Powerpoint Templates Page 1 Powerpoint Templates Calculus III Second Lecture Notes By Rubono Setiawan, S.Si.,M.Sc

Powerpoint TemplatesPage 1

Powerpoint Templates

Calculus III

Second Lecture Notes By

Rubono Setiawan, S.Si.,M.Sc.


Contents of this presentation

• 1.2. Equations of Lines and Curve


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


• 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


• 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 >


• 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


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


• 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 :


• These equations are called symmetric equations.

c

zz

b

yy

a

xx 000


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 ?


SKEW LINES

In general


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


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.



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------


• 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


• 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


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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Powerpoint Templates Page 1 Powerpoint Templates Calculus III Second Lecture Notes By Rubono Setiawan, S.Si.,M.Sc