If a curve is expressed in cylindrical coords., then is the coefficient of the basis vector that corresponds to the angular variable necessarily zero?
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Spherical surface $mathbf S : x^2 + y^2 + z^2 =1 ~$; Cylindrical suface $ mathbf C : x^2 + (y-0.5)^2 = 0.25 $
Let $~ mathbf S cap mathbf C = mathbf K (t)$. Then $ mathbf K(t) = left[ t sqrt 1-t^2 ~~~~ 1-t^2 ~~~~ t right]^mathsf T . $ Actually $ mathbf K$ parameterizes only half of $mathbf S cap mathbf C $, but for now, that is not important. The goal is to express $mathbf K$ in cylindrical coordinates. The cylindrical coordinates $mathbf r (mathbf x) $ corresponding to the cartesian coordinates $mathbf x$ are generally $$ mathbf r(mathbf x) = left[rho(mathbf x) ~~~~ phi(mathbf x) ~~~~z(mathbf x) right]^mathsf T = left [ sqrtx^2 + y^2 ~~~ arctan fracyx ~~~ z right ]^mathsf T .$$
Therefore $$mathbf r (mathbf K (t)) = left [ sqrt1- t^2 ~~~~ arctan fracsqrt1-t^2t ~~~~ t right ]^mathsf T . $$
Also:
$hat mathbf h_rho = left[ cos phi ~~ sin phi ~~ 0 right]^mathsf T$ ; $hat mathbf h_phi = left[ -sin phi ~~ cos phi ~~ 0 right]^mathsf T $ ; $hat mathbf h_z = left[ 0 ~~ 0 ~~ 1 right]^mathsf T $
$$cos (phi (mathbf K (t))) = cos left(arctan left( fracsqrt1-t^2t right) right) = frac1sqrt1+ left( fracsqrt1-t^2t right)^2 = t $$
$$sin (phi (mathbf K (t))) = fracsqrt1-t^2t ~~ cos left(arctan left( fracsqrt1-t^2t right) right) = sqrt1-t^2$$
Then
$$ left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right]_mathbf r (mathbf K(t)) = left[
beginmatrix
t & -sqrt1-t^2 & 0 \
sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right]
$$
The curve $mathbf K$ expressed in cylindrical coordinates therefore is:
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[
beginmatrix
sqrt1- t^2 \
arctan fracsqrt1-t^2t \
t \
endmatrix right ] = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[beginmatrix f_1(t) \ g(t) \ f_2 (t) endmatrix right].$$
Now here is the problem:
$g(t) = t^2 + t arctan left ( fracsqrt1-t^2t right) -1 neq 0$. I think that the coefficient of $hat mathbf h_phi$, in the correct expression of $~ mathbf S cap mathbf C $, in cylindrical coordinates, is necessarily zero. If somebody can tell me where I made an error that caused the result $g(t) neq 0$, or if they can prove or otherwise convince me that $g(t)$ is not necessarily zero, then I will set this question to a resolved state.
analytic-geometry cylindrical-coordinates
asked Aug 12 at 1:24
EricVonB
1139
add a comment |Â
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0
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Spherical surface $mathbf S : x^2 + y^2 + z^2 =1 ~$; Cylindrical suface $ mathbf C : x^2 + (y-0.5)^2 = 0.25 $
Let $~ mathbf S cap mathbf C = mathbf K (t)$. Then $ mathbf K(t) = left[ t sqrt 1-t^2 ~~~~ 1-t^2 ~~~~ t right]^mathsf T . $ Actually $ mathbf K$ parameterizes only half of $mathbf S cap mathbf C $, but for now, that is not important. The goal is to express $mathbf K$ in cylindrical coordinates. The cylindrical coordinates $mathbf r (mathbf x) $ corresponding to the cartesian coordinates $mathbf x$ are generally $$ mathbf r(mathbf x) = left[rho(mathbf x) ~~~~ phi(mathbf x) ~~~~z(mathbf x) right]^mathsf T = left [ sqrtx^2 + y^2 ~~~ arctan fracyx ~~~ z right ]^mathsf T .$$
Therefore $$mathbf r (mathbf K (t)) = left [ sqrt1- t^2 ~~~~ arctan fracsqrt1-t^2t ~~~~ t right ]^mathsf T . $$
Also:
$hat mathbf h_rho = left[ cos phi ~~ sin phi ~~ 0 right]^mathsf T$ ; $hat mathbf h_phi = left[ -sin phi ~~ cos phi ~~ 0 right]^mathsf T $ ; $hat mathbf h_z = left[ 0 ~~ 0 ~~ 1 right]^mathsf T $
$$cos (phi (mathbf K (t))) = cos left(arctan left( fracsqrt1-t^2t right) right) = frac1sqrt1+ left( fracsqrt1-t^2t right)^2 = t $$
$$sin (phi (mathbf K (t))) = fracsqrt1-t^2t ~~ cos left(arctan left( fracsqrt1-t^2t right) right) = sqrt1-t^2$$
Then
$$ left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right]_mathbf r (mathbf K(t)) = left[
beginmatrix
t & -sqrt1-t^2 & 0 \
sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right]
$$
The curve $mathbf K$ expressed in cylindrical coordinates therefore is:
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[
beginmatrix
sqrt1- t^2 \
arctan fracsqrt1-t^2t \
t \
endmatrix right ] = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[beginmatrix f_1(t) \ g(t) \ f_2 (t) endmatrix right].$$
Now here is the problem:
$g(t) = t^2 + t arctan left ( fracsqrt1-t^2t right) -1 neq 0$. I think that the coefficient of $hat mathbf h_phi$, in the correct expression of $~ mathbf S cap mathbf C $, in cylindrical coordinates, is necessarily zero. If somebody can tell me where I made an error that caused the result $g(t) neq 0$, or if they can prove or otherwise convince me that $g(t)$ is not necessarily zero, then I will set this question to a resolved state.
analytic-geometry cylindrical-coordinates
asked Aug 12 at 1:24
EricVonB
1139
1
Why in the world would you think that the coefficient of $hat mathbf h_phi$ would be $0$? If it were $0$, then $mathbf K$ would lie entirely in a single $rtext- z$ plane, which it obviously does not do.
– Paul Sinclair
Aug 12 at 14:40
Good point. Maybe the derivative of $mathbf K$ has non-zero coefficient of $hat mathbf h_phi $ therefore the curve itself spans some non-constant range of $phi$. I am not totally sure.
– EricVonB
Aug 14 at 13:40
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
Spherical surface $mathbf S : x^2 + y^2 + z^2 =1 ~$; Cylindrical suface $ mathbf C : x^2 + (y-0.5)^2 = 0.25 $
Let $~ mathbf S cap mathbf C = mathbf K (t)$. Then $ mathbf K(t) = left[ t sqrt 1-t^2 ~~~~ 1-t^2 ~~~~ t right]^mathsf T . $ Actually $ mathbf K$ parameterizes only half of $mathbf S cap mathbf C $, but for now, that is not important. The goal is to express $mathbf K$ in cylindrical coordinates. The cylindrical coordinates $mathbf r (mathbf x) $ corresponding to the cartesian coordinates $mathbf x$ are generally $$ mathbf r(mathbf x) = left[rho(mathbf x) ~~~~ phi(mathbf x) ~~~~z(mathbf x) right]^mathsf T = left [ sqrtx^2 + y^2 ~~~ arctan fracyx ~~~ z right ]^mathsf T .$$
Therefore $$mathbf r (mathbf K (t)) = left [ sqrt1- t^2 ~~~~ arctan fracsqrt1-t^2t ~~~~ t right ]^mathsf T . $$
Also:
$hat mathbf h_rho = left[ cos phi ~~ sin phi ~~ 0 right]^mathsf T$ ; $hat mathbf h_phi = left[ -sin phi ~~ cos phi ~~ 0 right]^mathsf T $ ; $hat mathbf h_z = left[ 0 ~~ 0 ~~ 1 right]^mathsf T $
$$cos (phi (mathbf K (t))) = cos left(arctan left( fracsqrt1-t^2t right) right) = frac1sqrt1+ left( fracsqrt1-t^2t right)^2 = t $$
$$sin (phi (mathbf K (t))) = fracsqrt1-t^2t ~~ cos left(arctan left( fracsqrt1-t^2t right) right) = sqrt1-t^2$$
Then
$$ left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right]_mathbf r (mathbf K(t)) = left[
beginmatrix
t & -sqrt1-t^2 & 0 \
sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right]
$$
The curve $mathbf K$ expressed in cylindrical coordinates therefore is:
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[
beginmatrix
sqrt1- t^2 \
arctan fracsqrt1-t^2t \
t \
endmatrix right ] = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[beginmatrix f_1(t) \ g(t) \ f_2 (t) endmatrix right].$$
Now here is the problem:
$g(t) = t^2 + t arctan left ( fracsqrt1-t^2t right) -1 neq 0$. I think that the coefficient of $hat mathbf h_phi$, in the correct expression of $~ mathbf S cap mathbf C $, in cylindrical coordinates, is necessarily zero. If somebody can tell me where I made an error that caused the result $g(t) neq 0$, or if they can prove or otherwise convince me that $g(t)$ is not necessarily zero, then I will set this question to a resolved state.
analytic-geometry cylindrical-coordinates
asked Aug 12 at 1:24
EricVonB
1139
Spherical surface $mathbf S : x^2 + y^2 + z^2 =1 ~$; Cylindrical suface $ mathbf C : x^2 + (y-0.5)^2 = 0.25 $
Let $~ mathbf S cap mathbf C = mathbf K (t)$. Then $ mathbf K(t) = left[ t sqrt 1-t^2 ~~~~ 1-t^2 ~~~~ t right]^mathsf T . $ Actually $ mathbf K$ parameterizes only half of $mathbf S cap mathbf C $, but for now, that is not important. The goal is to express $mathbf K$ in cylindrical coordinates. The cylindrical coordinates $mathbf r (mathbf x) $ corresponding to the cartesian coordinates $mathbf x$ are generally $$ mathbf r(mathbf x) = left[rho(mathbf x) ~~~~ phi(mathbf x) ~~~~z(mathbf x) right]^mathsf T = left [ sqrtx^2 + y^2 ~~~ arctan fracyx ~~~ z right ]^mathsf T .$$
Therefore $$mathbf r (mathbf K (t)) = left [ sqrt1- t^2 ~~~~ arctan fracsqrt1-t^2t ~~~~ t right ]^mathsf T . $$
Also:
$hat mathbf h_rho = left[ cos phi ~~ sin phi ~~ 0 right]^mathsf T$ ; $hat mathbf h_phi = left[ -sin phi ~~ cos phi ~~ 0 right]^mathsf T $ ; $hat mathbf h_z = left[ 0 ~~ 0 ~~ 1 right]^mathsf T $
$$cos (phi (mathbf K (t))) = cos left(arctan left( fracsqrt1-t^2t right) right) = frac1sqrt1+ left( fracsqrt1-t^2t right)^2 = t $$
$$sin (phi (mathbf K (t))) = fracsqrt1-t^2t ~~ cos left(arctan left( fracsqrt1-t^2t right) right) = sqrt1-t^2$$
Then
$$ left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right]_mathbf r (mathbf K(t)) = left[
beginmatrix
t & -sqrt1-t^2 & 0 \
sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right]
$$
The curve $mathbf K$ expressed in cylindrical coordinates therefore is:
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[
beginmatrix
sqrt1- t^2 \
arctan fracsqrt1-t^2t \
t \
endmatrix right ] = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[beginmatrix f_1(t) \ g(t) \ f_2 (t) endmatrix right].$$
Now here is the problem:
$g(t) = t^2 + t arctan left ( fracsqrt1-t^2t right) -1 neq 0$. I think that the coefficient of $hat mathbf h_phi$, in the correct expression of $~ mathbf S cap mathbf C $, in cylindrical coordinates, is necessarily zero. If somebody can tell me where I made an error that caused the result $g(t) neq 0$, or if they can prove or otherwise convince me that $g(t)$ is not necessarily zero, then I will set this question to a resolved state.
analytic-geometry cylindrical-coordinates
asked Aug 12 at 1:24
EricVonB
1139
edited Aug 12 at 1:32
asked Aug 12 at 1:24
EricVonB
1139
asked Aug 12 at 1:24
EricVonB
1139
asked Aug 12 at 1:24
EricVonB
1139
1139
1
Why in the world would you think that the coefficient of $hat mathbf h_phi$ would be $0$? If it were $0$, then $mathbf K$ would lie entirely in a single $rtext- z$ plane, which it obviously does not do.
– Paul Sinclair
Aug 12 at 14:40
Good point. Maybe the derivative of $mathbf K$ has non-zero coefficient of $hat mathbf h_phi $ therefore the curve itself spans some non-constant range of $phi$. I am not totally sure.
– EricVonB
Aug 14 at 13:40
add a comment |Â
1
Why in the world would you think that the coefficient of $hat mathbf h_phi$ would be $0$? If it were $0$, then $mathbf K$ would lie entirely in a single $rtext- z$ plane, which it obviously does not do.
– Paul Sinclair
Aug 12 at 14:40
Good point. Maybe the derivative of $mathbf K$ has non-zero coefficient of $hat mathbf h_phi $ therefore the curve itself spans some non-constant range of $phi$. I am not totally sure.
– EricVonB
Aug 14 at 13:40
1
1
Why in the world would you think that the coefficient of $hat mathbf h_phi$ would be $0$? If it were $0$, then $mathbf K$ would lie entirely in a single $rtext- z$ plane, which it obviously does not do.
– Paul Sinclair
Aug 12 at 14:40
Why in the world would you think that the coefficient of $hat mathbf h_phi$ would be $0$? If it were $0$, then $mathbf K$ would lie entirely in a single $rtext- z$ plane, which it obviously does not do.
– Paul Sinclair
Aug 12 at 14:40
Good point. Maybe the derivative of $mathbf K$ has non-zero coefficient of $hat mathbf h_phi $ therefore the curve itself spans some non-constant range of $phi$. I am not totally sure.
– EricVonB
Aug 14 at 13:40
Good point. Maybe the derivative of $mathbf K$ has non-zero coefficient of $hat mathbf h_phi $ therefore the curve itself spans some non-constant range of $phi$. I am not totally sure.
– EricVonB
Aug 14 at 13:40
add a comment |Â
1 Answer
1
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As stated in the original post, the goal is to express $mathbf K$ in cylindrical coordinates. $$mathbf K(t) = left[ t sqrt 1-t^2 ~~~~ 1-t^2 ~~~~ t right]^mathsf T ~~;~~ -1 le t le 1 $$
$$mathbf r(mathbf x) = left(rho(mathbf x) ~~~~ phi(mathbf x) ~~~~z(mathbf x) right) = left ( sqrtx^2 + y^2 ~~~ arctan fracyx ~~~ z right ) .$$
$$mathbf r (mathbf K (t)) = left ( sqrt1- t^2 ~~~~ arctan fracsqrt1-t^2t ~~~~ t right ) .$$
$$left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right]_mathbf r (mathbf K(t)) = left[
beginmatrix
t & -sqrt1-t^2 & 0 \
sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right]$$
The following statement is false.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[
beginmatrix
sqrt1- t^2 \
arctan fracsqrt1-t^2t \
t \
endmatrix right ].$$
The following statement is true.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[ beginmatrix t sqrt 1-t^2 \ 1-t^2 \ t endmatrix right] . $$
$mathbf K$ expressed in cylindrical coordinates is
$$ mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
sqrt1-t^2 \
0 \
t \
endmatrix
right] .$$
QED
answered Aug 14 at 12:26
EricVonB
1139
add a comment |Â
1 Answer
1
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
As stated in the original post, the goal is to express $mathbf K$ in cylindrical coordinates. $$mathbf K(t) = left[ t sqrt 1-t^2 ~~~~ 1-t^2 ~~~~ t right]^mathsf T ~~;~~ -1 le t le 1 $$
$$mathbf r(mathbf x) = left(rho(mathbf x) ~~~~ phi(mathbf x) ~~~~z(mathbf x) right) = left ( sqrtx^2 + y^2 ~~~ arctan fracyx ~~~ z right ) .$$
$$mathbf r (mathbf K (t)) = left ( sqrt1- t^2 ~~~~ arctan fracsqrt1-t^2t ~~~~ t right ) .$$
$$left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right]_mathbf r (mathbf K(t)) = left[
beginmatrix
t & -sqrt1-t^2 & 0 \
sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right]$$
The following statement is false.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[
beginmatrix
sqrt1- t^2 \
arctan fracsqrt1-t^2t \
t \
endmatrix right ].$$
The following statement is true.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[ beginmatrix t sqrt 1-t^2 \ 1-t^2 \ t endmatrix right] . $$
$mathbf K$ expressed in cylindrical coordinates is
$$ mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
sqrt1-t^2 \
0 \
t \
endmatrix
right] .$$
QED
answered Aug 14 at 12:26
EricVonB
1139
add a comment |Â
up vote
0
down vote
As stated in the original post, the goal is to express $mathbf K$ in cylindrical coordinates. $$mathbf K(t) = left[ t sqrt 1-t^2 ~~~~ 1-t^2 ~~~~ t right]^mathsf T ~~;~~ -1 le t le 1 $$
$$mathbf r(mathbf x) = left(rho(mathbf x) ~~~~ phi(mathbf x) ~~~~z(mathbf x) right) = left ( sqrtx^2 + y^2 ~~~ arctan fracyx ~~~ z right ) .$$
$$mathbf r (mathbf K (t)) = left ( sqrt1- t^2 ~~~~ arctan fracsqrt1-t^2t ~~~~ t right ) .$$
$$left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right]_mathbf r (mathbf K(t)) = left[
beginmatrix
t & -sqrt1-t^2 & 0 \
sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right]$$
The following statement is false.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[
beginmatrix
sqrt1- t^2 \
arctan fracsqrt1-t^2t \
t \
endmatrix right ].$$
The following statement is true.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[ beginmatrix t sqrt 1-t^2 \ 1-t^2 \ t endmatrix right] . $$
$mathbf K$ expressed in cylindrical coordinates is
$$ mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
sqrt1-t^2 \
0 \
t \
endmatrix
right] .$$
QED
answered Aug 14 at 12:26
EricVonB
1139
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As stated in the original post, the goal is to express $mathbf K$ in cylindrical coordinates. $$mathbf K(t) = left[ t sqrt 1-t^2 ~~~~ 1-t^2 ~~~~ t right]^mathsf T ~~;~~ -1 le t le 1 $$
$$mathbf r(mathbf x) = left(rho(mathbf x) ~~~~ phi(mathbf x) ~~~~z(mathbf x) right) = left ( sqrtx^2 + y^2 ~~~ arctan fracyx ~~~ z right ) .$$
$$mathbf r (mathbf K (t)) = left ( sqrt1- t^2 ~~~~ arctan fracsqrt1-t^2t ~~~~ t right ) .$$
$$left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right]_mathbf r (mathbf K(t)) = left[
beginmatrix
t & -sqrt1-t^2 & 0 \
sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right]$$
The following statement is false.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[
beginmatrix
sqrt1- t^2 \
arctan fracsqrt1-t^2t \
t \
endmatrix right ].$$
The following statement is true.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[ beginmatrix t sqrt 1-t^2 \ 1-t^2 \ t endmatrix right] . $$
$mathbf K$ expressed in cylindrical coordinates is
$$ mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
sqrt1-t^2 \
0 \
t \
endmatrix
right] .$$
QED
answered Aug 14 at 12:26
EricVonB
1139
As stated in the original post, the goal is to express $mathbf K$ in cylindrical coordinates. $$mathbf K(t) = left[ t sqrt 1-t^2 ~~~~ 1-t^2 ~~~~ t right]^mathsf T ~~;~~ -1 le t le 1 $$
$$mathbf r(mathbf x) = left(rho(mathbf x) ~~~~ phi(mathbf x) ~~~~z(mathbf x) right) = left ( sqrtx^2 + y^2 ~~~ arctan fracyx ~~~ z right ) .$$
$$mathbf r (mathbf K (t)) = left ( sqrt1- t^2 ~~~~ arctan fracsqrt1-t^2t ~~~~ t right ) .$$
$$left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right]_mathbf r (mathbf K(t)) = left[
beginmatrix
t & -sqrt1-t^2 & 0 \
sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right]$$
The following statement is false.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[
beginmatrix
sqrt1- t^2 \
arctan fracsqrt1-t^2t \
t \
endmatrix right ].$$
The following statement is true.
$$mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
t & sqrt1-t^2 & 0 \
-sqrt1-t^2 & t & 0 \
0 & 0 & 1 \
endmatrix
right] left[ beginmatrix t sqrt 1-t^2 \ 1-t^2 \ t endmatrix right] . $$
$mathbf K$ expressed in cylindrical coordinates is
$$ mathbf K (t) = left[
beginmatrix
hat mathbf h_rho & hat mathbf h_phi & hat mathbf h_z \
endmatrix right] left[
beginmatrix
sqrt1-t^2 \
0 \
t \
endmatrix
right] .$$
QED
answered Aug 14 at 12:26
EricVonB
1139
answered Aug 14 at 12:26
EricVonB
1139
answered Aug 14 at 12:26
EricVonB
1139
answered Aug 14 at 12:26
EricVonB
1139
1139
add a comment |Â
add a comment |Â
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1
Why in the world would you think that the coefficient of $hat mathbf h_phi$ would be $0$? If it were $0$, then $mathbf K$ would lie entirely in a single $rtext- z$ plane, which it obviously does not do.
– Paul Sinclair
Aug 12 at 14:40
Good point. Maybe the derivative of $mathbf K$ has non-zero coefficient of $hat mathbf h_phi $ therefore the curve itself spans some non-constant range of $phi$. I am not totally sure.
– EricVonB
Aug 14 at 13:40