Simple curve in the plane

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Given a parameterized plane curve G which is regular at point $t$. Prove that we can find a compact interval which contains $t$ and where the curve is simple.



The curve is regular so that the derivatives of $x(t)$ and $y(t)$ exist and are not equal to $0$ at the same time. That means that $x(t)$ and $y(t)$ are continuous at point $t$. But this doesn't help me a lot.




calculus




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    I think you assume that the curve is of class $C^1$. Assume for definiteness that $x'(t)ne0$. Then there is an open interval $(alpha,beta)ni t$ such that $x'(s)ne0$ for all $sin(alpha,beta)$. The mapping $(alpha,beta)ni smapsto x(s)in(x_1,x_2)$ has a $C^1$ inverse (call this inverse $s^-1$). So the interval of the curve corresponding to $sin(alpha,beta)$ is the graph of a $C^1$ function $(x_1,x_2)ni xmapsto y(s^-1(x))$.
    – user539887
    Aug 15 at 11:07














up vote
0
down vote

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Given a parameterized plane curve G which is regular at point $t$. Prove that we can find a compact interval which contains $t$ and where the curve is simple.



The curve is regular so that the derivatives of $x(t)$ and $y(t)$ exist and are not equal to $0$ at the same time. That means that $x(t)$ and $y(t)$ are continuous at point $t$. But this doesn't help me a lot.




calculus




share|cite|improve this question


















  • 1




    I think you assume that the curve is of class $C^1$. Assume for definiteness that $x'(t)ne0$. Then there is an open interval $(alpha,beta)ni t$ such that $x'(s)ne0$ for all $sin(alpha,beta)$. The mapping $(alpha,beta)ni smapsto x(s)in(x_1,x_2)$ has a $C^1$ inverse (call this inverse $s^-1$). So the interval of the curve corresponding to $sin(alpha,beta)$ is the graph of a $C^1$ function $(x_1,x_2)ni xmapsto y(s^-1(x))$.
    – user539887
    Aug 15 at 11:07












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given a parameterized plane curve G which is regular at point $t$. Prove that we can find a compact interval which contains $t$ and where the curve is simple.



The curve is regular so that the derivatives of $x(t)$ and $y(t)$ exist and are not equal to $0$ at the same time. That means that $x(t)$ and $y(t)$ are continuous at point $t$. But this doesn't help me a lot.




calculus




share|cite|improve this question














Given a parameterized plane curve G which is regular at point $t$. Prove that we can find a compact interval which contains $t$ and where the curve is simple.



The curve is regular so that the derivatives of $x(t)$ and $y(t)$ exist and are not equal to $0$ at the same time. That means that $x(t)$ and $y(t)$ are continuous at point $t$. But this doesn't help me a lot.





calculus





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edited Aug 15 at 11:26









Abcd

2,5031726




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asked Aug 15 at 10:52









spyer

978




978







  • 1




    I think you assume that the curve is of class $C^1$. Assume for definiteness that $x'(t)ne0$. Then there is an open interval $(alpha,beta)ni t$ such that $x'(s)ne0$ for all $sin(alpha,beta)$. The mapping $(alpha,beta)ni smapsto x(s)in(x_1,x_2)$ has a $C^1$ inverse (call this inverse $s^-1$). So the interval of the curve corresponding to $sin(alpha,beta)$ is the graph of a $C^1$ function $(x_1,x_2)ni xmapsto y(s^-1(x))$.
    – user539887
    Aug 15 at 11:07












  • 1




    I think you assume that the curve is of class $C^1$. Assume for definiteness that $x'(t)ne0$. Then there is an open interval $(alpha,beta)ni t$ such that $x'(s)ne0$ for all $sin(alpha,beta)$. The mapping $(alpha,beta)ni smapsto x(s)in(x_1,x_2)$ has a $C^1$ inverse (call this inverse $s^-1$). So the interval of the curve corresponding to $sin(alpha,beta)$ is the graph of a $C^1$ function $(x_1,x_2)ni xmapsto y(s^-1(x))$.
    – user539887
    Aug 15 at 11:07







1




1




I think you assume that the curve is of class $C^1$. Assume for definiteness that $x'(t)ne0$. Then there is an open interval $(alpha,beta)ni t$ such that $x'(s)ne0$ for all $sin(alpha,beta)$. The mapping $(alpha,beta)ni smapsto x(s)in(x_1,x_2)$ has a $C^1$ inverse (call this inverse $s^-1$). So the interval of the curve corresponding to $sin(alpha,beta)$ is the graph of a $C^1$ function $(x_1,x_2)ni xmapsto y(s^-1(x))$.
– user539887
Aug 15 at 11:07




I think you assume that the curve is of class $C^1$. Assume for definiteness that $x'(t)ne0$. Then there is an open interval $(alpha,beta)ni t$ such that $x'(s)ne0$ for all $sin(alpha,beta)$. The mapping $(alpha,beta)ni smapsto x(s)in(x_1,x_2)$ has a $C^1$ inverse (call this inverse $s^-1$). So the interval of the curve corresponding to $sin(alpha,beta)$ is the graph of a $C^1$ function $(x_1,x_2)ni xmapsto y(s^-1(x))$.
– user539887
Aug 15 at 11:07















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