these values don’t satisfy
the third equation. Therefore, there are no
values of t and s that satisfy the three equations.
Thus, L1 and L2 does not intersect. Hence, L1 and
L2 are skew lines.
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1.1.Equations of Lines and Curve
----Curve-----• Suppose that f, g, h are continous real function real valued
functions on an interval I. Then the set C of all point (x,y,z)
in a space, where
x= f(t) y = g(t) z= h(t)
........ (1)
and t is a varies throughout the interval I, is called a Space
curve .
• The equations in (1) are called parametric equations of C
and t is called a parameter.
• We can think of C as being traced out by moving particle
whose position at time t is (f (t ), g (t ), h (t) ).
• If we now consider the vector position
r (t)=< f(t),g(t),h(t)> or r(t)=f(t)i+g(t)j+h(t)k
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• Then r(t) is the position vector of the
point P(f(t),g(t),h(t) )on C . Thus, any
continuous vector function r defines a
space curve C that is traced out by the
tip of the moving vector r(t) , as
shown in Following figure
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• Plane curves also represented in vector
notation. For instance, the curve given
by the parametric equations
and
y  t 2  21 y  t  1
could also described by the vector
equation : r (t )  t 2  2t , t  1  t 2  2t i  (t  1) j
where i = <1,0> and j=<0,1>.
• Let’s we see the following example
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Example 3
1. Describe the curve defined by vector
function :
r (t) = < 1+t , 2+5t, - 1+ 6t >
2. Sketch the curve whose vector
equation is
r (t ) = cos t i + sin t j + t k
3. Find a vector equation and
parametric equations for the line
segment that joins the point P(1,3,-2)
to the point Q(2.- 1,3).
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