Textbook Notes (280,000)
CA (170,000)
UW (6,000)
MATH (80)
Chapter

# MATH104 Chapter Notes -Parametric Equation

Department
Mathematics
Course Code
MATH104
Professor
Steve Spencer

Page:
of 2
Lecture 1b
Vector Equation of a Line in R2
(pages 5-6)
The equation x2= (0.5)x1+ 1 deﬁnes the x2component of a point on the line
in terms the x1component. But when we graph the line, we think less of this
equation, and more of the following two facts: the x2-intercept is 1 and the
slope is 0.5. And so, when graphing, we may start at the point (0,1), and then
move “up one, right two” to ﬁnd a second point on the line (the point (2,2)),
and then connecting those points gives us our line. In fact, though, we could
have used any multiple of “up one, right two” to ﬁnd a second point on the
line. So, we can think of the line x2= (0.5)x1+ 1 as starting at the point (0,1),
and adding all scalar multiples of the directions “up one, right two”. Well, if
we decide to think of vectors as directions instead of points, then the direction
“up one, right two” corresponds to the vector 2
1. Then we can turn the
equation x2= (0.5)x1+ 1 into the equation x1
x2=0
1+t2
1for all
tR. Notice that picking t= 0 gives us our starting point of 0
1.
Deﬁnition: A line through ~p with direction vector ~
dis the set
{~p +t~
d|tR}
which has the vector equation
~x =~p +t~
d , t R
Note that in the case when ~p =~
0, we end up with the equation ~x =t~
d, which
is simply all scalar multiples of ~
d. This ends up being the equation of the line
through the origin that goes through ~
d. Another thing to note is that since the
direction vector ~
dcorresponds to the slope of a line, we see that two lines are
parallel if and only if their direction vectors are NON-ZERO MULTIPLES of
each other.
Sometimes it is useful to write the components of a vector equation separately.
For example, the equation x1
x2=0
1+t2
1is the same as saying
x1= 2tand x2= 1 + t(for the same t). In this case, tis called a ”parameter”,
and the equations are known as parametric equations.
1
Deﬁnition The parametric equation of the line ~x =~p +t~
dis the collection of
equations
x1=p1+td1
x2=p2+td2
tR
Now, if we solve both of these equations for t, we get
t=x1p1
d, t =x2p2
d2
But since t=t, we get
x1p1
d=x2p2
d2
and then solving for x2in terms of x1, we get equation x2=p2+d2
d1(x1p1).
This is the type of equation we started with!
Deﬁnition: The scalar form of the equation of a line is x2=p2+d2
d1(x1p1),
where p1
p2is a point on the line, and d1
d2is a direction vector for the line.
EXAMPLE The vector equation of a line passing through the point P(2,2)
with direction vector ~
d=2
3has vector equation
~x =2
2+t2
3tR
It has parametric equations
x1=2+2t
x2= 2 + 3ttR
Solving these parametric equations for tgives us
t=x1+ 2
2=x22
3
Solving for x2gives us the scalar form of the line: x2=3
2(x1+ 2) + 2, or
x2=3
2x1+ 5.
2