Mixing
Find The Curvature of the given curve
0
$begingroup$
Find the curvature of the curve :
$ x = t - sin{t} $ , $y = 1 - cos{t} $ , $z=4sin{dfrac{t}{2}}$.
I have been trying to find the curvature using the formula
$ {k_{1}}^2 = dfrac{{begin{vmatrix} x''& y'' \ x' & y' end{vmatrix}}^2 +{begin{vmatrix} y'' & z'' \ y' & z' end{vmatrix}}^2 + {begin{vmatrix} z'' & x'' \ z' & x' end{vmatrix}}^2}{({x'}^{2}+{y'}^{2}+{z'}^{2})^3}.$
differential-geometry
edited Jan 3 at 14:26
mathcounterexamples.net
25.8k21954
asked Jan 3 at 14:08
user505797user505797
33
$endgroup$
add a comment |
0
$begingroup$
Find the curvature of the curve :
$ x = t - sin{t} $ , $y = 1 - cos{t} $ , $z=4sin{dfrac{t}{2}}$.
I have been trying to find the curvature using the formula
$ {k_{1}}^2 = dfrac{{begin{vmatrix} x''& y'' \ x' & y' end{vmatrix}}^2 +{begin{vmatrix} y'' & z'' \ y' & z' end{vmatrix}}^2 + {begin{vmatrix} z'' & x'' \ z' & x' end{vmatrix}}^2}{({x'}^{2}+{y'}^{2}+{z'}^{2})^3}.$
differential-geometry
edited Jan 3 at 14:26
mathcounterexamples.net
25.8k21954
asked Jan 3 at 14:08
user505797user505797
33
$endgroup$
$begingroup$
What have you try? Thanks!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:09
$begingroup$
I was using the direct formula for the curvature given in the book A. Pogorelov Geometry, but not able to find precise answer. Shall I attach the image of my work, Sir?
$endgroup$
– user505797
Jan 3 at 14:11
$begingroup$
Try to apply the formula of the book and write done what you did!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:14
$begingroup$
I have added the formula Sir. It seems the answer is straight. But I'm lost in calculation.
$endgroup$
– user505797
Jan 3 at 14:25
$begingroup$
You have the formula already, so just plug in the expressions for $x, y, z$ and simplify (and just leave that as answer).
$endgroup$
– Arctic Char
Jan 3 at 15:12
add a comment |
0
0
0
1
$begingroup$
Find the curvature of the curve :
$ x = t - sin{t} $ , $y = 1 - cos{t} $ , $z=4sin{dfrac{t}{2}}$.
I have been trying to find the curvature using the formula
$ {k_{1}}^2 = dfrac{{begin{vmatrix} x''& y'' \ x' & y' end{vmatrix}}^2 +{begin{vmatrix} y'' & z'' \ y' & z' end{vmatrix}}^2 + {begin{vmatrix} z'' & x'' \ z' & x' end{vmatrix}}^2}{({x'}^{2}+{y'}^{2}+{z'}^{2})^3}.$
differential-geometry
edited Jan 3 at 14:26
mathcounterexamples.net
25.8k21954
asked Jan 3 at 14:08
user505797user505797
33
$endgroup$
Find the curvature of the curve :
$ x = t - sin{t} $ , $y = 1 - cos{t} $ , $z=4sin{dfrac{t}{2}}$.
I have been trying to find the curvature using the formula
$ {k_{1}}^2 = dfrac{{begin{vmatrix} x''& y'' \ x' & y' end{vmatrix}}^2 +{begin{vmatrix} y'' & z'' \ y' & z' end{vmatrix}}^2 + {begin{vmatrix} z'' & x'' \ z' & x' end{vmatrix}}^2}{({x'}^{2}+{y'}^{2}+{z'}^{2})^3}.$
differential-geometry
differential-geometry
edited Jan 3 at 14:26
mathcounterexamples.net
25.8k21954
asked Jan 3 at 14:08
user505797user505797
33
edited Jan 3 at 14:26
mathcounterexamples.net
25.8k21954
asked Jan 3 at 14:08
user505797user505797
33
edited Jan 3 at 14:26
mathcounterexamples.net
25.8k21954
edited Jan 3 at 14:26
mathcounterexamples.net
25.8k21954
edited Jan 3 at 14:26
mathcounterexamples.net
25.8k21954
25.8k21954
asked Jan 3 at 14:08
user505797user505797
33
asked Jan 3 at 14:08
user505797user505797
33
asked Jan 3 at 14:08
user505797user505797
33
33
$begingroup$
What have you try? Thanks!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:09
$begingroup$
I was using the direct formula for the curvature given in the book A. Pogorelov Geometry, but not able to find precise answer. Shall I attach the image of my work, Sir?
$endgroup$
– user505797
Jan 3 at 14:11
$begingroup$
Try to apply the formula of the book and write done what you did!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:14
$begingroup$
I have added the formula Sir. It seems the answer is straight. But I'm lost in calculation.
$endgroup$
– user505797
Jan 3 at 14:25
$begingroup$
You have the formula already, so just plug in the expressions for $x, y, z$ and simplify (and just leave that as answer).
$endgroup$
– Arctic Char
Jan 3 at 15:12
add a comment |
$begingroup$
What have you try? Thanks!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:09
$begingroup$
I was using the direct formula for the curvature given in the book A. Pogorelov Geometry, but not able to find precise answer. Shall I attach the image of my work, Sir?
$endgroup$
– user505797
Jan 3 at 14:11
$begingroup$
Try to apply the formula of the book and write done what you did!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:14
$begingroup$
I have added the formula Sir. It seems the answer is straight. But I'm lost in calculation.
$endgroup$
– user505797
Jan 3 at 14:25
$begingroup$
You have the formula already, so just plug in the expressions for $x, y, z$ and simplify (and just leave that as answer).
$endgroup$
– Arctic Char
Jan 3 at 15:12
$begingroup$
What have you try? Thanks!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:09
$begingroup$
What have you try? Thanks!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:09
$begingroup$
I was using the direct formula for the curvature given in the book A. Pogorelov Geometry, but not able to find precise answer. Shall I attach the image of my work, Sir?
$endgroup$
– user505797
Jan 3 at 14:11
$begingroup$
I was using the direct formula for the curvature given in the book A. Pogorelov Geometry, but not able to find precise answer. Shall I attach the image of my work, Sir?
$endgroup$
– user505797
Jan 3 at 14:11
$begingroup$
Try to apply the formula of the book and write done what you did!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:14
$begingroup$
Try to apply the formula of the book and write done what you did!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:14
$begingroup$
I have added the formula Sir. It seems the answer is straight. But I'm lost in calculation.
$endgroup$
– user505797
Jan 3 at 14:25
$begingroup$
I have added the formula Sir. It seems the answer is straight. But I'm lost in calculation.
$endgroup$
– user505797
Jan 3 at 14:25
$begingroup$
You have the formula already, so just plug in the expressions for $x, y, z$ and simplify (and just leave that as answer).
$endgroup$
– Arctic Char
Jan 3 at 15:12
$begingroup$
You have the formula already, so just plug in the expressions for $x, y, z$ and simplify (and just leave that as answer).
$endgroup$
– Arctic Char
Jan 3 at 15:12
add a comment |
0
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$begingroup$
What have you try? Thanks!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:09
$begingroup$
I was using the direct formula for the curvature given in the book A. Pogorelov Geometry, but not able to find precise answer. Shall I attach the image of my work, Sir?
$endgroup$
– user505797
Jan 3 at 14:11
$begingroup$
Try to apply the formula of the book and write done what you did!
$endgroup$
– mathcounterexamples.net
Jan 3 at 14:14
$begingroup$
I have added the formula Sir. It seems the answer is straight. But I'm lost in calculation.
$endgroup$
– user505797
Jan 3 at 14:25
$begingroup$
You have the formula already, so just plug in the expressions for $x, y, z$ and simplify (and just leave that as answer).
$endgroup$
– Arctic Char
Jan 3 at 15:12