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Parametrically Defined Curves: $f'$ and $g'$ Are Not Simultaneously Zero
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I can't find a clear, comprehensive explanation, on this site or elsewhere, for why parametrically defined curves frequently have the condition that the the derivatives of their points $x = f(t)$ and $y = g(t)$ cannot simultaneously be zero on the interval $[a, b]$.
Most of the explanations use language that assumes that the reader already understands the concept they're explaining, or the explanations make the meaningless claim that the curve must be "nice".
I would appreciate it if people could please take the time to explain, comprehensively (not rigorously), what is meant by this condition. If you're going to use words that are likely to be unfamiliar to someone who doesn't understand this concept, like "regular", then please take the time to define what it means.
real-analysis geometry derivatives curves parametrization
asked Jul 20 at 13:36
The Pointer
2,4542829
 |Â
show 5 more comments
up vote
3
down vote
favorite
I can't find a clear, comprehensive explanation, on this site or elsewhere, for why parametrically defined curves frequently have the condition that the the derivatives of their points $x = f(t)$ and $y = g(t)$ cannot simultaneously be zero on the interval $[a, b]$.
Most of the explanations use language that assumes that the reader already understands the concept they're explaining, or the explanations make the meaningless claim that the curve must be "nice".
I would appreciate it if people could please take the time to explain, comprehensively (not rigorously), what is meant by this condition. If you're going to use words that are likely to be unfamiliar to someone who doesn't understand this concept, like "regular", then please take the time to define what it means.
real-analysis geometry derivatives curves parametrization
asked Jul 20 at 13:36
The Pointer
2,4542829
It means the curve has a non zero speed for each value of the parameter. This means that the curve in an injective map so that any point on the curve defines a unique value of the parameter.
– copper.hat
Jul 20 at 14:12
@copper.hat So for example $tmapsto(cos(t),sin(t))$ is injective? Ok, you meant locally injective. But I really don't think this is exactly the point - the "bad" parametrization of the curve $y=|x|$ in my answer is injective...
– David C. Ullrich
Jul 20 at 14:14
@DavidC.Ullrich: I meant locally, it was more of a comment than a complete answer.
– copper.hat
Jul 20 at 14:21
@DavidC.Ullrich: I guess I would settle for rectifiable, but as your answer nicely demonstrates, this allows a lot of curves that don't fit with our intuition.
– copper.hat
Jul 20 at 14:27
@copper.hat In fact I'm pretty sure that any rectifiable curve has a $C^1$ parametrization...
– David C. Ullrich
Jul 20 at 14:39
 |Â
show 5 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I can't find a clear, comprehensive explanation, on this site or elsewhere, for why parametrically defined curves frequently have the condition that the the derivatives of their points $x = f(t)$ and $y = g(t)$ cannot simultaneously be zero on the interval $[a, b]$.
Most of the explanations use language that assumes that the reader already understands the concept they're explaining, or the explanations make the meaningless claim that the curve must be "nice".
I would appreciate it if people could please take the time to explain, comprehensively (not rigorously), what is meant by this condition. If you're going to use words that are likely to be unfamiliar to someone who doesn't understand this concept, like "regular", then please take the time to define what it means.
real-analysis geometry derivatives curves parametrization
asked Jul 20 at 13:36
The Pointer
2,4542829
I can't find a clear, comprehensive explanation, on this site or elsewhere, for why parametrically defined curves frequently have the condition that the the derivatives of their points $x = f(t)$ and $y = g(t)$ cannot simultaneously be zero on the interval $[a, b]$.
Most of the explanations use language that assumes that the reader already understands the concept they're explaining, or the explanations make the meaningless claim that the curve must be "nice".
I would appreciate it if people could please take the time to explain, comprehensively (not rigorously), what is meant by this condition. If you're going to use words that are likely to be unfamiliar to someone who doesn't understand this concept, like "regular", then please take the time to define what it means.
real-analysis geometry derivatives curves parametrization
asked Jul 20 at 13:36
The Pointer
2,4542829
asked Jul 20 at 13:36
The Pointer
2,4542829
asked Jul 20 at 13:36
The Pointer
2,4542829
asked Jul 20 at 13:36
The Pointer
2,4542829
2,4542829
It means the curve has a non zero speed for each value of the parameter. This means that the curve in an injective map so that any point on the curve defines a unique value of the parameter.
– copper.hat
Jul 20 at 14:12
@copper.hat So for example $tmapsto(cos(t),sin(t))$ is injective? Ok, you meant locally injective. But I really don't think this is exactly the point - the "bad" parametrization of the curve $y=|x|$ in my answer is injective...
– David C. Ullrich
Jul 20 at 14:14
@DavidC.Ullrich: I meant locally, it was more of a comment than a complete answer.
– copper.hat
Jul 20 at 14:21
@DavidC.Ullrich: I guess I would settle for rectifiable, but as your answer nicely demonstrates, this allows a lot of curves that don't fit with our intuition.
– copper.hat
Jul 20 at 14:27
@copper.hat In fact I'm pretty sure that any rectifiable curve has a $C^1$ parametrization...
– David C. Ullrich
Jul 20 at 14:39
 |Â
show 5 more comments
It means the curve has a non zero speed for each value of the parameter. This means that the curve in an injective map so that any point on the curve defines a unique value of the parameter.
– copper.hat
Jul 20 at 14:12
@copper.hat So for example $tmapsto(cos(t),sin(t))$ is injective? Ok, you meant locally injective. But I really don't think this is exactly the point - the "bad" parametrization of the curve $y=|x|$ in my answer is injective...
– David C. Ullrich
Jul 20 at 14:14
@DavidC.Ullrich: I meant locally, it was more of a comment than a complete answer.
– copper.hat
Jul 20 at 14:21
@DavidC.Ullrich: I guess I would settle for rectifiable, but as your answer nicely demonstrates, this allows a lot of curves that don't fit with our intuition.
– copper.hat
Jul 20 at 14:27
@copper.hat In fact I'm pretty sure that any rectifiable curve has a $C^1$ parametrization...
– David C. Ullrich
Jul 20 at 14:39
It means the curve has a non zero speed for each value of the parameter. This means that the curve in an injective map so that any point on the curve defines a unique value of the parameter.
– copper.hat
Jul 20 at 14:12
It means the curve has a non zero speed for each value of the parameter. This means that the curve in an injective map so that any point on the curve defines a unique value of the parameter.
– copper.hat
Jul 20 at 14:12
@copper.hat So for example $tmapsto(cos(t),sin(t))$ is injective? Ok, you meant locally injective. But I really don't think this is exactly the point - the "bad" parametrization of the curve $y=|x|$ in my answer is injective...
– David C. Ullrich
Jul 20 at 14:14
@copper.hat So for example $tmapsto(cos(t),sin(t))$ is injective? Ok, you meant locally injective. But I really don't think this is exactly the point - the "bad" parametrization of the curve $y=|x|$ in my answer is injective...
– David C. Ullrich
Jul 20 at 14:14
@DavidC.Ullrich: I meant locally, it was more of a comment than a complete answer.
– copper.hat
Jul 20 at 14:21
@DavidC.Ullrich: I meant locally, it was more of a comment than a complete answer.
– copper.hat
Jul 20 at 14:21
@DavidC.Ullrich: I guess I would settle for rectifiable, but as your answer nicely demonstrates, this allows a lot of curves that don't fit with our intuition.
– copper.hat
Jul 20 at 14:27
@DavidC.Ullrich: I guess I would settle for rectifiable, but as your answer nicely demonstrates, this allows a lot of curves that don't fit with our intuition.
– copper.hat
Jul 20 at 14:27
@copper.hat In fact I'm pretty sure that any rectifiable curve has a $C^1$ parametrization...
– David C. Ullrich
Jul 20 at 14:39
@copper.hat In fact I'm pretty sure that any rectifiable curve has a $C^1$ parametrization...
– David C. Ullrich
Jul 20 at 14:39
 |Â
show 5 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
The point is we want to say a curve is smooth, for example $C^1$ (continuously differentiable), if it has a smooth parametrization. And without that condition on non-vanishing derivatives that would admit "smooth" curves that we really don't want to call "smooth".
Example: Consider the curve in the plane defined by $y=|x|$. That has a corner at the origin - our definition of "$C^1$ curve" had better not include this curve, right?
But define $f,g:Bbb RtoBbb R$ by $g(t)=t^2$ and $$f(t)=begincases
t^2,&(tge0),
\-t^2,&(t<0).endcases$$Then $f$ and $g$ are both $C^1$, and $x=f(t)$, $y=g(t)$ is a parametrization of the curve $y=|x|$. Without the condition on non-vanishing derivatives this would make $y=|x|$ a $C^1$ curve. (With the condition on non-vanishing derivatives everything's fine - this parametization is disallowed because $f'(0)=g'(0)=0$.)
answered Jul 20 at 14:11
David C. Ullrich
54.2k33482
Thanks for the response. So what is the connection between this and the condition that the derivatives of the points of the curve cannot simultaneously be zero? For instance, you used the curve $y=|x|$ as an example, but this doesn't have derivatives equal to $0$ at any point; rather, it is only undefined at $x = 0$, since this is a corner. Furthermore, I don't see any reference at all to the concept that my question referenced, which is the points being simultaneously being zero. [...]
– The Pointer
Jul 20 at 14:25
Again, I appreciate the answer, but this was written by someone who understands the concept, for someone who understands the concept. I cannot derive anything useful from this, despite putting in the effort to understand it.
– The Pointer
Jul 20 at 14:26
@ThePointer The point is that although the curve $y=|x|$ is not differentiable, the paramteriization $(x(t),y(t))$ that I gave is differentiable! So we need to not allow that parametrization. If add the condition about non-vanishing derivatives we rule out that parametrization, since $x'(0)=y'(0)=0$.
– David C. Ullrich
Jul 20 at 14:30
@ThePointer: The point here is that the curve is continuously differentiable everywhere, but it has a kink when you just look at the range of the curve. Our natural inclination is to consider such curves as not being smooth even though they are.
– copper.hat
Jul 20 at 14:31
1
@ThePointer Right.
– David C. Ullrich
Jul 20 at 15:30
 |Â
show 9 more comments
up vote
0
down vote
Because when $f'=g'=0$, the direction of the curve is no more defined. At such a point, the curve could change direction abruptly.
Example.
answered Jul 20 at 14:39
Yves Daoust
111k665204
Singular point: mathworld.wolfram.com/SingularPoint.html
– The Pointer
Jul 20 at 14:44
Thanks for the response. So the reason for this condition is that we want the parameterised curve to be unidirectional?
– The Pointer
Jul 20 at 14:45
@ThePointer: what do you mean by unidirectional ?
– Yves Daoust
Jul 20 at 14:47
@ThePointer: "in one direction" means following a straight line. I doubt this is what you have in mind. But I have completely rephrased my answer.
– Yves Daoust
Jul 20 at 14:50
1
@ThePointer This is undesirable behavior because if we don't disallow this behavior we find the the curve $y=|x|$ is "differentiable", as I showed in my answer. We certainly don't want that - if $y=|x|$ is a "differentiable" curve then differentiable curves are nothing like what we think they should be.
– David C. Ullrich
Jul 20 at 15:11
 |Â
show 5 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The point is we want to say a curve is smooth, for example $C^1$ (continuously differentiable), if it has a smooth parametrization. And without that condition on non-vanishing derivatives that would admit "smooth" curves that we really don't want to call "smooth".
Example: Consider the curve in the plane defined by $y=|x|$. That has a corner at the origin - our definition of "$C^1$ curve" had better not include this curve, right?
But define $f,g:Bbb RtoBbb R$ by $g(t)=t^2$ and $$f(t)=begincases
t^2,&(tge0),
\-t^2,&(t<0).endcases$$Then $f$ and $g$ are both $C^1$, and $x=f(t)$, $y=g(t)$ is a parametrization of the curve $y=|x|$. Without the condition on non-vanishing derivatives this would make $y=|x|$ a $C^1$ curve. (With the condition on non-vanishing derivatives everything's fine - this parametization is disallowed because $f'(0)=g'(0)=0$.)
answered Jul 20 at 14:11
David C. Ullrich
54.2k33482
Thanks for the response. So what is the connection between this and the condition that the derivatives of the points of the curve cannot simultaneously be zero? For instance, you used the curve $y=|x|$ as an example, but this doesn't have derivatives equal to $0$ at any point; rather, it is only undefined at $x = 0$, since this is a corner. Furthermore, I don't see any reference at all to the concept that my question referenced, which is the points being simultaneously being zero. [...]
– The Pointer
Jul 20 at 14:25
Again, I appreciate the answer, but this was written by someone who understands the concept, for someone who understands the concept. I cannot derive anything useful from this, despite putting in the effort to understand it.
– The Pointer
Jul 20 at 14:26
@ThePointer The point is that although the curve $y=|x|$ is not differentiable, the paramteriization $(x(t),y(t))$ that I gave is differentiable! So we need to not allow that parametrization. If add the condition about non-vanishing derivatives we rule out that parametrization, since $x'(0)=y'(0)=0$.
– David C. Ullrich
Jul 20 at 14:30
@ThePointer: The point here is that the curve is continuously differentiable everywhere, but it has a kink when you just look at the range of the curve. Our natural inclination is to consider such curves as not being smooth even though they are.
– copper.hat
Jul 20 at 14:31
1
@ThePointer Right.
– David C. Ullrich
Jul 20 at 15:30
 |Â
show 9 more comments
up vote
3
down vote
accepted
The point is we want to say a curve is smooth, for example $C^1$ (continuously differentiable), if it has a smooth parametrization. And without that condition on non-vanishing derivatives that would admit "smooth" curves that we really don't want to call "smooth".
Example: Consider the curve in the plane defined by $y=|x|$. That has a corner at the origin - our definition of "$C^1$ curve" had better not include this curve, right?
But define $f,g:Bbb RtoBbb R$ by $g(t)=t^2$ and $$f(t)=begincases
t^2,&(tge0),
\-t^2,&(t<0).endcases$$Then $f$ and $g$ are both $C^1$, and $x=f(t)$, $y=g(t)$ is a parametrization of the curve $y=|x|$. Without the condition on non-vanishing derivatives this would make $y=|x|$ a $C^1$ curve. (With the condition on non-vanishing derivatives everything's fine - this parametization is disallowed because $f'(0)=g'(0)=0$.)
answered Jul 20 at 14:11
David C. Ullrich
54.2k33482
Thanks for the response. So what is the connection between this and the condition that the derivatives of the points of the curve cannot simultaneously be zero? For instance, you used the curve $y=|x|$ as an example, but this doesn't have derivatives equal to $0$ at any point; rather, it is only undefined at $x = 0$, since this is a corner. Furthermore, I don't see any reference at all to the concept that my question referenced, which is the points being simultaneously being zero. [...]
– The Pointer
Jul 20 at 14:25
Again, I appreciate the answer, but this was written by someone who understands the concept, for someone who understands the concept. I cannot derive anything useful from this, despite putting in the effort to understand it.
– The Pointer
Jul 20 at 14:26
@ThePointer The point is that although the curve $y=|x|$ is not differentiable, the paramteriization $(x(t),y(t))$ that I gave is differentiable! So we need to not allow that parametrization. If add the condition about non-vanishing derivatives we rule out that parametrization, since $x'(0)=y'(0)=0$.
– David C. Ullrich
Jul 20 at 14:30
@ThePointer: The point here is that the curve is continuously differentiable everywhere, but it has a kink when you just look at the range of the curve. Our natural inclination is to consider such curves as not being smooth even though they are.
– copper.hat
Jul 20 at 14:31
1
@ThePointer Right.
– David C. Ullrich
Jul 20 at 15:30
 |Â
show 9 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The point is we want to say a curve is smooth, for example $C^1$ (continuously differentiable), if it has a smooth parametrization. And without that condition on non-vanishing derivatives that would admit "smooth" curves that we really don't want to call "smooth".
Example: Consider the curve in the plane defined by $y=|x|$. That has a corner at the origin - our definition of "$C^1$ curve" had better not include this curve, right?
But define $f,g:Bbb RtoBbb R$ by $g(t)=t^2$ and $$f(t)=begincases
t^2,&(tge0),
\-t^2,&(t<0).endcases$$Then $f$ and $g$ are both $C^1$, and $x=f(t)$, $y=g(t)$ is a parametrization of the curve $y=|x|$. Without the condition on non-vanishing derivatives this would make $y=|x|$ a $C^1$ curve. (With the condition on non-vanishing derivatives everything's fine - this parametization is disallowed because $f'(0)=g'(0)=0$.)
answered Jul 20 at 14:11
David C. Ullrich
54.2k33482
The point is we want to say a curve is smooth, for example $C^1$ (continuously differentiable), if it has a smooth parametrization. And without that condition on non-vanishing derivatives that would admit "smooth" curves that we really don't want to call "smooth".
Example: Consider the curve in the plane defined by $y=|x|$. That has a corner at the origin - our definition of "$C^1$ curve" had better not include this curve, right?
But define $f,g:Bbb RtoBbb R$ by $g(t)=t^2$ and $$f(t)=begincases
t^2,&(tge0),
\-t^2,&(t<0).endcases$$Then $f$ and $g$ are both $C^1$, and $x=f(t)$, $y=g(t)$ is a parametrization of the curve $y=|x|$. Without the condition on non-vanishing derivatives this would make $y=|x|$ a $C^1$ curve. (With the condition on non-vanishing derivatives everything's fine - this parametization is disallowed because $f'(0)=g'(0)=0$.)
answered Jul 20 at 14:11
David C. Ullrich
54.2k33482
edited Jul 20 at 14:51
answered Jul 20 at 14:11
David C. Ullrich
54.2k33482
answered Jul 20 at 14:11
David C. Ullrich
54.2k33482
answered Jul 20 at 14:11
David C. Ullrich
54.2k33482
54.2k33482
Thanks for the response. So what is the connection between this and the condition that the derivatives of the points of the curve cannot simultaneously be zero? For instance, you used the curve $y=|x|$ as an example, but this doesn't have derivatives equal to $0$ at any point; rather, it is only undefined at $x = 0$, since this is a corner. Furthermore, I don't see any reference at all to the concept that my question referenced, which is the points being simultaneously being zero. [...]
– The Pointer
Jul 20 at 14:25
Again, I appreciate the answer, but this was written by someone who understands the concept, for someone who understands the concept. I cannot derive anything useful from this, despite putting in the effort to understand it.
– The Pointer
Jul 20 at 14:26
@ThePointer The point is that although the curve $y=|x|$ is not differentiable, the paramteriization $(x(t),y(t))$ that I gave is differentiable! So we need to not allow that parametrization. If add the condition about non-vanishing derivatives we rule out that parametrization, since $x'(0)=y'(0)=0$.
– David C. Ullrich
Jul 20 at 14:30
@ThePointer: The point here is that the curve is continuously differentiable everywhere, but it has a kink when you just look at the range of the curve. Our natural inclination is to consider such curves as not being smooth even though they are.
– copper.hat
Jul 20 at 14:31
1
@ThePointer Right.
– David C. Ullrich
Jul 20 at 15:30
 |Â
show 9 more comments
Thanks for the response. So what is the connection between this and the condition that the derivatives of the points of the curve cannot simultaneously be zero? For instance, you used the curve $y=|x|$ as an example, but this doesn't have derivatives equal to $0$ at any point; rather, it is only undefined at $x = 0$, since this is a corner. Furthermore, I don't see any reference at all to the concept that my question referenced, which is the points being simultaneously being zero. [...]
– The Pointer
Jul 20 at 14:25
Again, I appreciate the answer, but this was written by someone who understands the concept, for someone who understands the concept. I cannot derive anything useful from this, despite putting in the effort to understand it.
– The Pointer
Jul 20 at 14:26
@ThePointer The point is that although the curve $y=|x|$ is not differentiable, the paramteriization $(x(t),y(t))$ that I gave is differentiable! So we need to not allow that parametrization. If add the condition about non-vanishing derivatives we rule out that parametrization, since $x'(0)=y'(0)=0$.
– David C. Ullrich
Jul 20 at 14:30
@ThePointer: The point here is that the curve is continuously differentiable everywhere, but it has a kink when you just look at the range of the curve. Our natural inclination is to consider such curves as not being smooth even though they are.
– copper.hat
Jul 20 at 14:31
1
@ThePointer Right.
– David C. Ullrich
Jul 20 at 15:30
Thanks for the response. So what is the connection between this and the condition that the derivatives of the points of the curve cannot simultaneously be zero? For instance, you used the curve $y=|x|$ as an example, but this doesn't have derivatives equal to $0$ at any point; rather, it is only undefined at $x = 0$, since this is a corner. Furthermore, I don't see any reference at all to the concept that my question referenced, which is the points being simultaneously being zero. [...]
– The Pointer
Jul 20 at 14:25
Thanks for the response. So what is the connection between this and the condition that the derivatives of the points of the curve cannot simultaneously be zero? For instance, you used the curve $y=|x|$ as an example, but this doesn't have derivatives equal to $0$ at any point; rather, it is only undefined at $x = 0$, since this is a corner. Furthermore, I don't see any reference at all to the concept that my question referenced, which is the points being simultaneously being zero. [...]
– The Pointer
Jul 20 at 14:25
Again, I appreciate the answer, but this was written by someone who understands the concept, for someone who understands the concept. I cannot derive anything useful from this, despite putting in the effort to understand it.
– The Pointer
Jul 20 at 14:26
Again, I appreciate the answer, but this was written by someone who understands the concept, for someone who understands the concept. I cannot derive anything useful from this, despite putting in the effort to understand it.
– The Pointer
Jul 20 at 14:26
@ThePointer The point is that although the curve $y=|x|$ is not differentiable, the paramteriization $(x(t),y(t))$ that I gave is differentiable! So we need to not allow that parametrization. If add the condition about non-vanishing derivatives we rule out that parametrization, since $x'(0)=y'(0)=0$.
– David C. Ullrich
Jul 20 at 14:30
@ThePointer The point is that although the curve $y=|x|$ is not differentiable, the paramteriization $(x(t),y(t))$ that I gave is differentiable! So we need to not allow that parametrization. If add the condition about non-vanishing derivatives we rule out that parametrization, since $x'(0)=y'(0)=0$.
– David C. Ullrich
Jul 20 at 14:30
@ThePointer: The point here is that the curve is continuously differentiable everywhere, but it has a kink when you just look at the range of the curve. Our natural inclination is to consider such curves as not being smooth even though they are.
– copper.hat
Jul 20 at 14:31
@ThePointer: The point here is that the curve is continuously differentiable everywhere, but it has a kink when you just look at the range of the curve. Our natural inclination is to consider such curves as not being smooth even though they are.
– copper.hat
Jul 20 at 14:31
1
1
@ThePointer Right.
– David C. Ullrich
Jul 20 at 15:30
@ThePointer Right.
– David C. Ullrich
Jul 20 at 15:30
 |Â
show 9 more comments
up vote
0
down vote
Because when $f'=g'=0$, the direction of the curve is no more defined. At such a point, the curve could change direction abruptly.
Example.
answered Jul 20 at 14:39
Yves Daoust
111k665204
Singular point: mathworld.wolfram.com/SingularPoint.html
– The Pointer
Jul 20 at 14:44
Thanks for the response. So the reason for this condition is that we want the parameterised curve to be unidirectional?
– The Pointer
Jul 20 at 14:45
@ThePointer: what do you mean by unidirectional ?
– Yves Daoust
Jul 20 at 14:47
@ThePointer: "in one direction" means following a straight line. I doubt this is what you have in mind. But I have completely rephrased my answer.
– Yves Daoust
Jul 20 at 14:50
1
@ThePointer This is undesirable behavior because if we don't disallow this behavior we find the the curve $y=|x|$ is "differentiable", as I showed in my answer. We certainly don't want that - if $y=|x|$ is a "differentiable" curve then differentiable curves are nothing like what we think they should be.
– David C. Ullrich
Jul 20 at 15:11
 |Â
show 5 more comments
up vote
0
down vote
Because when $f'=g'=0$, the direction of the curve is no more defined. At such a point, the curve could change direction abruptly.
Example.
answered Jul 20 at 14:39
Yves Daoust
111k665204
Singular point: mathworld.wolfram.com/SingularPoint.html
– The Pointer
Jul 20 at 14:44
Thanks for the response. So the reason for this condition is that we want the parameterised curve to be unidirectional?
– The Pointer
Jul 20 at 14:45
@ThePointer: what do you mean by unidirectional ?
– Yves Daoust
Jul 20 at 14:47
@ThePointer: "in one direction" means following a straight line. I doubt this is what you have in mind. But I have completely rephrased my answer.
– Yves Daoust
Jul 20 at 14:50
1
@ThePointer This is undesirable behavior because if we don't disallow this behavior we find the the curve $y=|x|$ is "differentiable", as I showed in my answer. We certainly don't want that - if $y=|x|$ is a "differentiable" curve then differentiable curves are nothing like what we think they should be.
– David C. Ullrich
Jul 20 at 15:11
 |Â
show 5 more comments
up vote
0
down vote
up vote
0
down vote
Because when $f'=g'=0$, the direction of the curve is no more defined. At such a point, the curve could change direction abruptly.
Example.
answered Jul 20 at 14:39
Yves Daoust
111k665204
Because when $f'=g'=0$, the direction of the curve is no more defined. At such a point, the curve could change direction abruptly.
Example.
answered Jul 20 at 14:39
Yves Daoust
111k665204
edited Jul 20 at 14:51
answered Jul 20 at 14:39
Yves Daoust
111k665204
answered Jul 20 at 14:39
Yves Daoust
111k665204
answered Jul 20 at 14:39
Yves Daoust
111k665204
111k665204
Singular point: mathworld.wolfram.com/SingularPoint.html
– The Pointer
Jul 20 at 14:44
Thanks for the response. So the reason for this condition is that we want the parameterised curve to be unidirectional?
– The Pointer
Jul 20 at 14:45
@ThePointer: what do you mean by unidirectional ?
– Yves Daoust
Jul 20 at 14:47
@ThePointer: "in one direction" means following a straight line. I doubt this is what you have in mind. But I have completely rephrased my answer.
– Yves Daoust
Jul 20 at 14:50
1
@ThePointer This is undesirable behavior because if we don't disallow this behavior we find the the curve $y=|x|$ is "differentiable", as I showed in my answer. We certainly don't want that - if $y=|x|$ is a "differentiable" curve then differentiable curves are nothing like what we think they should be.
– David C. Ullrich
Jul 20 at 15:11
 |Â
show 5 more comments
Singular point: mathworld.wolfram.com/SingularPoint.html
– The Pointer
Jul 20 at 14:44
Thanks for the response. So the reason for this condition is that we want the parameterised curve to be unidirectional?
– The Pointer
Jul 20 at 14:45
@ThePointer: what do you mean by unidirectional ?
– Yves Daoust
Jul 20 at 14:47
@ThePointer: "in one direction" means following a straight line. I doubt this is what you have in mind. But I have completely rephrased my answer.
– Yves Daoust
Jul 20 at 14:50
1
@ThePointer This is undesirable behavior because if we don't disallow this behavior we find the the curve $y=|x|$ is "differentiable", as I showed in my answer. We certainly don't want that - if $y=|x|$ is a "differentiable" curve then differentiable curves are nothing like what we think they should be.
– David C. Ullrich
Jul 20 at 15:11
Singular point: mathworld.wolfram.com/SingularPoint.html
– The Pointer
Jul 20 at 14:44
Singular point: mathworld.wolfram.com/SingularPoint.html
– The Pointer
Jul 20 at 14:44
Thanks for the response. So the reason for this condition is that we want the parameterised curve to be unidirectional?
– The Pointer
Jul 20 at 14:45
Thanks for the response. So the reason for this condition is that we want the parameterised curve to be unidirectional?
– The Pointer
Jul 20 at 14:45
@ThePointer: what do you mean by unidirectional ?
– Yves Daoust
Jul 20 at 14:47
@ThePointer: what do you mean by unidirectional ?
– Yves Daoust
Jul 20 at 14:47
@ThePointer: "in one direction" means following a straight line. I doubt this is what you have in mind. But I have completely rephrased my answer.
– Yves Daoust
Jul 20 at 14:50
@ThePointer: "in one direction" means following a straight line. I doubt this is what you have in mind. But I have completely rephrased my answer.
– Yves Daoust
Jul 20 at 14:50
1
1
@ThePointer This is undesirable behavior because if we don't disallow this behavior we find the the curve $y=|x|$ is "differentiable", as I showed in my answer. We certainly don't want that - if $y=|x|$ is a "differentiable" curve then differentiable curves are nothing like what we think they should be.
– David C. Ullrich
Jul 20 at 15:11
@ThePointer This is undesirable behavior because if we don't disallow this behavior we find the the curve $y=|x|$ is "differentiable", as I showed in my answer. We certainly don't want that - if $y=|x|$ is a "differentiable" curve then differentiable curves are nothing like what we think they should be.
– David C. Ullrich
Jul 20 at 15:11
 |Â
show 5 more comments
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It means the curve has a non zero speed for each value of the parameter. This means that the curve in an injective map so that any point on the curve defines a unique value of the parameter.
– copper.hat
Jul 20 at 14:12
@copper.hat So for example $tmapsto(cos(t),sin(t))$ is injective? Ok, you meant locally injective. But I really don't think this is exactly the point - the "bad" parametrization of the curve $y=|x|$ in my answer is injective...
– David C. Ullrich
Jul 20 at 14:14
@DavidC.Ullrich: I meant locally, it was more of a comment than a complete answer.
– copper.hat
Jul 20 at 14:21
@DavidC.Ullrich: I guess I would settle for rectifiable, but as your answer nicely demonstrates, this allows a lot of curves that don't fit with our intuition.
– copper.hat
Jul 20 at 14:27
@copper.hat In fact I'm pretty sure that any rectifiable curve has a $C^1$ parametrization...
– David C. Ullrich
Jul 20 at 14:39