powerpoint templates page 1 powerpoint templates calculus iii second lecture notes by rubono...
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Powerpoint Templates
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 and )
OP0OP
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• If a is a vector with representation 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 .
PP0
•
• 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 . 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
Rt
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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>
Example 1
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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 :
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• These equations are called symmetric equations.
c
zz
b
yy
a
xx 000
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Example 21. 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 )ExampleShow that the lines L1 and L2, with parametric equations :
are skew lines !SolutionThe lines are not parallel, because the corresponding vectors < 1,3,-1> and <2,1,4> are not parallel ( Their components are not proportional ).
szsysx
tztytx
43,3,2
4,32,1
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SKEW LINES
If L1 and L2 had a point of intersection, there would be values of t and s, such that1+t = 2s-2+3t=3+s4-t=-3+4sBut 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
• 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
----Curve------
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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 could also described by the vector equation : where i = <1,0> and j=<0,1>.
• Let’s we see the following example
212 ty 1ty
jtittttttr )1(21,2)( 22
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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).