Parametrization of distance to non-unit circle/sphere with non-centered origin
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I attempt to parametrize the distance $z(theta;r,z_0)$ from the origin $(x,y)=(0,0)$ of my coordinate system to arbitrary points $(x,y)$ on a circle, as a function of the variable $theta$ (angle), the parameter $r$ (circle radius), and the parameter $z_0$ (minimum distance) defined by $z_0equiv z(theta=0)$. The circle has its unknown origin on the $x$ axis, so that $z(theta=pi/2)= z(theta=3pi/2)$.
I know the parametrization for a unit circle, $(x-z_0)^2+2(x-z_0)+y^2=0$, and the obvious relation $z^2=x^2+y^2$. However, I don't know what parametrization of the circle I can use for an arbitrary circle radius $r$. And I don't know how to express $x$ in terms of $theta$ for arbitrary $(r,z_0)$.
Therefore, do you have any suggestions how to determine $z(theta;r,z_0)$? I expect this function to look similar to the sine function, just "shallower" and non-negative.
Finally, my second goal is to parametrize the distance $z(theta,phi;r,z_0)$ to arbitrary points on a sphere, which requires the introduction of another angle $phi$. Do you have any suggestions how to proceed?
analytic-geometry polar-coordinates spherical-coordinates parametrization
edited Nov 23 at 1:06
Jean Marie
28.6k41849
asked Nov 23 at 1:03
LCF
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0
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I attempt to parametrize the distance $z(theta;r,z_0)$ from the origin $(x,y)=(0,0)$ of my coordinate system to arbitrary points $(x,y)$ on a circle, as a function of the variable $theta$ (angle), the parameter $r$ (circle radius), and the parameter $z_0$ (minimum distance) defined by $z_0equiv z(theta=0)$. The circle has its unknown origin on the $x$ axis, so that $z(theta=pi/2)= z(theta=3pi/2)$.
I know the parametrization for a unit circle, $(x-z_0)^2+2(x-z_0)+y^2=0$, and the obvious relation $z^2=x^2+y^2$. However, I don't know what parametrization of the circle I can use for an arbitrary circle radius $r$. And I don't know how to express $x$ in terms of $theta$ for arbitrary $(r,z_0)$.
Therefore, do you have any suggestions how to determine $z(theta;r,z_0)$? I expect this function to look similar to the sine function, just "shallower" and non-negative.
Finally, my second goal is to parametrize the distance $z(theta,phi;r,z_0)$ to arbitrary points on a sphere, which requires the introduction of another angle $phi$. Do you have any suggestions how to proceed?
analytic-geometry polar-coordinates spherical-coordinates parametrization
edited Nov 23 at 1:06
Jean Marie
28.6k41849
asked Nov 23 at 1:03
LCF
1327
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I attempt to parametrize the distance $z(theta;r,z_0)$ from the origin $(x,y)=(0,0)$ of my coordinate system to arbitrary points $(x,y)$ on a circle, as a function of the variable $theta$ (angle), the parameter $r$ (circle radius), and the parameter $z_0$ (minimum distance) defined by $z_0equiv z(theta=0)$. The circle has its unknown origin on the $x$ axis, so that $z(theta=pi/2)= z(theta=3pi/2)$.
I know the parametrization for a unit circle, $(x-z_0)^2+2(x-z_0)+y^2=0$, and the obvious relation $z^2=x^2+y^2$. However, I don't know what parametrization of the circle I can use for an arbitrary circle radius $r$. And I don't know how to express $x$ in terms of $theta$ for arbitrary $(r,z_0)$.
Therefore, do you have any suggestions how to determine $z(theta;r,z_0)$? I expect this function to look similar to the sine function, just "shallower" and non-negative.
Finally, my second goal is to parametrize the distance $z(theta,phi;r,z_0)$ to arbitrary points on a sphere, which requires the introduction of another angle $phi$. Do you have any suggestions how to proceed?
analytic-geometry polar-coordinates spherical-coordinates parametrization
edited Nov 23 at 1:06
Jean Marie
28.6k41849
asked Nov 23 at 1:03
LCF
1327
I attempt to parametrize the distance $z(theta;r,z_0)$ from the origin $(x,y)=(0,0)$ of my coordinate system to arbitrary points $(x,y)$ on a circle, as a function of the variable $theta$ (angle), the parameter $r$ (circle radius), and the parameter $z_0$ (minimum distance) defined by $z_0equiv z(theta=0)$. The circle has its unknown origin on the $x$ axis, so that $z(theta=pi/2)= z(theta=3pi/2)$.
I know the parametrization for a unit circle, $(x-z_0)^2+2(x-z_0)+y^2=0$, and the obvious relation $z^2=x^2+y^2$. However, I don't know what parametrization of the circle I can use for an arbitrary circle radius $r$. And I don't know how to express $x$ in terms of $theta$ for arbitrary $(r,z_0)$.
Therefore, do you have any suggestions how to determine $z(theta;r,z_0)$? I expect this function to look similar to the sine function, just "shallower" and non-negative.
Finally, my second goal is to parametrize the distance $z(theta,phi;r,z_0)$ to arbitrary points on a sphere, which requires the introduction of another angle $phi$. Do you have any suggestions how to proceed?
analytic-geometry polar-coordinates spherical-coordinates parametrization
analytic-geometry polar-coordinates spherical-coordinates parametrization
edited Nov 23 at 1:06
Jean Marie
28.6k41849
asked Nov 23 at 1:03
LCF
1327
edited Nov 23 at 1:06
Jean Marie
28.6k41849
asked Nov 23 at 1:03
LCF
1327
edited Nov 23 at 1:06
Jean Marie
28.6k41849
edited Nov 23 at 1:06
Jean Marie
28.6k41849
edited Nov 23 at 1:06
Jean Marie
28.6k41849
28.6k41849
asked Nov 23 at 1:03
LCF
1327
asked Nov 23 at 1:03
LCF
1327
asked Nov 23 at 1:03
LCF
1327
1327
add a comment |
add a comment |
1 Answer
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An arbitrary circle of radius $r$ with it's centre on the $x$-axis at $x_0$ given by
$$ (x - x_0)^2 + y^2 = r^2 $$
We can rewrite this as
$$ left(frac{x - x_0}{r}right)^2 + left(frac{y}{r}right)^2 = 1$$
So that we can let $cos(theta) = left(frac{x - x_0}{r}right)$ and $sin(theta) = left(frac{y}{r}right)$ to get $x = rcos(theta) + x_0, y = rsin(theta)$. Hence we get the vector parametrization
$$mathbf{r} = (rcos(theta) + x_0,rsin(theta))$$
Which satisfies $||mathbf{r}|| = z$.
If $x_0 < 0$, then the minimum distance takes place at $theta = 0$, so $mathbf{r}_0 = (r + x_0,0) Rightarrow z_0 = sqrt{(x_0 + r)^2} Rightarrow z_0 = |x_0 + r|$ And the result varies depending on whether $r > |x_0|$ or $r < |x_0|$. I'll just do the case where $r < |x_0|$. Then $x_0 + r < 0$ so that $x_0 = -(z_0 + r)$. So,
$$mathbf{r} = left(rcos(theta) - (z_0 + r),rsin(theta)right)$$
And hence,
$$ z^2 = left(rcos(theta) - (z_0 + r)right)^2 + left(rsin(theta)right)^2$$
as desired. I'll emphasize again that since the relation between $z_0, x_0, r,$ and $theta$ changes depending on $x_0$ and $r$, you'll get slightly different relations for the noted conditions above.
answered Nov 23 at 2:02
AlkaKadri
1,459411
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
An arbitrary circle of radius $r$ with it's centre on the $x$-axis at $x_0$ given by
$$ (x - x_0)^2 + y^2 = r^2 $$
We can rewrite this as
$$ left(frac{x - x_0}{r}right)^2 + left(frac{y}{r}right)^2 = 1$$
So that we can let $cos(theta) = left(frac{x - x_0}{r}right)$ and $sin(theta) = left(frac{y}{r}right)$ to get $x = rcos(theta) + x_0, y = rsin(theta)$. Hence we get the vector parametrization
$$mathbf{r} = (rcos(theta) + x_0,rsin(theta))$$
Which satisfies $||mathbf{r}|| = z$.
If $x_0 < 0$, then the minimum distance takes place at $theta = 0$, so $mathbf{r}_0 = (r + x_0,0) Rightarrow z_0 = sqrt{(x_0 + r)^2} Rightarrow z_0 = |x_0 + r|$ And the result varies depending on whether $r > |x_0|$ or $r < |x_0|$. I'll just do the case where $r < |x_0|$. Then $x_0 + r < 0$ so that $x_0 = -(z_0 + r)$. So,
$$mathbf{r} = left(rcos(theta) - (z_0 + r),rsin(theta)right)$$
And hence,
$$ z^2 = left(rcos(theta) - (z_0 + r)right)^2 + left(rsin(theta)right)^2$$
as desired. I'll emphasize again that since the relation between $z_0, x_0, r,$ and $theta$ changes depending on $x_0$ and $r$, you'll get slightly different relations for the noted conditions above.
answered Nov 23 at 2:02
AlkaKadri
1,459411
add a comment |
up vote
1
down vote
accepted
An arbitrary circle of radius $r$ with it's centre on the $x$-axis at $x_0$ given by
$$ (x - x_0)^2 + y^2 = r^2 $$
We can rewrite this as
$$ left(frac{x - x_0}{r}right)^2 + left(frac{y}{r}right)^2 = 1$$
So that we can let $cos(theta) = left(frac{x - x_0}{r}right)$ and $sin(theta) = left(frac{y}{r}right)$ to get $x = rcos(theta) + x_0, y = rsin(theta)$. Hence we get the vector parametrization
$$mathbf{r} = (rcos(theta) + x_0,rsin(theta))$$
Which satisfies $||mathbf{r}|| = z$.
If $x_0 < 0$, then the minimum distance takes place at $theta = 0$, so $mathbf{r}_0 = (r + x_0,0) Rightarrow z_0 = sqrt{(x_0 + r)^2} Rightarrow z_0 = |x_0 + r|$ And the result varies depending on whether $r > |x_0|$ or $r < |x_0|$. I'll just do the case where $r < |x_0|$. Then $x_0 + r < 0$ so that $x_0 = -(z_0 + r)$. So,
$$mathbf{r} = left(rcos(theta) - (z_0 + r),rsin(theta)right)$$
And hence,
$$ z^2 = left(rcos(theta) - (z_0 + r)right)^2 + left(rsin(theta)right)^2$$
as desired. I'll emphasize again that since the relation between $z_0, x_0, r,$ and $theta$ changes depending on $x_0$ and $r$, you'll get slightly different relations for the noted conditions above.
answered Nov 23 at 2:02
AlkaKadri
1,459411
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
An arbitrary circle of radius $r$ with it's centre on the $x$-axis at $x_0$ given by
$$ (x - x_0)^2 + y^2 = r^2 $$
We can rewrite this as
$$ left(frac{x - x_0}{r}right)^2 + left(frac{y}{r}right)^2 = 1$$
So that we can let $cos(theta) = left(frac{x - x_0}{r}right)$ and $sin(theta) = left(frac{y}{r}right)$ to get $x = rcos(theta) + x_0, y = rsin(theta)$. Hence we get the vector parametrization
$$mathbf{r} = (rcos(theta) + x_0,rsin(theta))$$
Which satisfies $||mathbf{r}|| = z$.
If $x_0 < 0$, then the minimum distance takes place at $theta = 0$, so $mathbf{r}_0 = (r + x_0,0) Rightarrow z_0 = sqrt{(x_0 + r)^2} Rightarrow z_0 = |x_0 + r|$ And the result varies depending on whether $r > |x_0|$ or $r < |x_0|$. I'll just do the case where $r < |x_0|$. Then $x_0 + r < 0$ so that $x_0 = -(z_0 + r)$. So,
$$mathbf{r} = left(rcos(theta) - (z_0 + r),rsin(theta)right)$$
And hence,
$$ z^2 = left(rcos(theta) - (z_0 + r)right)^2 + left(rsin(theta)right)^2$$
as desired. I'll emphasize again that since the relation between $z_0, x_0, r,$ and $theta$ changes depending on $x_0$ and $r$, you'll get slightly different relations for the noted conditions above.
answered Nov 23 at 2:02
AlkaKadri
1,459411
An arbitrary circle of radius $r$ with it's centre on the $x$-axis at $x_0$ given by
$$ (x - x_0)^2 + y^2 = r^2 $$
We can rewrite this as
$$ left(frac{x - x_0}{r}right)^2 + left(frac{y}{r}right)^2 = 1$$
So that we can let $cos(theta) = left(frac{x - x_0}{r}right)$ and $sin(theta) = left(frac{y}{r}right)$ to get $x = rcos(theta) + x_0, y = rsin(theta)$. Hence we get the vector parametrization
$$mathbf{r} = (rcos(theta) + x_0,rsin(theta))$$
Which satisfies $||mathbf{r}|| = z$.
If $x_0 < 0$, then the minimum distance takes place at $theta = 0$, so $mathbf{r}_0 = (r + x_0,0) Rightarrow z_0 = sqrt{(x_0 + r)^2} Rightarrow z_0 = |x_0 + r|$ And the result varies depending on whether $r > |x_0|$ or $r < |x_0|$. I'll just do the case where $r < |x_0|$. Then $x_0 + r < 0$ so that $x_0 = -(z_0 + r)$. So,
$$mathbf{r} = left(rcos(theta) - (z_0 + r),rsin(theta)right)$$
And hence,
$$ z^2 = left(rcos(theta) - (z_0 + r)right)^2 + left(rsin(theta)right)^2$$
as desired. I'll emphasize again that since the relation between $z_0, x_0, r,$ and $theta$ changes depending on $x_0$ and $r$, you'll get slightly different relations for the noted conditions above.
answered Nov 23 at 2:02
AlkaKadri
1,459411
answered Nov 23 at 2:02
AlkaKadri
1,459411
answered Nov 23 at 2:02
AlkaKadri
1,459411
answered Nov 23 at 2:02
AlkaKadri
1,459411
1,459411
add a comment |
add a comment |
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