What curve is described by $s = langle 2t,9sin t,9cos trangle$?
1
$begingroup$
What type of curve is described by the following?
$$s = langle 2t,9sin(t),9cos(t)rangle$$
Attempt
The $j$ and $k$ components of the curve describe a circle of radius $3$ in the $j-k$ plane and the $i$ component is linear. How can the type of curve be determined from this?
calculus geometry
edited Jan 15 at 17:47
Blue
49.3k870157
asked Jan 15 at 16:57
SjosephSjoseph
201111
$endgroup$
add a comment |
1
$begingroup$
What type of curve is described by the following?
$$s = langle 2t,9sin(t),9cos(t)rangle$$
Attempt
The $j$ and $k$ components of the curve describe a circle of radius $3$ in the $j-k$ plane and the $i$ component is linear. How can the type of curve be determined from this?
calculus geometry
edited Jan 15 at 17:47
Blue
49.3k870157
asked Jan 15 at 16:57
SjosephSjoseph
201111
$endgroup$
1
$begingroup$
I think you'll find it's a helix lined up on the $x$ axis.
$endgroup$
– Adrian Keister
Jan 15 at 17:00
$begingroup$
Thanks! Do you know this from experience or is there a nice way to figure that out?
$endgroup$
– Sjoseph
Jan 15 at 17:01
2
$begingroup$
Both. The $y$ and $z$ components describe a circle, by themselves. The $x$ coordinate is just going to march out steadily. If you imagine that in your mind, it comes out to a helix.
$endgroup$
– Adrian Keister
Jan 15 at 17:02
2
$begingroup$
From experience it is quickly recognizable as a helix, but you can also figure it out by noting that the projection of the curve on the $yz$-plane is a circle of radius 9 centered at the origin and the $x$-component climbs out of the $yz$-plane at a constant rate.
$endgroup$
– THW
Jan 15 at 17:04
add a comment |
1
1
1
$begingroup$
What type of curve is described by the following?
$$s = langle 2t,9sin(t),9cos(t)rangle$$
Attempt
The $j$ and $k$ components of the curve describe a circle of radius $3$ in the $j-k$ plane and the $i$ component is linear. How can the type of curve be determined from this?
calculus geometry
edited Jan 15 at 17:47
Blue
49.3k870157
asked Jan 15 at 16:57
SjosephSjoseph
201111
$endgroup$
What type of curve is described by the following?
$$s = langle 2t,9sin(t),9cos(t)rangle$$
Attempt
The $j$ and $k$ components of the curve describe a circle of radius $3$ in the $j-k$ plane and the $i$ component is linear. How can the type of curve be determined from this?
calculus geometry
calculus geometry
edited Jan 15 at 17:47
Blue
49.3k870157
asked Jan 15 at 16:57
SjosephSjoseph
201111
edited Jan 15 at 17:47
Blue
49.3k870157
asked Jan 15 at 16:57
SjosephSjoseph
201111
edited Jan 15 at 17:47
Blue
49.3k870157
edited Jan 15 at 17:47
Blue
49.3k870157
edited Jan 15 at 17:47
Blue
49.3k870157
49.3k870157
asked Jan 15 at 16:57
SjosephSjoseph
201111
asked Jan 15 at 16:57
SjosephSjoseph
201111
asked Jan 15 at 16:57
SjosephSjoseph
201111
201111
1
$begingroup$
I think you'll find it's a helix lined up on the $x$ axis.
$endgroup$
– Adrian Keister
Jan 15 at 17:00
$begingroup$
Thanks! Do you know this from experience or is there a nice way to figure that out?
$endgroup$
– Sjoseph
Jan 15 at 17:01
2
$begingroup$
Both. The $y$ and $z$ components describe a circle, by themselves. The $x$ coordinate is just going to march out steadily. If you imagine that in your mind, it comes out to a helix.
$endgroup$
– Adrian Keister
Jan 15 at 17:02
2
$begingroup$
From experience it is quickly recognizable as a helix, but you can also figure it out by noting that the projection of the curve on the $yz$-plane is a circle of radius 9 centered at the origin and the $x$-component climbs out of the $yz$-plane at a constant rate.
$endgroup$
– THW
Jan 15 at 17:04
add a comment |
1
$begingroup$
I think you'll find it's a helix lined up on the $x$ axis.
$endgroup$
– Adrian Keister
Jan 15 at 17:00
$begingroup$
Thanks! Do you know this from experience or is there a nice way to figure that out?
$endgroup$
– Sjoseph
Jan 15 at 17:01
2
$begingroup$
Both. The $y$ and $z$ components describe a circle, by themselves. The $x$ coordinate is just going to march out steadily. If you imagine that in your mind, it comes out to a helix.
$endgroup$
– Adrian Keister
Jan 15 at 17:02
2
$begingroup$
From experience it is quickly recognizable as a helix, but you can also figure it out by noting that the projection of the curve on the $yz$-plane is a circle of radius 9 centered at the origin and the $x$-component climbs out of the $yz$-plane at a constant rate.
$endgroup$
– THW
Jan 15 at 17:04
1
1
$begingroup$
I think you'll find it's a helix lined up on the $x$ axis.
$endgroup$
– Adrian Keister
Jan 15 at 17:00
$begingroup$
I think you'll find it's a helix lined up on the $x$ axis.
$endgroup$
– Adrian Keister
Jan 15 at 17:00
$begingroup$
Thanks! Do you know this from experience or is there a nice way to figure that out?
$endgroup$
– Sjoseph
Jan 15 at 17:01
$begingroup$
Thanks! Do you know this from experience or is there a nice way to figure that out?
$endgroup$
– Sjoseph
Jan 15 at 17:01
2
2
$begingroup$
Both. The $y$ and $z$ components describe a circle, by themselves. The $x$ coordinate is just going to march out steadily. If you imagine that in your mind, it comes out to a helix.
$endgroup$
– Adrian Keister
Jan 15 at 17:02
$begingroup$
Both. The $y$ and $z$ components describe a circle, by themselves. The $x$ coordinate is just going to march out steadily. If you imagine that in your mind, it comes out to a helix.
$endgroup$
– Adrian Keister
Jan 15 at 17:02
2
2
$begingroup$
From experience it is quickly recognizable as a helix, but you can also figure it out by noting that the projection of the curve on the $yz$-plane is a circle of radius 9 centered at the origin and the $x$-component climbs out of the $yz$-plane at a constant rate.
$endgroup$
– THW
Jan 15 at 17:04
$begingroup$
From experience it is quickly recognizable as a helix, but you can also figure it out by noting that the projection of the curve on the $yz$-plane is a circle of radius 9 centered at the origin and the $x$-component climbs out of the $yz$-plane at a constant rate.
$endgroup$
– THW
Jan 15 at 17:04
add a comment |
1 Answer
1
active
oldest
votes
4
$begingroup$
This may be of help
Built using a simple python script
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
ax = fig.gca(projection = '3d')
t = np.linspace(0, 4 * np.pi, num = 200)
x = 2 * t
y = 9 * np.sin(t)
z = 9 * np.cos(t)
ax.plot(x, y, z)
plt.show()
answered Jan 15 at 17:04
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Great - thanks for this!
$endgroup$
– Sjoseph
Jan 15 at 17:05
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
$begingroup$
This may be of help
Built using a simple python script
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
ax = fig.gca(projection = '3d')
t = np.linspace(0, 4 * np.pi, num = 200)
x = 2 * t
y = 9 * np.sin(t)
z = 9 * np.cos(t)
ax.plot(x, y, z)
plt.show()
answered Jan 15 at 17:04
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Great - thanks for this!
$endgroup$
– Sjoseph
Jan 15 at 17:05
add a comment |
4
$begingroup$
This may be of help
Built using a simple python script
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
ax = fig.gca(projection = '3d')
t = np.linspace(0, 4 * np.pi, num = 200)
x = 2 * t
y = 9 * np.sin(t)
z = 9 * np.cos(t)
ax.plot(x, y, z)
plt.show()
answered Jan 15 at 17:04
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Great - thanks for this!
$endgroup$
– Sjoseph
Jan 15 at 17:05
add a comment |
4
4
4
$begingroup$
This may be of help
Built using a simple python script
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
ax = fig.gca(projection = '3d')
t = np.linspace(0, 4 * np.pi, num = 200)
x = 2 * t
y = 9 * np.sin(t)
z = 9 * np.cos(t)
ax.plot(x, y, z)
plt.show()
answered Jan 15 at 17:04
caveraccaverac
14.8k31130
$endgroup$
This may be of help
Built using a simple python script
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
import matplotlib.pyplot as plt
ax = fig.gca(projection = '3d')
t = np.linspace(0, 4 * np.pi, num = 200)
x = 2 * t
y = 9 * np.sin(t)
z = 9 * np.cos(t)
ax.plot(x, y, z)
plt.show()
answered Jan 15 at 17:04
caveraccaverac
14.8k31130
answered Jan 15 at 17:04
caveraccaverac
14.8k31130
answered Jan 15 at 17:04
caveraccaverac
14.8k31130
answered Jan 15 at 17:04
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
Great - thanks for this!
$endgroup$
– Sjoseph
Jan 15 at 17:05
add a comment |
$begingroup$
Great - thanks for this!
$endgroup$
– Sjoseph
Jan 15 at 17:05
$begingroup$
Great - thanks for this!
$endgroup$
– Sjoseph
Jan 15 at 17:05
$begingroup$
Great - thanks for this!
$endgroup$
– Sjoseph
Jan 15 at 17:05
add a comment |
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1
$begingroup$
I think you'll find it's a helix lined up on the $x$ axis.
$endgroup$
– Adrian Keister
Jan 15 at 17:00
$begingroup$
Thanks! Do you know this from experience or is there a nice way to figure that out?
$endgroup$
– Sjoseph
Jan 15 at 17:01
2
$begingroup$
Both. The $y$ and $z$ components describe a circle, by themselves. The $x$ coordinate is just going to march out steadily. If you imagine that in your mind, it comes out to a helix.
$endgroup$
– Adrian Keister
Jan 15 at 17:02
2
$begingroup$
From experience it is quickly recognizable as a helix, but you can also figure it out by noting that the projection of the curve on the $yz$-plane is a circle of radius 9 centered at the origin and the $x$-component climbs out of the $yz$-plane at a constant rate.
$endgroup$
– THW
Jan 15 at 17:04