Mixing
Finding acceleration of a curve in three dimensions
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An aeroplane flying on a great circle, it was shown that its velocity in spherical coordinates is
$$ mathbf v(t) = frac sin gamma sin t sqrt1-sin^2 gamma cos^2 t mathbfe_theta + frac cos gammasqrt1-sin^2 gamma cos^2 t mathbfe_phi$$
where $γ$ is the angle of tilt from the equatorial plane. Also, we assumed $r$ to be $1$. Find $mathbfa(t)$ and prove it is $bf proportional$ to $mathbfe_r$
Try
Now, taking derivatives in spherical coordinates is difficult. I know that the position in rectangular coord is given by
$$ r = cos γ cos t mathbfi + sin t mathbfj + sin γ cos t mathbfk , $$
So, in rectangular coords,
$$ a(t) = - cos gamma cos t mathbfi - sin t mathbfj - sin gamma cos t mathbfk = - r $$
Do I have to convert this back to spherical? is this a correct approach?
calculus
asked Jul 31 at 5:21
Jimmy Sabater
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An aeroplane flying on a great circle, it was shown that its velocity in spherical coordinates is
$$ mathbf v(t) = frac sin gamma sin t sqrt1-sin^2 gamma cos^2 t mathbfe_theta + frac cos gammasqrt1-sin^2 gamma cos^2 t mathbfe_phi$$
where $γ$ is the angle of tilt from the equatorial plane. Also, we assumed $r$ to be $1$. Find $mathbfa(t)$ and prove it is $bf proportional$ to $mathbfe_r$
Try
Now, taking derivatives in spherical coordinates is difficult. I know that the position in rectangular coord is given by
$$ r = cos γ cos t mathbfi + sin t mathbfj + sin γ cos t mathbfk , $$
So, in rectangular coords,
$$ a(t) = - cos gamma cos t mathbfi - sin t mathbfj - sin gamma cos t mathbfk = - r $$
Do I have to convert this back to spherical? is this a correct approach?
calculus
asked Jul 31 at 5:21
Jimmy Sabater
1,02810
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
An aeroplane flying on a great circle, it was shown that its velocity in spherical coordinates is
$$ mathbf v(t) = frac sin gamma sin t sqrt1-sin^2 gamma cos^2 t mathbfe_theta + frac cos gammasqrt1-sin^2 gamma cos^2 t mathbfe_phi$$
where $γ$ is the angle of tilt from the equatorial plane. Also, we assumed $r$ to be $1$. Find $mathbfa(t)$ and prove it is $bf proportional$ to $mathbfe_r$
Try
Now, taking derivatives in spherical coordinates is difficult. I know that the position in rectangular coord is given by
$$ r = cos γ cos t mathbfi + sin t mathbfj + sin γ cos t mathbfk , $$
So, in rectangular coords,
$$ a(t) = - cos gamma cos t mathbfi - sin t mathbfj - sin gamma cos t mathbfk = - r $$
Do I have to convert this back to spherical? is this a correct approach?
calculus
asked Jul 31 at 5:21
Jimmy Sabater
1,02810
An aeroplane flying on a great circle, it was shown that its velocity in spherical coordinates is
$$ mathbf v(t) = frac sin gamma sin t sqrt1-sin^2 gamma cos^2 t mathbfe_theta + frac cos gammasqrt1-sin^2 gamma cos^2 t mathbfe_phi$$
where $γ$ is the angle of tilt from the equatorial plane. Also, we assumed $r$ to be $1$. Find $mathbfa(t)$ and prove it is $bf proportional$ to $mathbfe_r$
Try
Now, taking derivatives in spherical coordinates is difficult. I know that the position in rectangular coord is given by
$$ r = cos γ cos t mathbfi + sin t mathbfj + sin γ cos t mathbfk , $$
So, in rectangular coords,
$$ a(t) = - cos gamma cos t mathbfi - sin t mathbfj - sin gamma cos t mathbfk = - r $$
Do I have to convert this back to spherical? is this a correct approach?
calculus
asked Jul 31 at 5:21
Jimmy Sabater
1,02810
asked Jul 31 at 5:21
Jimmy Sabater
1,02810
asked Jul 31 at 5:21
Jimmy Sabater
1,02810
asked Jul 31 at 5:21
Jimmy Sabater
1,02810
1,02810
add a comment |Â
add a comment |Â
1 Answer
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If your $bf r(t)$ as written is correct then your try is fine. It remains the question why you started to work with $bf e_r$, $bf e_phi$, and $bf e_theta$ to begin with. I have never seen your first formula before.
One easily computes $|bf v(t)|^2equiv1$, hence the plane is flying with constant speed $1$ on a circle of radius $1$. We therefore may as well look at the motion
$$tmapsto bf z(t)=cos t>bf e_1+sin t>bf e_2$$
in the plane of said great circle, whereby $bf e_1$ and $bf e_2$ are fixed orthonormal vectors in this plane. One computes $bf v(t)=dotbf z(t)=-sin t>bf e_1+cos t>bf e_2$ and
$$bf a(t)=ddotbf z(t)=-cos t>bf e_1-sin t>bf e_2=-bf z(t) .$$
This says that the acceleration $bf a(t)$, considered as a vector in the tangent space $T_bf z(t)$, is at all times directed to the center of the circle. In your moving coordinate system this means that $bf a(t)$ is $>=-bf e_rin T_bf r$ at all times.
answered Jul 31 at 7:26
Christian Blatter
163k7107305
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If your $bf r(t)$ as written is correct then your try is fine. It remains the question why you started to work with $bf e_r$, $bf e_phi$, and $bf e_theta$ to begin with. I have never seen your first formula before.
One easily computes $|bf v(t)|^2equiv1$, hence the plane is flying with constant speed $1$ on a circle of radius $1$. We therefore may as well look at the motion
$$tmapsto bf z(t)=cos t>bf e_1+sin t>bf e_2$$
in the plane of said great circle, whereby $bf e_1$ and $bf e_2$ are fixed orthonormal vectors in this plane. One computes $bf v(t)=dotbf z(t)=-sin t>bf e_1+cos t>bf e_2$ and
$$bf a(t)=ddotbf z(t)=-cos t>bf e_1-sin t>bf e_2=-bf z(t) .$$
This says that the acceleration $bf a(t)$, considered as a vector in the tangent space $T_bf z(t)$, is at all times directed to the center of the circle. In your moving coordinate system this means that $bf a(t)$ is $>=-bf e_rin T_bf r$ at all times.
answered Jul 31 at 7:26
Christian Blatter
163k7107305
add a comment |Â
up vote
0
down vote
If your $bf r(t)$ as written is correct then your try is fine. It remains the question why you started to work with $bf e_r$, $bf e_phi$, and $bf e_theta$ to begin with. I have never seen your first formula before.
One easily computes $|bf v(t)|^2equiv1$, hence the plane is flying with constant speed $1$ on a circle of radius $1$. We therefore may as well look at the motion
$$tmapsto bf z(t)=cos t>bf e_1+sin t>bf e_2$$
in the plane of said great circle, whereby $bf e_1$ and $bf e_2$ are fixed orthonormal vectors in this plane. One computes $bf v(t)=dotbf z(t)=-sin t>bf e_1+cos t>bf e_2$ and
$$bf a(t)=ddotbf z(t)=-cos t>bf e_1-sin t>bf e_2=-bf z(t) .$$
This says that the acceleration $bf a(t)$, considered as a vector in the tangent space $T_bf z(t)$, is at all times directed to the center of the circle. In your moving coordinate system this means that $bf a(t)$ is $>=-bf e_rin T_bf r$ at all times.
answered Jul 31 at 7:26
Christian Blatter
163k7107305
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If your $bf r(t)$ as written is correct then your try is fine. It remains the question why you started to work with $bf e_r$, $bf e_phi$, and $bf e_theta$ to begin with. I have never seen your first formula before.
One easily computes $|bf v(t)|^2equiv1$, hence the plane is flying with constant speed $1$ on a circle of radius $1$. We therefore may as well look at the motion
$$tmapsto bf z(t)=cos t>bf e_1+sin t>bf e_2$$
in the plane of said great circle, whereby $bf e_1$ and $bf e_2$ are fixed orthonormal vectors in this plane. One computes $bf v(t)=dotbf z(t)=-sin t>bf e_1+cos t>bf e_2$ and
$$bf a(t)=ddotbf z(t)=-cos t>bf e_1-sin t>bf e_2=-bf z(t) .$$
This says that the acceleration $bf a(t)$, considered as a vector in the tangent space $T_bf z(t)$, is at all times directed to the center of the circle. In your moving coordinate system this means that $bf a(t)$ is $>=-bf e_rin T_bf r$ at all times.
answered Jul 31 at 7:26
Christian Blatter
163k7107305
If your $bf r(t)$ as written is correct then your try is fine. It remains the question why you started to work with $bf e_r$, $bf e_phi$, and $bf e_theta$ to begin with. I have never seen your first formula before.
One easily computes $|bf v(t)|^2equiv1$, hence the plane is flying with constant speed $1$ on a circle of radius $1$. We therefore may as well look at the motion
$$tmapsto bf z(t)=cos t>bf e_1+sin t>bf e_2$$
in the plane of said great circle, whereby $bf e_1$ and $bf e_2$ are fixed orthonormal vectors in this plane. One computes $bf v(t)=dotbf z(t)=-sin t>bf e_1+cos t>bf e_2$ and
$$bf a(t)=ddotbf z(t)=-cos t>bf e_1-sin t>bf e_2=-bf z(t) .$$
This says that the acceleration $bf a(t)$, considered as a vector in the tangent space $T_bf z(t)$, is at all times directed to the center of the circle. In your moving coordinate system this means that $bf a(t)$ is $>=-bf e_rin T_bf r$ at all times.
answered Jul 31 at 7:26
Christian Blatter
163k7107305
edited Aug 1 at 10:16
answered Jul 31 at 7:26
Christian Blatter
163k7107305
answered Jul 31 at 7:26
Christian Blatter
163k7107305
answered Jul 31 at 7:26
Christian Blatter
163k7107305
163k7107305
add a comment |Â
add a comment |Â
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