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The conditions for parameterisation
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I have proved that $gamma(t) = (1-cost, tant-sint)$ satisfies the equation for the conchoid $(x-1)^2(x^2+y^2)=x^2$. But is there any reason why this is not a parameterisation? How do I have to restrict the parameter $t$ to get a parameterisation for each branch of the curve?
The graph of the conchoids is produced here, which is the union of two disjoint connected curves.
What I can think about is regarding the domain of $t$. In this problem, considering $sint$, $cost$ and $tant$, $t$ shouldn't be equal to $fracpi2+kpi$, where $k$ is an integer. Hence, neither $sint$ nor $cost$ can reach their maxima and minima. So $1-cost≠1$, while $tant-sint$ tends to be infinity and negative infinity from different directions, which seems satisfying the graph of the given conchoid.
I am a bit confused here. Thanks in advance. :)
differential-geometry parametrization
asked Jul 30 at 6:20
Evelyn Venne
143
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I have proved that $gamma(t) = (1-cost, tant-sint)$ satisfies the equation for the conchoid $(x-1)^2(x^2+y^2)=x^2$. But is there any reason why this is not a parameterisation? How do I have to restrict the parameter $t$ to get a parameterisation for each branch of the curve?
The graph of the conchoids is produced here, which is the union of two disjoint connected curves.
What I can think about is regarding the domain of $t$. In this problem, considering $sint$, $cost$ and $tant$, $t$ shouldn't be equal to $fracpi2+kpi$, where $k$ is an integer. Hence, neither $sint$ nor $cost$ can reach their maxima and minima. So $1-cost≠1$, while $tant-sint$ tends to be infinity and negative infinity from different directions, which seems satisfying the graph of the given conchoid.
I am a bit confused here. Thanks in advance. :)
differential-geometry parametrization
asked Jul 30 at 6:20
Evelyn Venne
143
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have proved that $gamma(t) = (1-cost, tant-sint)$ satisfies the equation for the conchoid $(x-1)^2(x^2+y^2)=x^2$. But is there any reason why this is not a parameterisation? How do I have to restrict the parameter $t$ to get a parameterisation for each branch of the curve?
The graph of the conchoids is produced here, which is the union of two disjoint connected curves.
What I can think about is regarding the domain of $t$. In this problem, considering $sint$, $cost$ and $tant$, $t$ shouldn't be equal to $fracpi2+kpi$, where $k$ is an integer. Hence, neither $sint$ nor $cost$ can reach their maxima and minima. So $1-cost≠1$, while $tant-sint$ tends to be infinity and negative infinity from different directions, which seems satisfying the graph of the given conchoid.
I am a bit confused here. Thanks in advance. :)
differential-geometry parametrization
asked Jul 30 at 6:20
Evelyn Venne
143
I have proved that $gamma(t) = (1-cost, tant-sint)$ satisfies the equation for the conchoid $(x-1)^2(x^2+y^2)=x^2$. But is there any reason why this is not a parameterisation? How do I have to restrict the parameter $t$ to get a parameterisation for each branch of the curve?
The graph of the conchoids is produced here, which is the union of two disjoint connected curves.
What I can think about is regarding the domain of $t$. In this problem, considering $sint$, $cost$ and $tant$, $t$ shouldn't be equal to $fracpi2+kpi$, where $k$ is an integer. Hence, neither $sint$ nor $cost$ can reach their maxima and minima. So $1-cost≠1$, while $tant-sint$ tends to be infinity and negative infinity from different directions, which seems satisfying the graph of the given conchoid.
I am a bit confused here. Thanks in advance. :)
differential-geometry parametrization
asked Jul 30 at 6:20
Evelyn Venne
143
asked Jul 30 at 6:20
Evelyn Venne
143
asked Jul 30 at 6:20
Evelyn Venne
143
asked Jul 30 at 6:20
Evelyn Venne
143
143
add a comment |Â
add a comment |Â
1 Answer
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Usually, by a plane curve we mean a continuous function $gamma colon I to mathbb R^2$ where $I subseteq mathbb R$ is an interval.
If $gamma$ is defined by $gamma(t) = (1 - cos t, tan t - sin t)$, there are two natural choices for the interval:
- if we choose $I_1 = left ( - frac pi 2, frac pi 2 right )$ we obtain the branch of the conchoid containing the singular point $(0, 0)$, because as $t$ goes from $- frac pi 2$ to $0$ the abscissa $(1 - cos t)$ goes from $1$ to $0$, and then as $t$ goes from $0$ to $frac pi 2$ the abscissa goes from $0$ to $1$;
- if we choose $I_2 = left ( frac pi 2, frac 3 2 pi right )$ by a similar reasoning we obtain the other branch.
Any other interval having length $pi$ will give you the same points given by $I_1$ or $I_2$ because $gamma$ has period $2 pi$ and as you already said $frac pi 2 + k pi$ must be excluded from the domain for any $k in mathbb Z$.
answered Jul 30 at 8:54
Luca Bressan
3,8322935
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Usually, by a plane curve we mean a continuous function $gamma colon I to mathbb R^2$ where $I subseteq mathbb R$ is an interval.
If $gamma$ is defined by $gamma(t) = (1 - cos t, tan t - sin t)$, there are two natural choices for the interval:
- if we choose $I_1 = left ( - frac pi 2, frac pi 2 right )$ we obtain the branch of the conchoid containing the singular point $(0, 0)$, because as $t$ goes from $- frac pi 2$ to $0$ the abscissa $(1 - cos t)$ goes from $1$ to $0$, and then as $t$ goes from $0$ to $frac pi 2$ the abscissa goes from $0$ to $1$;
- if we choose $I_2 = left ( frac pi 2, frac 3 2 pi right )$ by a similar reasoning we obtain the other branch.
Any other interval having length $pi$ will give you the same points given by $I_1$ or $I_2$ because $gamma$ has period $2 pi$ and as you already said $frac pi 2 + k pi$ must be excluded from the domain for any $k in mathbb Z$.
answered Jul 30 at 8:54
Luca Bressan
3,8322935
add a comment |Â
up vote
0
down vote
accepted
Usually, by a plane curve we mean a continuous function $gamma colon I to mathbb R^2$ where $I subseteq mathbb R$ is an interval.
If $gamma$ is defined by $gamma(t) = (1 - cos t, tan t - sin t)$, there are two natural choices for the interval:
- if we choose $I_1 = left ( - frac pi 2, frac pi 2 right )$ we obtain the branch of the conchoid containing the singular point $(0, 0)$, because as $t$ goes from $- frac pi 2$ to $0$ the abscissa $(1 - cos t)$ goes from $1$ to $0$, and then as $t$ goes from $0$ to $frac pi 2$ the abscissa goes from $0$ to $1$;
- if we choose $I_2 = left ( frac pi 2, frac 3 2 pi right )$ by a similar reasoning we obtain the other branch.
Any other interval having length $pi$ will give you the same points given by $I_1$ or $I_2$ because $gamma$ has period $2 pi$ and as you already said $frac pi 2 + k pi$ must be excluded from the domain for any $k in mathbb Z$.
answered Jul 30 at 8:54
Luca Bressan
3,8322935
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Usually, by a plane curve we mean a continuous function $gamma colon I to mathbb R^2$ where $I subseteq mathbb R$ is an interval.
If $gamma$ is defined by $gamma(t) = (1 - cos t, tan t - sin t)$, there are two natural choices for the interval:
- if we choose $I_1 = left ( - frac pi 2, frac pi 2 right )$ we obtain the branch of the conchoid containing the singular point $(0, 0)$, because as $t$ goes from $- frac pi 2$ to $0$ the abscissa $(1 - cos t)$ goes from $1$ to $0$, and then as $t$ goes from $0$ to $frac pi 2$ the abscissa goes from $0$ to $1$;
- if we choose $I_2 = left ( frac pi 2, frac 3 2 pi right )$ by a similar reasoning we obtain the other branch.
Any other interval having length $pi$ will give you the same points given by $I_1$ or $I_2$ because $gamma$ has period $2 pi$ and as you already said $frac pi 2 + k pi$ must be excluded from the domain for any $k in mathbb Z$.
answered Jul 30 at 8:54
Luca Bressan
3,8322935
Usually, by a plane curve we mean a continuous function $gamma colon I to mathbb R^2$ where $I subseteq mathbb R$ is an interval.
If $gamma$ is defined by $gamma(t) = (1 - cos t, tan t - sin t)$, there are two natural choices for the interval:
- if we choose $I_1 = left ( - frac pi 2, frac pi 2 right )$ we obtain the branch of the conchoid containing the singular point $(0, 0)$, because as $t$ goes from $- frac pi 2$ to $0$ the abscissa $(1 - cos t)$ goes from $1$ to $0$, and then as $t$ goes from $0$ to $frac pi 2$ the abscissa goes from $0$ to $1$;
- if we choose $I_2 = left ( frac pi 2, frac 3 2 pi right )$ by a similar reasoning we obtain the other branch.
Any other interval having length $pi$ will give you the same points given by $I_1$ or $I_2$ because $gamma$ has period $2 pi$ and as you already said $frac pi 2 + k pi$ must be excluded from the domain for any $k in mathbb Z$.
answered Jul 30 at 8:54
Luca Bressan
3,8322935
answered Jul 30 at 8:54
Luca Bressan
3,8322935
answered Jul 30 at 8:54
Luca Bressan
3,8322935
answered Jul 30 at 8:54
Luca Bressan
3,8322935
3,8322935
add a comment |Â
add a comment |Â
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