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Parametrizing the curve of intersection between a elliptic cylinder and a sphere
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How can I form the parameterization of a curve of intersection given a sphere $x^2+y^2+z^2 =1$ and an elliptic cylinder $2x^2 + z^2 = 1$ ?
calculus multivariable-calculus spheres parametrization
asked Jan 14 at 22:36
ZaltahZaltah
74
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add a comment |
0
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How can I form the parameterization of a curve of intersection given a sphere $x^2+y^2+z^2 =1$ and an elliptic cylinder $2x^2 + z^2 = 1$ ?
calculus multivariable-calculus spheres parametrization
asked Jan 14 at 22:36
ZaltahZaltah
74
$endgroup$
add a comment |
0
0
0
$begingroup$
How can I form the parameterization of a curve of intersection given a sphere $x^2+y^2+z^2 =1$ and an elliptic cylinder $2x^2 + z^2 = 1$ ?
calculus multivariable-calculus spheres parametrization
asked Jan 14 at 22:36
ZaltahZaltah
74
$endgroup$
How can I form the parameterization of a curve of intersection given a sphere $x^2+y^2+z^2 =1$ and an elliptic cylinder $2x^2 + z^2 = 1$ ?
calculus multivariable-calculus spheres parametrization
calculus multivariable-calculus spheres parametrization
asked Jan 14 at 22:36
ZaltahZaltah
74
asked Jan 14 at 22:36
ZaltahZaltah
74
edited Jan 14 at 22:47
Zaltah
asked Jan 14 at 22:36
ZaltahZaltah
74
asked Jan 14 at 22:36
ZaltahZaltah
74
asked Jan 14 at 22:36
ZaltahZaltah
74
74
add a comment |
add a comment |
1 Answer
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Thanks to @WillJagy$${x={cos tover sqrt 2}\y={cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$and $${x={cos tover sqrt 2}\y=-{cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$
Comment
My original answer suggested just one of those branches while there are two. The images below suggest why:
answered Jan 14 at 22:43
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
1
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there is also a piece with $y=-x$
$endgroup$
– Will Jagy
Jan 14 at 22:44
$begingroup$
I think it just determines the direction of recursion on the intersection but I agree with you. A parametrization is not unique even as with the unit circle we have $$x=sin u\y=cos u$$or$$x=sin u\y=-cos u$$
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:45
$begingroup$
$( frac{- cos t}{sqrt 2} , ; frac{ cos t}{sqrt 2} , ; sin t )$ also solves, there are just two intersection points, $t = pm pi/2$ They are two ellipses, they cross twice
$endgroup$
– Will Jagy
Jan 14 at 22:49
$begingroup$
Even with the answer I am not sure how you are supposed to calculate this problem.
$endgroup$
– Zaltah
Jan 14 at 22:56
$begingroup$
You are quite right. I edited the answer. Thanks....
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:59
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
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votes
1
$begingroup$
Thanks to @WillJagy$${x={cos tover sqrt 2}\y={cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$and $${x={cos tover sqrt 2}\y=-{cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$
Comment
My original answer suggested just one of those branches while there are two. The images below suggest why:
answered Jan 14 at 22:43
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
1
$begingroup$
there is also a piece with $y=-x$
$endgroup$
– Will Jagy
Jan 14 at 22:44
$begingroup$
I think it just determines the direction of recursion on the intersection but I agree with you. A parametrization is not unique even as with the unit circle we have $$x=sin u\y=cos u$$or$$x=sin u\y=-cos u$$
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:45
$begingroup$
$( frac{- cos t}{sqrt 2} , ; frac{ cos t}{sqrt 2} , ; sin t )$ also solves, there are just two intersection points, $t = pm pi/2$ They are two ellipses, they cross twice
$endgroup$
– Will Jagy
Jan 14 at 22:49
$begingroup$
Even with the answer I am not sure how you are supposed to calculate this problem.
$endgroup$
– Zaltah
Jan 14 at 22:56
$begingroup$
You are quite right. I edited the answer. Thanks....
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:59
add a comment |
1
$begingroup$
Thanks to @WillJagy$${x={cos tover sqrt 2}\y={cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$and $${x={cos tover sqrt 2}\y=-{cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$
Comment
My original answer suggested just one of those branches while there are two. The images below suggest why:
answered Jan 14 at 22:43
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
1
$begingroup$
there is also a piece with $y=-x$
$endgroup$
– Will Jagy
Jan 14 at 22:44
$begingroup$
I think it just determines the direction of recursion on the intersection but I agree with you. A parametrization is not unique even as with the unit circle we have $$x=sin u\y=cos u$$or$$x=sin u\y=-cos u$$
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:45
$begingroup$
$( frac{- cos t}{sqrt 2} , ; frac{ cos t}{sqrt 2} , ; sin t )$ also solves, there are just two intersection points, $t = pm pi/2$ They are two ellipses, they cross twice
$endgroup$
– Will Jagy
Jan 14 at 22:49
$begingroup$
Even with the answer I am not sure how you are supposed to calculate this problem.
$endgroup$
– Zaltah
Jan 14 at 22:56
$begingroup$
You are quite right. I edited the answer. Thanks....
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:59
add a comment |
1
1
1
$begingroup$
Thanks to @WillJagy$${x={cos tover sqrt 2}\y={cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$and $${x={cos tover sqrt 2}\y=-{cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$
Comment
My original answer suggested just one of those branches while there are two. The images below suggest why:
answered Jan 14 at 22:43
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Thanks to @WillJagy$${x={cos tover sqrt 2}\y={cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$and $${x={cos tover sqrt 2}\y=-{cos tover sqrt 2}quad ,quad tin Bbb R\z=sin t}$$
Comment
My original answer suggested just one of those branches while there are two. The images below suggest why:
answered Jan 14 at 22:43
Mostafa AyazMostafa Ayaz
15.6k3939
edited Jan 14 at 22:59
answered Jan 14 at 22:43
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 22:43
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 22:43
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
1
$begingroup$
there is also a piece with $y=-x$
$endgroup$
– Will Jagy
Jan 14 at 22:44
$begingroup$
I think it just determines the direction of recursion on the intersection but I agree with you. A parametrization is not unique even as with the unit circle we have $$x=sin u\y=cos u$$or$$x=sin u\y=-cos u$$
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:45
$begingroup$
$( frac{- cos t}{sqrt 2} , ; frac{ cos t}{sqrt 2} , ; sin t )$ also solves, there are just two intersection points, $t = pm pi/2$ They are two ellipses, they cross twice
$endgroup$
– Will Jagy
Jan 14 at 22:49
$begingroup$
Even with the answer I am not sure how you are supposed to calculate this problem.
$endgroup$
– Zaltah
Jan 14 at 22:56
$begingroup$
You are quite right. I edited the answer. Thanks....
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:59
add a comment |
1
$begingroup$
there is also a piece with $y=-x$
$endgroup$
– Will Jagy
Jan 14 at 22:44
$begingroup$
I think it just determines the direction of recursion on the intersection but I agree with you. A parametrization is not unique even as with the unit circle we have $$x=sin u\y=cos u$$or$$x=sin u\y=-cos u$$
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:45
$begingroup$
$( frac{- cos t}{sqrt 2} , ; frac{ cos t}{sqrt 2} , ; sin t )$ also solves, there are just two intersection points, $t = pm pi/2$ They are two ellipses, they cross twice
$endgroup$
– Will Jagy
Jan 14 at 22:49
$begingroup$
Even with the answer I am not sure how you are supposed to calculate this problem.
$endgroup$
– Zaltah
Jan 14 at 22:56
$begingroup$
You are quite right. I edited the answer. Thanks....
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:59
1
1
$begingroup$
there is also a piece with $y=-x$
$endgroup$
– Will Jagy
Jan 14 at 22:44
$begingroup$
there is also a piece with $y=-x$
$endgroup$
– Will Jagy
Jan 14 at 22:44
$begingroup$
I think it just determines the direction of recursion on the intersection but I agree with you. A parametrization is not unique even as with the unit circle we have $$x=sin u\y=cos u$$or$$x=sin u\y=-cos u$$
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:45
$begingroup$
I think it just determines the direction of recursion on the intersection but I agree with you. A parametrization is not unique even as with the unit circle we have $$x=sin u\y=cos u$$or$$x=sin u\y=-cos u$$
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:45
$begingroup$
$( frac{- cos t}{sqrt 2} , ; frac{ cos t}{sqrt 2} , ; sin t )$ also solves, there are just two intersection points, $t = pm pi/2$ They are two ellipses, they cross twice
$endgroup$
– Will Jagy
Jan 14 at 22:49
$begingroup$
$( frac{- cos t}{sqrt 2} , ; frac{ cos t}{sqrt 2} , ; sin t )$ also solves, there are just two intersection points, $t = pm pi/2$ They are two ellipses, they cross twice
$endgroup$
– Will Jagy
Jan 14 at 22:49
$begingroup$
Even with the answer I am not sure how you are supposed to calculate this problem.
$endgroup$
– Zaltah
Jan 14 at 22:56
$begingroup$
Even with the answer I am not sure how you are supposed to calculate this problem.
$endgroup$
– Zaltah
Jan 14 at 22:56
$begingroup$
You are quite right. I edited the answer. Thanks....
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:59
$begingroup$
You are quite right. I edited the answer. Thanks....
$endgroup$
– Mostafa Ayaz
Jan 14 at 22:59
add a comment |
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