Parametric equation for a space curve
10
$begingroup$
With reference to the following image:
the blue curve has trivially a parametrization:
$$(x, y, z) = (costheta, , sintheta, , 0) ; ; ; text{with} ; theta in [0,,2pi)$$
I would like to determine the parametric equations of the red curve, very badly drawn in Paint, where I mean a sinusoidal curve along the blue circumference.
Although I thought about it a lot, I still could not figure out how to derive these parametric equation. Any ideas?
curves parametrization
edited Jan 2 at 12:42
dmtri
1,5322521
asked Jan 2 at 12:05
TeMTeM
459316
$endgroup$
add a comment |
10
$begingroup$
With reference to the following image:
the blue curve has trivially a parametrization:
$$(x, y, z) = (costheta, , sintheta, , 0) ; ; ; text{with} ; theta in [0,,2pi)$$
I would like to determine the parametric equations of the red curve, very badly drawn in Paint, where I mean a sinusoidal curve along the blue circumference.
Although I thought about it a lot, I still could not figure out how to derive these parametric equation. Any ideas?
curves parametrization
edited Jan 2 at 12:42
dmtri
1,5322521
asked Jan 2 at 12:05
TeMTeM
459316
$endgroup$
3
$begingroup$
It looks like a curve of the form $$ r(theta) = r_0 + A cos{(omega theta )} $$ But it's not very clear what you mean with the hand-drawn curve. Is it supposed to come out of the $z=0$-plane?
$endgroup$
– Matti P.
Jan 2 at 12:13
2
$begingroup$
@MattiP.: Yes, it must exit the $z = 0$ plane and have the circumference as the average line.
$endgroup$
– TeM
Jan 2 at 12:34
add a comment |
10
10
10
$begingroup$
With reference to the following image:
the blue curve has trivially a parametrization:
$$(x, y, z) = (costheta, , sintheta, , 0) ; ; ; text{with} ; theta in [0,,2pi)$$
I would like to determine the parametric equations of the red curve, very badly drawn in Paint, where I mean a sinusoidal curve along the blue circumference.
Although I thought about it a lot, I still could not figure out how to derive these parametric equation. Any ideas?
curves parametrization
edited Jan 2 at 12:42
dmtri
1,5322521
asked Jan 2 at 12:05
TeMTeM
459316
$endgroup$
With reference to the following image:
the blue curve has trivially a parametrization:
$$(x, y, z) = (costheta, , sintheta, , 0) ; ; ; text{with} ; theta in [0,,2pi)$$
I would like to determine the parametric equations of the red curve, very badly drawn in Paint, where I mean a sinusoidal curve along the blue circumference.
Although I thought about it a lot, I still could not figure out how to derive these parametric equation. Any ideas?
curves parametrization
curves parametrization
edited Jan 2 at 12:42
dmtri
1,5322521
asked Jan 2 at 12:05
TeMTeM
459316
edited Jan 2 at 12:42
dmtri
1,5322521
asked Jan 2 at 12:05
TeMTeM
459316
edited Jan 2 at 12:42
dmtri
1,5322521
edited Jan 2 at 12:42
dmtri
1,5322521
edited Jan 2 at 12:42
dmtri
1,5322521
1,5322521
asked Jan 2 at 12:05
TeMTeM
459316
asked Jan 2 at 12:05
TeMTeM
459316
asked Jan 2 at 12:05
TeMTeM
459316
459316
3
$begingroup$
It looks like a curve of the form $$ r(theta) = r_0 + A cos{(omega theta )} $$ But it's not very clear what you mean with the hand-drawn curve. Is it supposed to come out of the $z=0$-plane?
$endgroup$
– Matti P.
Jan 2 at 12:13
2
$begingroup$
@MattiP.: Yes, it must exit the $z = 0$ plane and have the circumference as the average line.
$endgroup$
– TeM
Jan 2 at 12:34
add a comment |
3
$begingroup$
It looks like a curve of the form $$ r(theta) = r_0 + A cos{(omega theta )} $$ But it's not very clear what you mean with the hand-drawn curve. Is it supposed to come out of the $z=0$-plane?
$endgroup$
– Matti P.
Jan 2 at 12:13
2
$begingroup$
@MattiP.: Yes, it must exit the $z = 0$ plane and have the circumference as the average line.
$endgroup$
– TeM
Jan 2 at 12:34
3
3
$begingroup$
It looks like a curve of the form $$ r(theta) = r_0 + A cos{(omega theta )} $$ But it's not very clear what you mean with the hand-drawn curve. Is it supposed to come out of the $z=0$-plane?
$endgroup$
– Matti P.
Jan 2 at 12:13
$begingroup$
It looks like a curve of the form $$ r(theta) = r_0 + A cos{(omega theta )} $$ But it's not very clear what you mean with the hand-drawn curve. Is it supposed to come out of the $z=0$-plane?
$endgroup$
– Matti P.
Jan 2 at 12:13
2
2
$begingroup$
@MattiP.: Yes, it must exit the $z = 0$ plane and have the circumference as the average line.
$endgroup$
– TeM
Jan 2 at 12:34
$begingroup$
@MattiP.: Yes, it must exit the $z = 0$ plane and have the circumference as the average line.
$endgroup$
– TeM
Jan 2 at 12:34
add a comment |
1 Answer
1
active
oldest
votes
12
$begingroup$
You can try$$thetamapstoleft(costheta,sintheta,frac{cos(8theta)}8right),$$for instance.
answered Jan 2 at 12:13
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
It's exactly what I wanted, thank you! Could you tell me how you managed to understand it?
$endgroup$
– TeM
Jan 2 at 12:41
3
$begingroup$
The $z$ coordinate had to be a waving line again, and so I thought about $cos(8theta)$, but then the waves would go too high and too low. That's why I divided by $8$.
$endgroup$
– José Carlos Santos
Jan 2 at 12:45
$begingroup$
Perfect! Before closing I also wanted to ask how to get the parametric equations of the red curve if it lay in the $z = 0$ plane. Is it better that I open a new question or change the request?
$endgroup$
– TeM
Jan 2 at 12:55
3
$begingroup$
No need for that. Just consider:$$thetamapstoleft(costheta+frac{cos(8theta)}8,sintheta,0right).$$
$endgroup$
– José Carlos Santos
Jan 2 at 13:02
1
$begingroup$
@TeM If you want the red curve in the Z=0 plane, then you probably want to oscillate in the direction of the unit normal vector rather than in the x direction. Here's a comparison: math3d.org/cCvATTaY
$endgroup$
– Chris Chudzicki
Jan 2 at 19:00
|
show 1 more comment
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
12
$begingroup$
You can try$$thetamapstoleft(costheta,sintheta,frac{cos(8theta)}8right),$$for instance.
answered Jan 2 at 12:13
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
It's exactly what I wanted, thank you! Could you tell me how you managed to understand it?
$endgroup$
– TeM
Jan 2 at 12:41
3
$begingroup$
The $z$ coordinate had to be a waving line again, and so I thought about $cos(8theta)$, but then the waves would go too high and too low. That's why I divided by $8$.
$endgroup$
– José Carlos Santos
Jan 2 at 12:45
$begingroup$
Perfect! Before closing I also wanted to ask how to get the parametric equations of the red curve if it lay in the $z = 0$ plane. Is it better that I open a new question or change the request?
$endgroup$
– TeM
Jan 2 at 12:55
3
$begingroup$
No need for that. Just consider:$$thetamapstoleft(costheta+frac{cos(8theta)}8,sintheta,0right).$$
$endgroup$
– José Carlos Santos
Jan 2 at 13:02
1
$begingroup$
@TeM If you want the red curve in the Z=0 plane, then you probably want to oscillate in the direction of the unit normal vector rather than in the x direction. Here's a comparison: math3d.org/cCvATTaY
$endgroup$
– Chris Chudzicki
Jan 2 at 19:00
|
show 1 more comment
12
$begingroup$
You can try$$thetamapstoleft(costheta,sintheta,frac{cos(8theta)}8right),$$for instance.
answered Jan 2 at 12:13
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
$begingroup$
It's exactly what I wanted, thank you! Could you tell me how you managed to understand it?
$endgroup$
– TeM
Jan 2 at 12:41
3
$begingroup$
The $z$ coordinate had to be a waving line again, and so I thought about $cos(8theta)$, but then the waves would go too high and too low. That's why I divided by $8$.
$endgroup$
– José Carlos Santos
Jan 2 at 12:45
$begingroup$
Perfect! Before closing I also wanted to ask how to get the parametric equations of the red curve if it lay in the $z = 0$ plane. Is it better that I open a new question or change the request?
$endgroup$
– TeM
Jan 2 at 12:55
3
$begingroup$
No need for that. Just consider:$$thetamapstoleft(costheta+frac{cos(8theta)}8,sintheta,0right).$$
$endgroup$
– José Carlos Santos
Jan 2 at 13:02
1
$begingroup$
@TeM If you want the red curve in the Z=0 plane, then you probably want to oscillate in the direction of the unit normal vector rather than in the x direction. Here's a comparison: math3d.org/cCvATTaY
$endgroup$
– Chris Chudzicki
Jan 2 at 19:00
|
show 1 more comment
12
12
12
$begingroup$
You can try$$thetamapstoleft(costheta,sintheta,frac{cos(8theta)}8right),$$for instance.
answered Jan 2 at 12:13
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
You can try$$thetamapstoleft(costheta,sintheta,frac{cos(8theta)}8right),$$for instance.
answered Jan 2 at 12:13
José Carlos SantosJosé Carlos Santos
165k22132235
edited Jan 2 at 12:38
answered Jan 2 at 12:13
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 2 at 12:13
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 2 at 12:13
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
$begingroup$
It's exactly what I wanted, thank you! Could you tell me how you managed to understand it?
$endgroup$
– TeM
Jan 2 at 12:41
3
$begingroup$
The $z$ coordinate had to be a waving line again, and so I thought about $cos(8theta)$, but then the waves would go too high and too low. That's why I divided by $8$.
$endgroup$
– José Carlos Santos
Jan 2 at 12:45
$begingroup$
Perfect! Before closing I also wanted to ask how to get the parametric equations of the red curve if it lay in the $z = 0$ plane. Is it better that I open a new question or change the request?
$endgroup$
– TeM
Jan 2 at 12:55
3
$begingroup$
No need for that. Just consider:$$thetamapstoleft(costheta+frac{cos(8theta)}8,sintheta,0right).$$
$endgroup$
– José Carlos Santos
Jan 2 at 13:02
1
$begingroup$
@TeM If you want the red curve in the Z=0 plane, then you probably want to oscillate in the direction of the unit normal vector rather than in the x direction. Here's a comparison: math3d.org/cCvATTaY
$endgroup$
– Chris Chudzicki
Jan 2 at 19:00
|
show 1 more comment
$begingroup$
It's exactly what I wanted, thank you! Could you tell me how you managed to understand it?
$endgroup$
– TeM
Jan 2 at 12:41
3
$begingroup$
The $z$ coordinate had to be a waving line again, and so I thought about $cos(8theta)$, but then the waves would go too high and too low. That's why I divided by $8$.
$endgroup$
– José Carlos Santos
Jan 2 at 12:45
$begingroup$
Perfect! Before closing I also wanted to ask how to get the parametric equations of the red curve if it lay in the $z = 0$ plane. Is it better that I open a new question or change the request?
$endgroup$
– TeM
Jan 2 at 12:55
3
$begingroup$
No need for that. Just consider:$$thetamapstoleft(costheta+frac{cos(8theta)}8,sintheta,0right).$$
$endgroup$
– José Carlos Santos
Jan 2 at 13:02
1
$begingroup$
@TeM If you want the red curve in the Z=0 plane, then you probably want to oscillate in the direction of the unit normal vector rather than in the x direction. Here's a comparison: math3d.org/cCvATTaY
$endgroup$
– Chris Chudzicki
Jan 2 at 19:00
$begingroup$
It's exactly what I wanted, thank you! Could you tell me how you managed to understand it?
$endgroup$
– TeM
Jan 2 at 12:41
$begingroup$
It's exactly what I wanted, thank you! Could you tell me how you managed to understand it?
$endgroup$
– TeM
Jan 2 at 12:41
3
3
$begingroup$
The $z$ coordinate had to be a waving line again, and so I thought about $cos(8theta)$, but then the waves would go too high and too low. That's why I divided by $8$.
$endgroup$
– José Carlos Santos
Jan 2 at 12:45
$begingroup$
The $z$ coordinate had to be a waving line again, and so I thought about $cos(8theta)$, but then the waves would go too high and too low. That's why I divided by $8$.
$endgroup$
– José Carlos Santos
Jan 2 at 12:45
$begingroup$
Perfect! Before closing I also wanted to ask how to get the parametric equations of the red curve if it lay in the $z = 0$ plane. Is it better that I open a new question or change the request?
$endgroup$
– TeM
Jan 2 at 12:55
$begingroup$
Perfect! Before closing I also wanted to ask how to get the parametric equations of the red curve if it lay in the $z = 0$ plane. Is it better that I open a new question or change the request?
$endgroup$
– TeM
Jan 2 at 12:55
3
3
$begingroup$
No need for that. Just consider:$$thetamapstoleft(costheta+frac{cos(8theta)}8,sintheta,0right).$$
$endgroup$
– José Carlos Santos
Jan 2 at 13:02
$begingroup$
No need for that. Just consider:$$thetamapstoleft(costheta+frac{cos(8theta)}8,sintheta,0right).$$
$endgroup$
– José Carlos Santos
Jan 2 at 13:02
1
1
$begingroup$
@TeM If you want the red curve in the Z=0 plane, then you probably want to oscillate in the direction of the unit normal vector rather than in the x direction. Here's a comparison: math3d.org/cCvATTaY
$endgroup$
– Chris Chudzicki
Jan 2 at 19:00
$begingroup$
@TeM If you want the red curve in the Z=0 plane, then you probably want to oscillate in the direction of the unit normal vector rather than in the x direction. Here's a comparison: math3d.org/cCvATTaY
$endgroup$
– Chris Chudzicki
Jan 2 at 19:00
|
show 1 more comment
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3
$begingroup$
It looks like a curve of the form $$ r(theta) = r_0 + A cos{(omega theta )} $$ But it's not very clear what you mean with the hand-drawn curve. Is it supposed to come out of the $z=0$-plane?
$endgroup$
– Matti P.
Jan 2 at 12:13
2
$begingroup$
@MattiP.: Yes, it must exit the $z = 0$ plane and have the circumference as the average line.
$endgroup$
– TeM
Jan 2 at 12:34