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When is a brachistochrone not a tautochrone?
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I thought I understood why the cycloid is both a brachistochrone (path of shortest time) and tautochrone (path of equal time for different starting points and the same end point) for the problem of a bead sliding, without friction, along a wire from $(x_1, y_1)$ to $(x_2, y_2)$, given $y_2 < y_1$ (with gravity pointing down along the negative $y$-axis).
But then someone I know asked about the case of $x_1=x_2$, where the cycloid becomes a vertical line. This must be a brachistochrone, but clearly isn't a tautochrone because objects take longer to fall to the same point from different heights. So where does the brachistochrone = tautochrone relation break down in this case?
What if $x_2 = x_1 + epsilon$ for $epsilon ll y_2-y_1$? The brachistochrone would still be a cycloid, but that's practically straight down: would it still be a tautochrone?
calculus-of-variations curves
asked Jul 16 at 8:46
Tom
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I thought I understood why the cycloid is both a brachistochrone (path of shortest time) and tautochrone (path of equal time for different starting points and the same end point) for the problem of a bead sliding, without friction, along a wire from $(x_1, y_1)$ to $(x_2, y_2)$, given $y_2 < y_1$ (with gravity pointing down along the negative $y$-axis).
But then someone I know asked about the case of $x_1=x_2$, where the cycloid becomes a vertical line. This must be a brachistochrone, but clearly isn't a tautochrone because objects take longer to fall to the same point from different heights. So where does the brachistochrone = tautochrone relation break down in this case?
What if $x_2 = x_1 + epsilon$ for $epsilon ll y_2-y_1$? The brachistochrone would still be a cycloid, but that's practically straight down: would it still be a tautochrone?
calculus-of-variations curves
asked Jul 16 at 8:46
Tom
1736
I think it's a tautochrone when the end point is the minimum of the cycloid. I haven't checked it in any way, but it seems reasonable to me.
– Arthur
Jul 16 at 8:53
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up vote
1
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up vote
1
down vote
favorite
I thought I understood why the cycloid is both a brachistochrone (path of shortest time) and tautochrone (path of equal time for different starting points and the same end point) for the problem of a bead sliding, without friction, along a wire from $(x_1, y_1)$ to $(x_2, y_2)$, given $y_2 < y_1$ (with gravity pointing down along the negative $y$-axis).
But then someone I know asked about the case of $x_1=x_2$, where the cycloid becomes a vertical line. This must be a brachistochrone, but clearly isn't a tautochrone because objects take longer to fall to the same point from different heights. So where does the brachistochrone = tautochrone relation break down in this case?
What if $x_2 = x_1 + epsilon$ for $epsilon ll y_2-y_1$? The brachistochrone would still be a cycloid, but that's practically straight down: would it still be a tautochrone?
calculus-of-variations curves
asked Jul 16 at 8:46
Tom
1736
I thought I understood why the cycloid is both a brachistochrone (path of shortest time) and tautochrone (path of equal time for different starting points and the same end point) for the problem of a bead sliding, without friction, along a wire from $(x_1, y_1)$ to $(x_2, y_2)$, given $y_2 < y_1$ (with gravity pointing down along the negative $y$-axis).
But then someone I know asked about the case of $x_1=x_2$, where the cycloid becomes a vertical line. This must be a brachistochrone, but clearly isn't a tautochrone because objects take longer to fall to the same point from different heights. So where does the brachistochrone = tautochrone relation break down in this case?
What if $x_2 = x_1 + epsilon$ for $epsilon ll y_2-y_1$? The brachistochrone would still be a cycloid, but that's practically straight down: would it still be a tautochrone?
calculus-of-variations curves
asked Jul 16 at 8:46
Tom
1736
edited Jul 16 at 8:52
asked Jul 16 at 8:46
Tom
1736
asked Jul 16 at 8:46
Tom
1736
asked Jul 16 at 8:46
Tom
1736
1736
I think it's a tautochrone when the end point is the minimum of the cycloid. I haven't checked it in any way, but it seems reasonable to me.
– Arthur
Jul 16 at 8:53
add a comment |Â
I think it's a tautochrone when the end point is the minimum of the cycloid. I haven't checked it in any way, but it seems reasonable to me.
– Arthur
Jul 16 at 8:53
I think it's a tautochrone when the end point is the minimum of the cycloid. I haven't checked it in any way, but it seems reasonable to me.
– Arthur
Jul 16 at 8:53
I think it's a tautochrone when the end point is the minimum of the cycloid. I haven't checked it in any way, but it seems reasonable to me.
– Arthur
Jul 16 at 8:53
add a comment |Â
1 Answer
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A brachistochrone always includes the cusp of the cycloid (not surprisingly, since the tangent becomes vertical there and this is the fastest way to accelerate initially), whereas the tautochrone always includes the minimum point (it is not isochronous to any other point, as can be seen by examining the integral for the descent time given on MathWorld with a more general angle than $pi$).
Therefore only the totality of the first descending arc is both a brachistochrone between its endpoints and a tautochrone to its bottom endpoint. Such a cycloid exists joining $(x_1,y_1)$ to $(x_2,y_2)$ when $fracx_2-x_1y_1-y_2 = fracpi2$, as may be seen by examining the parametric equations $frac12k^2(theta-sintheta,costheta-1)$. Any brachistochrone that contains a minimum is a tautochrone to its minimum (from either side, of course). On the other hand, any tautochrone that contains a cusp is a brachistochrone from that cusp to any other point on it.
answered Jul 16 at 12:59
Chappers
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A brachistochrone always includes the cusp of the cycloid (not surprisingly, since the tangent becomes vertical there and this is the fastest way to accelerate initially), whereas the tautochrone always includes the minimum point (it is not isochronous to any other point, as can be seen by examining the integral for the descent time given on MathWorld with a more general angle than $pi$).
Therefore only the totality of the first descending arc is both a brachistochrone between its endpoints and a tautochrone to its bottom endpoint. Such a cycloid exists joining $(x_1,y_1)$ to $(x_2,y_2)$ when $fracx_2-x_1y_1-y_2 = fracpi2$, as may be seen by examining the parametric equations $frac12k^2(theta-sintheta,costheta-1)$. Any brachistochrone that contains a minimum is a tautochrone to its minimum (from either side, of course). On the other hand, any tautochrone that contains a cusp is a brachistochrone from that cusp to any other point on it.
answered Jul 16 at 12:59
Chappers
55k74191
add a comment |Â
up vote
2
down vote
accepted
A brachistochrone always includes the cusp of the cycloid (not surprisingly, since the tangent becomes vertical there and this is the fastest way to accelerate initially), whereas the tautochrone always includes the minimum point (it is not isochronous to any other point, as can be seen by examining the integral for the descent time given on MathWorld with a more general angle than $pi$).
Therefore only the totality of the first descending arc is both a brachistochrone between its endpoints and a tautochrone to its bottom endpoint. Such a cycloid exists joining $(x_1,y_1)$ to $(x_2,y_2)$ when $fracx_2-x_1y_1-y_2 = fracpi2$, as may be seen by examining the parametric equations $frac12k^2(theta-sintheta,costheta-1)$. Any brachistochrone that contains a minimum is a tautochrone to its minimum (from either side, of course). On the other hand, any tautochrone that contains a cusp is a brachistochrone from that cusp to any other point on it.
answered Jul 16 at 12:59
Chappers
55k74191
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A brachistochrone always includes the cusp of the cycloid (not surprisingly, since the tangent becomes vertical there and this is the fastest way to accelerate initially), whereas the tautochrone always includes the minimum point (it is not isochronous to any other point, as can be seen by examining the integral for the descent time given on MathWorld with a more general angle than $pi$).
Therefore only the totality of the first descending arc is both a brachistochrone between its endpoints and a tautochrone to its bottom endpoint. Such a cycloid exists joining $(x_1,y_1)$ to $(x_2,y_2)$ when $fracx_2-x_1y_1-y_2 = fracpi2$, as may be seen by examining the parametric equations $frac12k^2(theta-sintheta,costheta-1)$. Any brachistochrone that contains a minimum is a tautochrone to its minimum (from either side, of course). On the other hand, any tautochrone that contains a cusp is a brachistochrone from that cusp to any other point on it.
answered Jul 16 at 12:59
Chappers
55k74191
A brachistochrone always includes the cusp of the cycloid (not surprisingly, since the tangent becomes vertical there and this is the fastest way to accelerate initially), whereas the tautochrone always includes the minimum point (it is not isochronous to any other point, as can be seen by examining the integral for the descent time given on MathWorld with a more general angle than $pi$).
Therefore only the totality of the first descending arc is both a brachistochrone between its endpoints and a tautochrone to its bottom endpoint. Such a cycloid exists joining $(x_1,y_1)$ to $(x_2,y_2)$ when $fracx_2-x_1y_1-y_2 = fracpi2$, as may be seen by examining the parametric equations $frac12k^2(theta-sintheta,costheta-1)$. Any brachistochrone that contains a minimum is a tautochrone to its minimum (from either side, of course). On the other hand, any tautochrone that contains a cusp is a brachistochrone from that cusp to any other point on it.
answered Jul 16 at 12:59
Chappers
55k74191
answered Jul 16 at 12:59
Chappers
55k74191
answered Jul 16 at 12:59
Chappers
55k74191
answered Jul 16 at 12:59
Chappers
55k74191
55k74191
add a comment |Â
add a comment |Â
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I think it's a tautochrone when the end point is the minimum of the cycloid. I haven't checked it in any way, but it seems reasonable to me.
– Arthur
Jul 16 at 8:53