Is this a correct parametrization of a rectangle on the complex plane?
$begingroup$
$z = 3 + i(2t - 1), t in [0,1)
\
z = 3 - 6(t-1) + i, t in [1,2)
\
z = -3 + i(1 - 2(t-2)), t in [2,3)
\
z = 6(t-3) - 3 - i, t in [3,4]$
I parameterized a rectangle with vertices at (-3,-i),(-3,i),(3,i), and (3,-i) in the above manner. However, I am having odd results later on in the problem I am working on and I am wondering if it is stemming from a faulty parameterized. Is this valid, and even if it is, is there an even simpler way that I could do it?
parametric
asked Feb 11 '16 at 17:31
StudentStudent
318210
$endgroup$
add a comment |
$begingroup$
$z = 3 + i(2t - 1), t in [0,1)
\
z = 3 - 6(t-1) + i, t in [1,2)
\
z = -3 + i(1 - 2(t-2)), t in [2,3)
\
z = 6(t-3) - 3 - i, t in [3,4]$
I parameterized a rectangle with vertices at (-3,-i),(-3,i),(3,i), and (3,-i) in the above manner. However, I am having odd results later on in the problem I am working on and I am wondering if it is stemming from a faulty parameterized. Is this valid, and even if it is, is there an even simpler way that I could do it?
parametric
asked Feb 11 '16 at 17:31
StudentStudent
318210
$endgroup$
add a comment |
$begingroup$
$z = 3 + i(2t - 1), t in [0,1)
\
z = 3 - 6(t-1) + i, t in [1,2)
\
z = -3 + i(1 - 2(t-2)), t in [2,3)
\
z = 6(t-3) - 3 - i, t in [3,4]$
I parameterized a rectangle with vertices at (-3,-i),(-3,i),(3,i), and (3,-i) in the above manner. However, I am having odd results later on in the problem I am working on and I am wondering if it is stemming from a faulty parameterized. Is this valid, and even if it is, is there an even simpler way that I could do it?
parametric
asked Feb 11 '16 at 17:31
StudentStudent
318210
$endgroup$
$z = 3 + i(2t - 1), t in [0,1)
\
z = 3 - 6(t-1) + i, t in [1,2)
\
z = -3 + i(1 - 2(t-2)), t in [2,3)
\
z = 6(t-3) - 3 - i, t in [3,4]$
I parameterized a rectangle with vertices at (-3,-i),(-3,i),(3,i), and (3,-i) in the above manner. However, I am having odd results later on in the problem I am working on and I am wondering if it is stemming from a faulty parameterized. Is this valid, and even if it is, is there an even simpler way that I could do it?
parametric
parametric
asked Feb 11 '16 at 17:31
StudentStudent
318210
asked Feb 11 '16 at 17:31
StudentStudent
318210
asked Feb 11 '16 at 17:31
StudentStudent
318210
asked Feb 11 '16 at 17:31
StudentStudent
318210
asked Feb 11 '16 at 17:31
StudentStudent
318210
318210
add a comment |
add a comment |
1 Answer
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$begingroup$
You have indeed described a path around a rectangle. As $t$ ranges from $0$ to $4$, the point $z(t)$ moves from $3-i$ to $3+i$ to $-3+i$ to $-3-i$ finally arriving back at $3-i$. The motion is piecewise linear.
If you write it as
$$z(t)=begin{cases}tz_0+(1-t)z_1,& 0leq t leq1\
t'z_1+(1-t')z_2,&0leq t'leq1&(t'=t-1)\
t''z_2+(1-t'')z_3,&0leq t'leq1&(t''=t'-1)\
t'''z_3+(1-t''')z_0,&0leq t''leq1&(t'''=t''-1)
end{cases}$$
then it is immediately clear how the point progresses around the path (here the $z_k$ are the vertices of the path).
answered Feb 11 '16 at 17:39
MPWMPW
30.4k12157
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
You have indeed described a path around a rectangle. As $t$ ranges from $0$ to $4$, the point $z(t)$ moves from $3-i$ to $3+i$ to $-3+i$ to $-3-i$ finally arriving back at $3-i$. The motion is piecewise linear.
If you write it as
$$z(t)=begin{cases}tz_0+(1-t)z_1,& 0leq t leq1\
t'z_1+(1-t')z_2,&0leq t'leq1&(t'=t-1)\
t''z_2+(1-t'')z_3,&0leq t'leq1&(t''=t'-1)\
t'''z_3+(1-t''')z_0,&0leq t''leq1&(t'''=t''-1)
end{cases}$$
then it is immediately clear how the point progresses around the path (here the $z_k$ are the vertices of the path).
answered Feb 11 '16 at 17:39
MPWMPW
30.4k12157
$endgroup$
add a comment |
$begingroup$
You have indeed described a path around a rectangle. As $t$ ranges from $0$ to $4$, the point $z(t)$ moves from $3-i$ to $3+i$ to $-3+i$ to $-3-i$ finally arriving back at $3-i$. The motion is piecewise linear.
If you write it as
$$z(t)=begin{cases}tz_0+(1-t)z_1,& 0leq t leq1\
t'z_1+(1-t')z_2,&0leq t'leq1&(t'=t-1)\
t''z_2+(1-t'')z_3,&0leq t'leq1&(t''=t'-1)\
t'''z_3+(1-t''')z_0,&0leq t''leq1&(t'''=t''-1)
end{cases}$$
then it is immediately clear how the point progresses around the path (here the $z_k$ are the vertices of the path).
answered Feb 11 '16 at 17:39
MPWMPW
30.4k12157
$endgroup$
add a comment |
$begingroup$
You have indeed described a path around a rectangle. As $t$ ranges from $0$ to $4$, the point $z(t)$ moves from $3-i$ to $3+i$ to $-3+i$ to $-3-i$ finally arriving back at $3-i$. The motion is piecewise linear.
If you write it as
$$z(t)=begin{cases}tz_0+(1-t)z_1,& 0leq t leq1\
t'z_1+(1-t')z_2,&0leq t'leq1&(t'=t-1)\
t''z_2+(1-t'')z_3,&0leq t'leq1&(t''=t'-1)\
t'''z_3+(1-t''')z_0,&0leq t''leq1&(t'''=t''-1)
end{cases}$$
then it is immediately clear how the point progresses around the path (here the $z_k$ are the vertices of the path).
answered Feb 11 '16 at 17:39
MPWMPW
30.4k12157
$endgroup$
You have indeed described a path around a rectangle. As $t$ ranges from $0$ to $4$, the point $z(t)$ moves from $3-i$ to $3+i$ to $-3+i$ to $-3-i$ finally arriving back at $3-i$. The motion is piecewise linear.
If you write it as
$$z(t)=begin{cases}tz_0+(1-t)z_1,& 0leq t leq1\
t'z_1+(1-t')z_2,&0leq t'leq1&(t'=t-1)\
t''z_2+(1-t'')z_3,&0leq t'leq1&(t''=t'-1)\
t'''z_3+(1-t''')z_0,&0leq t''leq1&(t'''=t''-1)
end{cases}$$
then it is immediately clear how the point progresses around the path (here the $z_k$ are the vertices of the path).
answered Feb 11 '16 at 17:39
MPWMPW
30.4k12157
edited Feb 11 '16 at 17:50
answered Feb 11 '16 at 17:39
MPWMPW
30.4k12157
answered Feb 11 '16 at 17:39
MPWMPW
30.4k12157
answered Feb 11 '16 at 17:39
MPWMPW
30.4k12157
30.4k12157
add a comment |
add a comment |
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