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Is this curve rectifiable?
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I am given the task to show, that the curve:
$f:[0,1]tomathbbR^2$, $f(x)=begincases (0,0)spacetextif x=0\ (x,x^2cos(fracpix))spacetextif x > 0endcases$
$f$ is differentiable and $f$ is not
rectifiable.
But I actually think, that it might be rectifiable.
To show, that the curve is differentiable is simple. I have to show, that the coordinate-functions are differentiable.
I just go over $g(x)=x^2cos(fracpix)$ and show that is differentiable in $x_0=0$. Obviously it is differentiable for $x>0$.
$lim_hto 0 fracg(x_0+h)-g(x_0)h=lim_hto 0 frach^2cos(fracpih)-0h=lim_hto 0 hcos(fracpih)leqlim_hto 0 h=0$
Therefor differentiable in $x_0=0$.
I am concerned, that $f$ might be rectifiable since the it is $x^2cos(fracpicolorredx)$ and not $x^2cos(fracpicolorredx^2)$.
Else I would just take the standard sequence $a_k=frac1sqrt2k$, which gives me $frac12kcos(2kpi)=frac12k$ and we are left with a harmonic series, which diverges.
But in the given case this is not possible.
We can not choose such a sequence which converges to 0 slow enough, and are left with something divergent.
Can you back this up? Or is this curve indeed not rectifiable?
If it is rectifiable it would be interesting to know its length.
curves
asked Aug 5 at 23:20
Cornman
2,60721128
add a comment |Â
up vote
0
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favorite
I am given the task to show, that the curve:
$f:[0,1]tomathbbR^2$, $f(x)=begincases (0,0)spacetextif x=0\ (x,x^2cos(fracpix))spacetextif x > 0endcases$
$f$ is differentiable and $f$ is not
rectifiable.
But I actually think, that it might be rectifiable.
To show, that the curve is differentiable is simple. I have to show, that the coordinate-functions are differentiable.
I just go over $g(x)=x^2cos(fracpix)$ and show that is differentiable in $x_0=0$. Obviously it is differentiable for $x>0$.
$lim_hto 0 fracg(x_0+h)-g(x_0)h=lim_hto 0 frach^2cos(fracpih)-0h=lim_hto 0 hcos(fracpih)leqlim_hto 0 h=0$
Therefor differentiable in $x_0=0$.
I am concerned, that $f$ might be rectifiable since the it is $x^2cos(fracpicolorredx)$ and not $x^2cos(fracpicolorredx^2)$.
Else I would just take the standard sequence $a_k=frac1sqrt2k$, which gives me $frac12kcos(2kpi)=frac12k$ and we are left with a harmonic series, which diverges.
But in the given case this is not possible.
We can not choose such a sequence which converges to 0 slow enough, and are left with something divergent.
Can you back this up? Or is this curve indeed not rectifiable?
If it is rectifiable it would be interesting to know its length.
curves
asked Aug 5 at 23:20
Cornman
2,60721128
2
I think you are right. Since the integral $int_0^1 sqrt1+(2xcos(fracpix)+pi sin(fracpix))^2 dx$ is finite, this should be the length of the curve. (about $2.586$)
– user35359
Aug 6 at 1:29
You can not integrate this by hand, can you?
– Cornman
Aug 6 at 14:35
1
Not sure, but I wouldn't know how and Mathematica doesn't give me a closed form solution either.
– user35359
Aug 6 at 15:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given the task to show, that the curve:
$f:[0,1]tomathbbR^2$, $f(x)=begincases (0,0)spacetextif x=0\ (x,x^2cos(fracpix))spacetextif x > 0endcases$
$f$ is differentiable and $f$ is not
rectifiable.
But I actually think, that it might be rectifiable.
To show, that the curve is differentiable is simple. I have to show, that the coordinate-functions are differentiable.
I just go over $g(x)=x^2cos(fracpix)$ and show that is differentiable in $x_0=0$. Obviously it is differentiable for $x>0$.
$lim_hto 0 fracg(x_0+h)-g(x_0)h=lim_hto 0 frach^2cos(fracpih)-0h=lim_hto 0 hcos(fracpih)leqlim_hto 0 h=0$
Therefor differentiable in $x_0=0$.
I am concerned, that $f$ might be rectifiable since the it is $x^2cos(fracpicolorredx)$ and not $x^2cos(fracpicolorredx^2)$.
Else I would just take the standard sequence $a_k=frac1sqrt2k$, which gives me $frac12kcos(2kpi)=frac12k$ and we are left with a harmonic series, which diverges.
But in the given case this is not possible.
We can not choose such a sequence which converges to 0 slow enough, and are left with something divergent.
Can you back this up? Or is this curve indeed not rectifiable?
If it is rectifiable it would be interesting to know its length.
curves
asked Aug 5 at 23:20
Cornman
2,60721128
I am given the task to show, that the curve:
$f:[0,1]tomathbbR^2$, $f(x)=begincases (0,0)spacetextif x=0\ (x,x^2cos(fracpix))spacetextif x > 0endcases$
$f$ is differentiable and $f$ is not
rectifiable.
But I actually think, that it might be rectifiable.
To show, that the curve is differentiable is simple. I have to show, that the coordinate-functions are differentiable.
I just go over $g(x)=x^2cos(fracpix)$ and show that is differentiable in $x_0=0$. Obviously it is differentiable for $x>0$.
$lim_hto 0 fracg(x_0+h)-g(x_0)h=lim_hto 0 frach^2cos(fracpih)-0h=lim_hto 0 hcos(fracpih)leqlim_hto 0 h=0$
Therefor differentiable in $x_0=0$.
I am concerned, that $f$ might be rectifiable since the it is $x^2cos(fracpicolorredx)$ and not $x^2cos(fracpicolorredx^2)$.
Else I would just take the standard sequence $a_k=frac1sqrt2k$, which gives me $frac12kcos(2kpi)=frac12k$ and we are left with a harmonic series, which diverges.
But in the given case this is not possible.
We can not choose such a sequence which converges to 0 slow enough, and are left with something divergent.
Can you back this up? Or is this curve indeed not rectifiable?
If it is rectifiable it would be interesting to know its length.
curves
asked Aug 5 at 23:20
Cornman
2,60721128
edited Aug 6 at 1:29
asked Aug 5 at 23:20
Cornman
2,60721128
asked Aug 5 at 23:20
Cornman
2,60721128
asked Aug 5 at 23:20
Cornman
2,60721128
2,60721128
2
I think you are right. Since the integral $int_0^1 sqrt1+(2xcos(fracpix)+pi sin(fracpix))^2 dx$ is finite, this should be the length of the curve. (about $2.586$)
– user35359
Aug 6 at 1:29
You can not integrate this by hand, can you?
– Cornman
Aug 6 at 14:35
1
Not sure, but I wouldn't know how and Mathematica doesn't give me a closed form solution either.
– user35359
Aug 6 at 15:33
add a comment |Â
2
I think you are right. Since the integral $int_0^1 sqrt1+(2xcos(fracpix)+pi sin(fracpix))^2 dx$ is finite, this should be the length of the curve. (about $2.586$)
– user35359
Aug 6 at 1:29
You can not integrate this by hand, can you?
– Cornman
Aug 6 at 14:35
1
Not sure, but I wouldn't know how and Mathematica doesn't give me a closed form solution either.
– user35359
Aug 6 at 15:33
2
2
I think you are right. Since the integral $int_0^1 sqrt1+(2xcos(fracpix)+pi sin(fracpix))^2 dx$ is finite, this should be the length of the curve. (about $2.586$)
– user35359
Aug 6 at 1:29
I think you are right. Since the integral $int_0^1 sqrt1+(2xcos(fracpix)+pi sin(fracpix))^2 dx$ is finite, this should be the length of the curve. (about $2.586$)
– user35359
Aug 6 at 1:29
You can not integrate this by hand, can you?
– Cornman
Aug 6 at 14:35
You can not integrate this by hand, can you?
– Cornman
Aug 6 at 14:35
1
1
Not sure, but I wouldn't know how and Mathematica doesn't give me a closed form solution either.
– user35359
Aug 6 at 15:33
Not sure, but I wouldn't know how and Mathematica doesn't give me a closed form solution either.
– user35359
Aug 6 at 15:33
add a comment |Â
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2
I think you are right. Since the integral $int_0^1 sqrt1+(2xcos(fracpix)+pi sin(fracpix))^2 dx$ is finite, this should be the length of the curve. (about $2.586$)
– user35359
Aug 6 at 1:29
You can not integrate this by hand, can you?
– Cornman
Aug 6 at 14:35
1
Not sure, but I wouldn't know how and Mathematica doesn't give me a closed form solution either.
– user35359
Aug 6 at 15:33