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Showing that a given curve is a circumference
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Consider the curve $beta(s)=(frac45 cos(s),1-sin(s), frac-35 cos(s))$. Prove that the trace of $beta$ is a circumference.
My approach:
$(*)$ If there exists a point $p in mathbbR^3$ such that for any point $beta(s)$ the distance between $beta(s)$ and $p$ is constant, then the trace of $beta(s)$ is either a circumference or a sphere (is this true?).
I have already proved that $beta$ is planar, showing that its torsion is constantly zero, but I have failed to prove $(*).$ Any hints or different approaches?
differential-geometry
edited Jul 15 at 0:12
Michael Hardy
204k23186463
asked Jul 15 at 0:06
Yagger
5321315
add a comment |Â
up vote
1
down vote
favorite
Consider the curve $beta(s)=(frac45 cos(s),1-sin(s), frac-35 cos(s))$. Prove that the trace of $beta$ is a circumference.
My approach:
$(*)$ If there exists a point $p in mathbbR^3$ such that for any point $beta(s)$ the distance between $beta(s)$ and $p$ is constant, then the trace of $beta(s)$ is either a circumference or a sphere (is this true?).
I have already proved that $beta$ is planar, showing that its torsion is constantly zero, but I have failed to prove $(*).$ Any hints or different approaches?
differential-geometry
edited Jul 15 at 0:12
Michael Hardy
204k23186463
asked Jul 15 at 0:06
Yagger
5321315
Try guessing what the center point might be. There are clues in the parameterization of the curve. Failing that, you could find the center from three noncolinear points on the curve.
– amd
Jul 15 at 0:14
What would $ 1over 2 (beta(0)+beta(pi))$ correspond to?
– copper.hat
Jul 15 at 0:14
By "a circumference" do you mean specifically of a circle?
– Michael Hardy
Jul 15 at 0:18
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider the curve $beta(s)=(frac45 cos(s),1-sin(s), frac-35 cos(s))$. Prove that the trace of $beta$ is a circumference.
My approach:
$(*)$ If there exists a point $p in mathbbR^3$ such that for any point $beta(s)$ the distance between $beta(s)$ and $p$ is constant, then the trace of $beta(s)$ is either a circumference or a sphere (is this true?).
I have already proved that $beta$ is planar, showing that its torsion is constantly zero, but I have failed to prove $(*).$ Any hints or different approaches?
differential-geometry
edited Jul 15 at 0:12
Michael Hardy
204k23186463
asked Jul 15 at 0:06
Yagger
5321315
Consider the curve $beta(s)=(frac45 cos(s),1-sin(s), frac-35 cos(s))$. Prove that the trace of $beta$ is a circumference.
My approach:
$(*)$ If there exists a point $p in mathbbR^3$ such that for any point $beta(s)$ the distance between $beta(s)$ and $p$ is constant, then the trace of $beta(s)$ is either a circumference or a sphere (is this true?).
I have already proved that $beta$ is planar, showing that its torsion is constantly zero, but I have failed to prove $(*).$ Any hints or different approaches?
differential-geometry
edited Jul 15 at 0:12
Michael Hardy
204k23186463
asked Jul 15 at 0:06
Yagger
5321315
edited Jul 15 at 0:12
Michael Hardy
204k23186463
edited Jul 15 at 0:12
Michael Hardy
204k23186463
edited Jul 15 at 0:12
Michael Hardy
204k23186463
204k23186463
asked Jul 15 at 0:06
Yagger
5321315
asked Jul 15 at 0:06
Yagger
5321315
asked Jul 15 at 0:06
Yagger
5321315
5321315
Try guessing what the center point might be. There are clues in the parameterization of the curve. Failing that, you could find the center from three noncolinear points on the curve.
– amd
Jul 15 at 0:14
What would $ 1over 2 (beta(0)+beta(pi))$ correspond to?
– copper.hat
Jul 15 at 0:14
By "a circumference" do you mean specifically of a circle?
– Michael Hardy
Jul 15 at 0:18
add a comment |Â
Try guessing what the center point might be. There are clues in the parameterization of the curve. Failing that, you could find the center from three noncolinear points on the curve.
– amd
Jul 15 at 0:14
What would $ 1over 2 (beta(0)+beta(pi))$ correspond to?
– copper.hat
Jul 15 at 0:14
By "a circumference" do you mean specifically of a circle?
– Michael Hardy
Jul 15 at 0:18
Try guessing what the center point might be. There are clues in the parameterization of the curve. Failing that, you could find the center from three noncolinear points on the curve.
– amd
Jul 15 at 0:14
Try guessing what the center point might be. There are clues in the parameterization of the curve. Failing that, you could find the center from three noncolinear points on the curve.
– amd
Jul 15 at 0:14
What would $ 1over 2 (beta(0)+beta(pi))$ correspond to?
– copper.hat
Jul 15 at 0:14
What would $ 1over 2 (beta(0)+beta(pi))$ correspond to?
– copper.hat
Jul 15 at 0:14
By "a circumference" do you mean specifically of a circle?
– Michael Hardy
Jul 15 at 0:18
By "a circumference" do you mean specifically of a circle?
– Michael Hardy
Jul 15 at 0:18
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
$$ left(frac45 cos s,1-sin s, frac-35 cos sright) tag 1 $$
The set of all points of the form $big(x,y,(-3/4)xbig)$ for $x,yinmathbb R$ is a plane.
Let $x = dfrac 4 5 cos s$ and let $y= 1-sin s,$ and we see that as $s$ varies, all points of the curve $(1)$ lie within that plane.
As $s$ varies, the average value of $(1)$ is $(0,1,0),$ so if $(1)$ is a circle then that must be the center. Show that the distance from a point of the form $(1)$ to $(0,1,0)$ does not depend on $s.$
answered Jul 15 at 0:27
Michael Hardy
204k23186463
add a comment |Â
up vote
1
down vote
Model the circle as $p+ (cos t) v_1+(sin t) v_2$, where $v_1 bot v_2$ and
$|v_1|= |v_2|$.
Once you compute $p$, it is not hard to guess a suitable $v_1, v_2$ for the
given $beta$.
Note that the period is $2 pi$.
Compute $p=1 over 2 (beta(0)+ beta(pi))= (0,1,0)$. We can read off $v_1= (4 over 4, 0, - 3 over 5)$ and $v_2 = (0,-1,0)$.
answered Jul 15 at 0:17
copper.hat
122k557156
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$ left(frac45 cos s,1-sin s, frac-35 cos sright) tag 1 $$
The set of all points of the form $big(x,y,(-3/4)xbig)$ for $x,yinmathbb R$ is a plane.
Let $x = dfrac 4 5 cos s$ and let $y= 1-sin s,$ and we see that as $s$ varies, all points of the curve $(1)$ lie within that plane.
As $s$ varies, the average value of $(1)$ is $(0,1,0),$ so if $(1)$ is a circle then that must be the center. Show that the distance from a point of the form $(1)$ to $(0,1,0)$ does not depend on $s.$
answered Jul 15 at 0:27
Michael Hardy
204k23186463
add a comment |Â
up vote
1
down vote
accepted
$$ left(frac45 cos s,1-sin s, frac-35 cos sright) tag 1 $$
The set of all points of the form $big(x,y,(-3/4)xbig)$ for $x,yinmathbb R$ is a plane.
Let $x = dfrac 4 5 cos s$ and let $y= 1-sin s,$ and we see that as $s$ varies, all points of the curve $(1)$ lie within that plane.
As $s$ varies, the average value of $(1)$ is $(0,1,0),$ so if $(1)$ is a circle then that must be the center. Show that the distance from a point of the form $(1)$ to $(0,1,0)$ does not depend on $s.$
answered Jul 15 at 0:27
Michael Hardy
204k23186463
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$ left(frac45 cos s,1-sin s, frac-35 cos sright) tag 1 $$
The set of all points of the form $big(x,y,(-3/4)xbig)$ for $x,yinmathbb R$ is a plane.
Let $x = dfrac 4 5 cos s$ and let $y= 1-sin s,$ and we see that as $s$ varies, all points of the curve $(1)$ lie within that plane.
As $s$ varies, the average value of $(1)$ is $(0,1,0),$ so if $(1)$ is a circle then that must be the center. Show that the distance from a point of the form $(1)$ to $(0,1,0)$ does not depend on $s.$
answered Jul 15 at 0:27
Michael Hardy
204k23186463
$$ left(frac45 cos s,1-sin s, frac-35 cos sright) tag 1 $$
The set of all points of the form $big(x,y,(-3/4)xbig)$ for $x,yinmathbb R$ is a plane.
Let $x = dfrac 4 5 cos s$ and let $y= 1-sin s,$ and we see that as $s$ varies, all points of the curve $(1)$ lie within that plane.
As $s$ varies, the average value of $(1)$ is $(0,1,0),$ so if $(1)$ is a circle then that must be the center. Show that the distance from a point of the form $(1)$ to $(0,1,0)$ does not depend on $s.$
answered Jul 15 at 0:27
Michael Hardy
204k23186463
answered Jul 15 at 0:27
Michael Hardy
204k23186463
answered Jul 15 at 0:27
Michael Hardy
204k23186463
answered Jul 15 at 0:27
Michael Hardy
204k23186463
204k23186463
add a comment |Â
add a comment |Â
up vote
1
down vote
Model the circle as $p+ (cos t) v_1+(sin t) v_2$, where $v_1 bot v_2$ and
$|v_1|= |v_2|$.
Once you compute $p$, it is not hard to guess a suitable $v_1, v_2$ for the
given $beta$.
Note that the period is $2 pi$.
Compute $p=1 over 2 (beta(0)+ beta(pi))= (0,1,0)$. We can read off $v_1= (4 over 4, 0, - 3 over 5)$ and $v_2 = (0,-1,0)$.
answered Jul 15 at 0:17
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
Model the circle as $p+ (cos t) v_1+(sin t) v_2$, where $v_1 bot v_2$ and
$|v_1|= |v_2|$.
Once you compute $p$, it is not hard to guess a suitable $v_1, v_2$ for the
given $beta$.
Note that the period is $2 pi$.
Compute $p=1 over 2 (beta(0)+ beta(pi))= (0,1,0)$. We can read off $v_1= (4 over 4, 0, - 3 over 5)$ and $v_2 = (0,-1,0)$.
answered Jul 15 at 0:17
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Model the circle as $p+ (cos t) v_1+(sin t) v_2$, where $v_1 bot v_2$ and
$|v_1|= |v_2|$.
Once you compute $p$, it is not hard to guess a suitable $v_1, v_2$ for the
given $beta$.
Note that the period is $2 pi$.
Compute $p=1 over 2 (beta(0)+ beta(pi))= (0,1,0)$. We can read off $v_1= (4 over 4, 0, - 3 over 5)$ and $v_2 = (0,-1,0)$.
answered Jul 15 at 0:17
copper.hat
122k557156
Model the circle as $p+ (cos t) v_1+(sin t) v_2$, where $v_1 bot v_2$ and
$|v_1|= |v_2|$.
Once you compute $p$, it is not hard to guess a suitable $v_1, v_2$ for the
given $beta$.
Note that the period is $2 pi$.
Compute $p=1 over 2 (beta(0)+ beta(pi))= (0,1,0)$. We can read off $v_1= (4 over 4, 0, - 3 over 5)$ and $v_2 = (0,-1,0)$.
answered Jul 15 at 0:17
copper.hat
122k557156
answered Jul 15 at 0:17
copper.hat
122k557156
answered Jul 15 at 0:17
copper.hat
122k557156
answered Jul 15 at 0:17
copper.hat
122k557156
122k557156
add a comment |Â
add a comment |Â
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Try guessing what the center point might be. There are clues in the parameterization of the curve. Failing that, you could find the center from three noncolinear points on the curve.
– amd
Jul 15 at 0:14
What would $ 1over 2 (beta(0)+beta(pi))$ correspond to?
– copper.hat
Jul 15 at 0:14
By "a circumference" do you mean specifically of a circle?
– Michael Hardy
Jul 15 at 0:18