Orthogonal trajectories of circles centered on x-axis The Next CEO of Stack OverflowHow to find the orthogonal trajectories of the family of all the circles through the points $(1,1)$ and $(-1,-1)$?How can I find the orthogonal trajectories of curvesOrthogonal trajectories . eliminating parameterFinding Orthogonal Trajectoriesfinding the orthogonal trajectoriesTo Find Orthogonal Trajectories Equation.Orthogonal Trajectories of a CircleDE: Orthogonal trajectories to an elipsewhat are parameter and constants in orthogonal trajectories?Orthogonal trajectories of the family of hyperbolas
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Orthogonal trajectories of circles centered on x-axis
The Next CEO of Stack OverflowHow to find the orthogonal trajectories of the family of all the circles through the points $(1,1)$ and $(-1,-1)$?How can I find the orthogonal trajectories of curvesOrthogonal trajectories . eliminating parameterFinding Orthogonal Trajectoriesfinding the orthogonal trajectoriesTo Find Orthogonal Trajectories Equation.Orthogonal Trajectories of a CircleDE: Orthogonal trajectories to an elipsewhat are parameter and constants in orthogonal trajectories?Orthogonal trajectories of the family of hyperbolas
$begingroup$
Find ( $c$ is an arbitrary constant ) orthogonal trajectories of circles:
$$ (x-c)^2+ y^2= 1 $$
integration ordinary-differential-equations plane-curves
edited 2 days ago
J. M. is not a mathematician
61.5k5152290
asked Mar 27 at 18:53
NarasimhamNarasimham
21.1k62258
$endgroup$
add a comment |
$begingroup$
Find ( $c$ is an arbitrary constant ) orthogonal trajectories of circles:
$$ (x-c)^2+ y^2= 1 $$
integration ordinary-differential-equations plane-curves
edited 2 days ago
J. M. is not a mathematician
61.5k5152290
asked Mar 27 at 18:53
NarasimhamNarasimham
21.1k62258
$endgroup$
$begingroup$
What have you tried?
$endgroup$
– Shubham Johri
Mar 27 at 18:55
$begingroup$
Got $ x=int sqrt1/y^2-1 dy$
$endgroup$
– Narasimham
Mar 27 at 19:11
$begingroup$
Here is an old cartoon I made that may be of interest.
$endgroup$
– J. M. is not a mathematician
2 days ago
add a comment |
$begingroup$
Find ( $c$ is an arbitrary constant ) orthogonal trajectories of circles:
$$ (x-c)^2+ y^2= 1 $$
integration ordinary-differential-equations plane-curves
edited 2 days ago
J. M. is not a mathematician
61.5k5152290
asked Mar 27 at 18:53
NarasimhamNarasimham
21.1k62258
$endgroup$
Find ( $c$ is an arbitrary constant ) orthogonal trajectories of circles:
$$ (x-c)^2+ y^2= 1 $$
integration ordinary-differential-equations plane-curves
integration ordinary-differential-equations plane-curves
edited 2 days ago
J. M. is not a mathematician
61.5k5152290
asked Mar 27 at 18:53
NarasimhamNarasimham
21.1k62258
edited 2 days ago
J. M. is not a mathematician
61.5k5152290
asked Mar 27 at 18:53
NarasimhamNarasimham
21.1k62258
edited 2 days ago
J. M. is not a mathematician
61.5k5152290
edited 2 days ago
J. M. is not a mathematician
61.5k5152290
edited 2 days ago
J. M. is not a mathematician
61.5k5152290
61.5k5152290
asked Mar 27 at 18:53
NarasimhamNarasimham
21.1k62258
asked Mar 27 at 18:53
NarasimhamNarasimham
21.1k62258
asked Mar 27 at 18:53
NarasimhamNarasimham
21.1k62258
21.1k62258
$begingroup$
What have you tried?
$endgroup$
– Shubham Johri
Mar 27 at 18:55
$begingroup$
Got $ x=int sqrt1/y^2-1 dy$
$endgroup$
– Narasimham
Mar 27 at 19:11
$begingroup$
Here is an old cartoon I made that may be of interest.
$endgroup$
– J. M. is not a mathematician
2 days ago
add a comment |
$begingroup$
What have you tried?
$endgroup$
– Shubham Johri
Mar 27 at 18:55
$begingroup$
Got $ x=int sqrt1/y^2-1 dy$
$endgroup$
– Narasimham
Mar 27 at 19:11
$begingroup$
Here is an old cartoon I made that may be of interest.
$endgroup$
– J. M. is not a mathematician
2 days ago
$begingroup$
What have you tried?
$endgroup$
– Shubham Johri
Mar 27 at 18:55
$begingroup$
What have you tried?
$endgroup$
– Shubham Johri
Mar 27 at 18:55
$begingroup$
Got $ x=int sqrt1/y^2-1 dy$
$endgroup$
– Narasimham
Mar 27 at 19:11
$begingroup$
Got $ x=int sqrt1/y^2-1 dy$
$endgroup$
– Narasimham
Mar 27 at 19:11
$begingroup$
Here is an old cartoon I made that may be of interest.
$endgroup$
– J. M. is not a mathematician
2 days ago
$begingroup$
Here is an old cartoon I made that may be of interest.
$endgroup$
– J. M. is not a mathematician
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$pm x+k=int dysqrtfrac1y^2-1$$Put $y=costheta$,$$pm x+k=-int sinthetatantheta~dtheta=-intfrac1-cos^2thetacosthetadtheta\=sintheta-ln|sectheta+tantheta|\=sqrt1-y^2-lnleft|1/y+sqrt1/y^2-1right|$$
answered Mar 27 at 19:06
Shubham JohriShubham Johri
5,477818
$endgroup$
$begingroup$
How to get $y=f(x)?$
$endgroup$
– Narasimham
Mar 27 at 19:16
$begingroup$
@Narasimham I'm afraid an explicit expression for $y$ using elementary functions is not possible
$endgroup$
– Shubham Johri
Mar 27 at 19:18
$begingroup$
How does $pm$ sign work? Are there two solutions? We know it to be a tractrix.
$endgroup$
– Narasimham
2 days ago
$begingroup$
@Narasimham The differential equation of the family of circles is $(y')^2=1/y^2-1implies y'=pmsqrt1/y^2-1$
$endgroup$
– Shubham Johri
2 days ago
$begingroup$
Is known as ...if $(y^'=fracdydx = tan phi ),rightarrow y = cos phi $
$endgroup$
– Narasimham
2 days ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$pm x+k=int dysqrtfrac1y^2-1$$Put $y=costheta$,$$pm x+k=-int sinthetatantheta~dtheta=-intfrac1-cos^2thetacosthetadtheta\=sintheta-ln|sectheta+tantheta|\=sqrt1-y^2-lnleft|1/y+sqrt1/y^2-1right|$$
answered Mar 27 at 19:06
Shubham JohriShubham Johri
5,477818
$endgroup$
$begingroup$
How to get $y=f(x)?$
$endgroup$
– Narasimham
Mar 27 at 19:16
$begingroup$
@Narasimham I'm afraid an explicit expression for $y$ using elementary functions is not possible
$endgroup$
– Shubham Johri
Mar 27 at 19:18
$begingroup$
How does $pm$ sign work? Are there two solutions? We know it to be a tractrix.
$endgroup$
– Narasimham
2 days ago
$begingroup$
@Narasimham The differential equation of the family of circles is $(y')^2=1/y^2-1implies y'=pmsqrt1/y^2-1$
$endgroup$
– Shubham Johri
2 days ago
$begingroup$
Is known as ...if $(y^'=fracdydx = tan phi ),rightarrow y = cos phi $
$endgroup$
– Narasimham
2 days ago
add a comment |
$begingroup$
$$pm x+k=int dysqrtfrac1y^2-1$$Put $y=costheta$,$$pm x+k=-int sinthetatantheta~dtheta=-intfrac1-cos^2thetacosthetadtheta\=sintheta-ln|sectheta+tantheta|\=sqrt1-y^2-lnleft|1/y+sqrt1/y^2-1right|$$
answered Mar 27 at 19:06
Shubham JohriShubham Johri
5,477818
$endgroup$
$begingroup$
How to get $y=f(x)?$
$endgroup$
– Narasimham
Mar 27 at 19:16
$begingroup$
@Narasimham I'm afraid an explicit expression for $y$ using elementary functions is not possible
$endgroup$
– Shubham Johri
Mar 27 at 19:18
$begingroup$
How does $pm$ sign work? Are there two solutions? We know it to be a tractrix.
$endgroup$
– Narasimham
2 days ago
$begingroup$
@Narasimham The differential equation of the family of circles is $(y')^2=1/y^2-1implies y'=pmsqrt1/y^2-1$
$endgroup$
– Shubham Johri
2 days ago
$begingroup$
Is known as ...if $(y^'=fracdydx = tan phi ),rightarrow y = cos phi $
$endgroup$
– Narasimham
2 days ago
add a comment |
$begingroup$
$$pm x+k=int dysqrtfrac1y^2-1$$Put $y=costheta$,$$pm x+k=-int sinthetatantheta~dtheta=-intfrac1-cos^2thetacosthetadtheta\=sintheta-ln|sectheta+tantheta|\=sqrt1-y^2-lnleft|1/y+sqrt1/y^2-1right|$$
answered Mar 27 at 19:06
Shubham JohriShubham Johri
5,477818
$endgroup$
$$pm x+k=int dysqrtfrac1y^2-1$$Put $y=costheta$,$$pm x+k=-int sinthetatantheta~dtheta=-intfrac1-cos^2thetacosthetadtheta\=sintheta-ln|sectheta+tantheta|\=sqrt1-y^2-lnleft|1/y+sqrt1/y^2-1right|$$
answered Mar 27 at 19:06
Shubham JohriShubham Johri
5,477818
answered Mar 27 at 19:06
Shubham JohriShubham Johri
5,477818
answered Mar 27 at 19:06
Shubham JohriShubham Johri
5,477818
answered Mar 27 at 19:06
Shubham JohriShubham Johri
5,477818
5,477818
$begingroup$
How to get $y=f(x)?$
$endgroup$
– Narasimham
Mar 27 at 19:16
$begingroup$
@Narasimham I'm afraid an explicit expression for $y$ using elementary functions is not possible
$endgroup$
– Shubham Johri
Mar 27 at 19:18
$begingroup$
How does $pm$ sign work? Are there two solutions? We know it to be a tractrix.
$endgroup$
– Narasimham
2 days ago
$begingroup$
@Narasimham The differential equation of the family of circles is $(y')^2=1/y^2-1implies y'=pmsqrt1/y^2-1$
$endgroup$
– Shubham Johri
2 days ago
$begingroup$
Is known as ...if $(y^'=fracdydx = tan phi ),rightarrow y = cos phi $
$endgroup$
– Narasimham
2 days ago
add a comment |
$begingroup$
How to get $y=f(x)?$
$endgroup$
– Narasimham
Mar 27 at 19:16
$begingroup$
@Narasimham I'm afraid an explicit expression for $y$ using elementary functions is not possible
$endgroup$
– Shubham Johri
Mar 27 at 19:18
$begingroup$
How does $pm$ sign work? Are there two solutions? We know it to be a tractrix.
$endgroup$
– Narasimham
2 days ago
$begingroup$
@Narasimham The differential equation of the family of circles is $(y')^2=1/y^2-1implies y'=pmsqrt1/y^2-1$
$endgroup$
– Shubham Johri
2 days ago
$begingroup$
Is known as ...if $(y^'=fracdydx = tan phi ),rightarrow y = cos phi $
$endgroup$
– Narasimham
2 days ago
$begingroup$
How to get $y=f(x)?$
$endgroup$
– Narasimham
Mar 27 at 19:16
$begingroup$
How to get $y=f(x)?$
$endgroup$
– Narasimham
Mar 27 at 19:16
$begingroup$
@Narasimham I'm afraid an explicit expression for $y$ using elementary functions is not possible
$endgroup$
– Shubham Johri
Mar 27 at 19:18
$begingroup$
@Narasimham I'm afraid an explicit expression for $y$ using elementary functions is not possible
$endgroup$
– Shubham Johri
Mar 27 at 19:18
$begingroup$
How does $pm$ sign work? Are there two solutions? We know it to be a tractrix.
$endgroup$
– Narasimham
2 days ago
$begingroup$
How does $pm$ sign work? Are there two solutions? We know it to be a tractrix.
$endgroup$
– Narasimham
2 days ago
$begingroup$
@Narasimham The differential equation of the family of circles is $(y')^2=1/y^2-1implies y'=pmsqrt1/y^2-1$
$endgroup$
– Shubham Johri
2 days ago
$begingroup$
@Narasimham The differential equation of the family of circles is $(y')^2=1/y^2-1implies y'=pmsqrt1/y^2-1$
$endgroup$
– Shubham Johri
2 days ago
$begingroup$
Is known as ...if $(y^'=fracdydx = tan phi ),rightarrow y = cos phi $
$endgroup$
– Narasimham
2 days ago
$begingroup$
Is known as ...if $(y^'=fracdydx = tan phi ),rightarrow y = cos phi $
$endgroup$
– Narasimham
2 days ago
add a comment |
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ArzQti11vspJerg5UIvLOfg6xlboe9h2fdHaA OYX
$begingroup$
What have you tried?
$endgroup$
– Shubham Johri
Mar 27 at 18:55
$begingroup$
Got $ x=int sqrt1/y^2-1 dy$
$endgroup$
– Narasimham
Mar 27 at 19:11
$begingroup$
Here is an old cartoon I made that may be of interest.
$endgroup$
– J. M. is not a mathematician
2 days ago