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Curvature of curve defined in terms of unusual function
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My textbook, Oneill's Elementary Differential Geometry, defines a curve in terms of the function $f$:
$$f(t)=begincases0, &textfor tleq0\
e^-frac1t^2, &textfor tgt0endcases$$
The curve is defined $$a(t)=t,f(t),f(-t)$$
The book states that the curvature is only zero at $t=0$. But when I calculate it, I get that the curvature is proportional to the length of $$a'(t)times a''(t)$$
I get the first and second components of this product equal to zero. For the third component, for $tgt0$
I get $$4frace^-1/t^2t^6-6frace^-1/t^2t^4$$
Setting this equal to zero, I get $$t=sqrt2/3$$
So my result is at variance with the textbook. Have I made an error? If so, where?
differential-geometry
asked Jul 25 at 17:45
Gene Naden
203
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0
down vote
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My textbook, Oneill's Elementary Differential Geometry, defines a curve in terms of the function $f$:
$$f(t)=begincases0, &textfor tleq0\
e^-frac1t^2, &textfor tgt0endcases$$
The curve is defined $$a(t)=t,f(t),f(-t)$$
The book states that the curvature is only zero at $t=0$. But when I calculate it, I get that the curvature is proportional to the length of $$a'(t)times a''(t)$$
I get the first and second components of this product equal to zero. For the third component, for $tgt0$
I get $$4frace^-1/t^2t^6-6frace^-1/t^2t^4$$
Setting this equal to zero, I get $$t=sqrt2/3$$
So my result is at variance with the textbook. Have I made an error? If so, where?
differential-geometry
asked Jul 25 at 17:45
Gene Naden
203
I haven't checked your calculation, but remember that if $t>0$, then $f(-t)=0$, and if $t<0$, then $f(t)=0$.
– Ted Shifrin
Jul 25 at 22:56
Right, so for t>0,
– Gene Naden
Jul 25 at 23:54
Right, so for t>0, $alphaprime=0,frace^-1/t^2t^6,0-frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:02
I meant to say $alpha^prime=1,frac2e^-1/t^2t^3,0$ and $alphaprimeprime=0,4frace^-1/t^2t^6-6frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:11
Ah, yes, of course, you're right. If you graph the curve, you have an inflection point, hence a point with $kappa=0$. The interesting thing about this curve is not its curvature, but that it has torsion $tau=0$ everywhere except at $t=0$, and yet it is not planar.
– Ted Shifrin
Jul 26 at 0:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My textbook, Oneill's Elementary Differential Geometry, defines a curve in terms of the function $f$:
$$f(t)=begincases0, &textfor tleq0\
e^-frac1t^2, &textfor tgt0endcases$$
The curve is defined $$a(t)=t,f(t),f(-t)$$
The book states that the curvature is only zero at $t=0$. But when I calculate it, I get that the curvature is proportional to the length of $$a'(t)times a''(t)$$
I get the first and second components of this product equal to zero. For the third component, for $tgt0$
I get $$4frace^-1/t^2t^6-6frace^-1/t^2t^4$$
Setting this equal to zero, I get $$t=sqrt2/3$$
So my result is at variance with the textbook. Have I made an error? If so, where?
differential-geometry
asked Jul 25 at 17:45
Gene Naden
203
My textbook, Oneill's Elementary Differential Geometry, defines a curve in terms of the function $f$:
$$f(t)=begincases0, &textfor tleq0\
e^-frac1t^2, &textfor tgt0endcases$$
The curve is defined $$a(t)=t,f(t),f(-t)$$
The book states that the curvature is only zero at $t=0$. But when I calculate it, I get that the curvature is proportional to the length of $$a'(t)times a''(t)$$
I get the first and second components of this product equal to zero. For the third component, for $tgt0$
I get $$4frace^-1/t^2t^6-6frace^-1/t^2t^4$$
Setting this equal to zero, I get $$t=sqrt2/3$$
So my result is at variance with the textbook. Have I made an error? If so, where?
differential-geometry
asked Jul 25 at 17:45
Gene Naden
203
edited Jul 25 at 18:32
asked Jul 25 at 17:45
Gene Naden
203
asked Jul 25 at 17:45
Gene Naden
203
asked Jul 25 at 17:45
Gene Naden
203
203
I haven't checked your calculation, but remember that if $t>0$, then $f(-t)=0$, and if $t<0$, then $f(t)=0$.
– Ted Shifrin
Jul 25 at 22:56
Right, so for t>0,
– Gene Naden
Jul 25 at 23:54
Right, so for t>0, $alphaprime=0,frace^-1/t^2t^6,0-frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:02
I meant to say $alpha^prime=1,frac2e^-1/t^2t^3,0$ and $alphaprimeprime=0,4frace^-1/t^2t^6-6frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:11
Ah, yes, of course, you're right. If you graph the curve, you have an inflection point, hence a point with $kappa=0$. The interesting thing about this curve is not its curvature, but that it has torsion $tau=0$ everywhere except at $t=0$, and yet it is not planar.
– Ted Shifrin
Jul 26 at 0:28
add a comment |Â
I haven't checked your calculation, but remember that if $t>0$, then $f(-t)=0$, and if $t<0$, then $f(t)=0$.
– Ted Shifrin
Jul 25 at 22:56
Right, so for t>0,
– Gene Naden
Jul 25 at 23:54
Right, so for t>0, $alphaprime=0,frace^-1/t^2t^6,0-frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:02
I meant to say $alpha^prime=1,frac2e^-1/t^2t^3,0$ and $alphaprimeprime=0,4frace^-1/t^2t^6-6frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:11
Ah, yes, of course, you're right. If you graph the curve, you have an inflection point, hence a point with $kappa=0$. The interesting thing about this curve is not its curvature, but that it has torsion $tau=0$ everywhere except at $t=0$, and yet it is not planar.
– Ted Shifrin
Jul 26 at 0:28
I haven't checked your calculation, but remember that if $t>0$, then $f(-t)=0$, and if $t<0$, then $f(t)=0$.
– Ted Shifrin
Jul 25 at 22:56
I haven't checked your calculation, but remember that if $t>0$, then $f(-t)=0$, and if $t<0$, then $f(t)=0$.
– Ted Shifrin
Jul 25 at 22:56
Right, so for t>0,
– Gene Naden
Jul 25 at 23:54
Right, so for t>0,
– Gene Naden
Jul 25 at 23:54
Right, so for t>0, $alphaprime=0,frace^-1/t^2t^6,0-frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:02
Right, so for t>0, $alphaprime=0,frace^-1/t^2t^6,0-frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:02
I meant to say $alpha^prime=1,frac2e^-1/t^2t^3,0$ and $alphaprimeprime=0,4frace^-1/t^2t^6-6frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:11
I meant to say $alpha^prime=1,frac2e^-1/t^2t^3,0$ and $alphaprimeprime=0,4frace^-1/t^2t^6-6frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:11
Ah, yes, of course, you're right. If you graph the curve, you have an inflection point, hence a point with $kappa=0$. The interesting thing about this curve is not its curvature, but that it has torsion $tau=0$ everywhere except at $t=0$, and yet it is not planar.
– Ted Shifrin
Jul 26 at 0:28
Ah, yes, of course, you're right. If you graph the curve, you have an inflection point, hence a point with $kappa=0$. The interesting thing about this curve is not its curvature, but that it has torsion $tau=0$ everywhere except at $t=0$, and yet it is not planar.
– Ted Shifrin
Jul 26 at 0:28
add a comment |Â
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I haven't checked your calculation, but remember that if $t>0$, then $f(-t)=0$, and if $t<0$, then $f(t)=0$.
– Ted Shifrin
Jul 25 at 22:56
Right, so for t>0,
– Gene Naden
Jul 25 at 23:54
Right, so for t>0, $alphaprime=0,frace^-1/t^2t^6,0-frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:02
I meant to say $alpha^prime=1,frac2e^-1/t^2t^3,0$ and $alphaprimeprime=0,4frace^-1/t^2t^6-6frace^-1/t^2t^4,0$
– Gene Naden
Jul 26 at 0:11
Ah, yes, of course, you're right. If you graph the curve, you have an inflection point, hence a point with $kappa=0$. The interesting thing about this curve is not its curvature, but that it has torsion $tau=0$ everywhere except at $t=0$, and yet it is not planar.
– Ted Shifrin
Jul 26 at 0:28