How do you find out if a polar curve retraces itself?
I was solving a question where you were supposed to find the area of petal formed by a polar curve. But the answer I got was greater than the actual answer by a factor of 2. Apparenty few polar curves retrace themselves, how do you find out wherther a polar curve retraces itself?
calculus graphing-functions
asked Nov 29 '18 at 13:48
Nikhil KrishnaNikhil Krishna
1
add a comment |
I was solving a question where you were supposed to find the area of petal formed by a polar curve. But the answer I got was greater than the actual answer by a factor of 2. Apparenty few polar curves retrace themselves, how do you find out wherther a polar curve retraces itself?
calculus graphing-functions
asked Nov 29 '18 at 13:48
Nikhil KrishnaNikhil Krishna
1
add a comment |
I was solving a question where you were supposed to find the area of petal formed by a polar curve. But the answer I got was greater than the actual answer by a factor of 2. Apparenty few polar curves retrace themselves, how do you find out wherther a polar curve retraces itself?
calculus graphing-functions
asked Nov 29 '18 at 13:48
Nikhil KrishnaNikhil Krishna
1
I was solving a question where you were supposed to find the area of petal formed by a polar curve. But the answer I got was greater than the actual answer by a factor of 2. Apparenty few polar curves retrace themselves, how do you find out wherther a polar curve retraces itself?
calculus graphing-functions
calculus graphing-functions
asked Nov 29 '18 at 13:48
Nikhil KrishnaNikhil Krishna
1
asked Nov 29 '18 at 13:48
Nikhil KrishnaNikhil Krishna
1
asked Nov 29 '18 at 13:48
Nikhil KrishnaNikhil Krishna
1
asked Nov 29 '18 at 13:48
Nikhil KrishnaNikhil Krishna
1
asked Nov 29 '18 at 13:48
Nikhil KrishnaNikhil Krishna
1
1
add a comment |
add a comment |
1 Answer
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If your function is $r(theta)$, with the polar radius expressed as a function of the polar angle, then it retraces itself if $r(theta) = r(theta + 2pi)$ (or possibly if $r(theta) = -r(theta + pi)$).
answered Nov 29 '18 at 13:54
ArthurArthur
111k7107189
The question I did satisfies the first condition you mentioned, but can you explain the reason?
– Nikhil Krishna
Nov 29 '18 at 14:00
1
@NikhilKrishna Because an angle of $theta$ and an angle of $theta + 2pi$ gives the same ray out from the origin. So if the value of $r$ is the same, then that means that the graph from the two (technically different) angles will overlap exactly. The second condition is basically the same thing, except an angle of $theta$ is exactly in the opposite direction of an angle of $theta + pi$, so if the value of $r$ is completely opposite as well, that means the graph once again overlaps.
– Arthur
Nov 29 '18 at 14:02
add a comment |
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1 Answer
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1 Answer
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If your function is $r(theta)$, with the polar radius expressed as a function of the polar angle, then it retraces itself if $r(theta) = r(theta + 2pi)$ (or possibly if $r(theta) = -r(theta + pi)$).
answered Nov 29 '18 at 13:54
ArthurArthur
111k7107189
The question I did satisfies the first condition you mentioned, but can you explain the reason?
– Nikhil Krishna
Nov 29 '18 at 14:00
1
@NikhilKrishna Because an angle of $theta$ and an angle of $theta + 2pi$ gives the same ray out from the origin. So if the value of $r$ is the same, then that means that the graph from the two (technically different) angles will overlap exactly. The second condition is basically the same thing, except an angle of $theta$ is exactly in the opposite direction of an angle of $theta + pi$, so if the value of $r$ is completely opposite as well, that means the graph once again overlaps.
– Arthur
Nov 29 '18 at 14:02
add a comment |
If your function is $r(theta)$, with the polar radius expressed as a function of the polar angle, then it retraces itself if $r(theta) = r(theta + 2pi)$ (or possibly if $r(theta) = -r(theta + pi)$).
answered Nov 29 '18 at 13:54
ArthurArthur
111k7107189
The question I did satisfies the first condition you mentioned, but can you explain the reason?
– Nikhil Krishna
Nov 29 '18 at 14:00
1
@NikhilKrishna Because an angle of $theta$ and an angle of $theta + 2pi$ gives the same ray out from the origin. So if the value of $r$ is the same, then that means that the graph from the two (technically different) angles will overlap exactly. The second condition is basically the same thing, except an angle of $theta$ is exactly in the opposite direction of an angle of $theta + pi$, so if the value of $r$ is completely opposite as well, that means the graph once again overlaps.
– Arthur
Nov 29 '18 at 14:02
add a comment |
If your function is $r(theta)$, with the polar radius expressed as a function of the polar angle, then it retraces itself if $r(theta) = r(theta + 2pi)$ (or possibly if $r(theta) = -r(theta + pi)$).
answered Nov 29 '18 at 13:54
ArthurArthur
111k7107189
If your function is $r(theta)$, with the polar radius expressed as a function of the polar angle, then it retraces itself if $r(theta) = r(theta + 2pi)$ (or possibly if $r(theta) = -r(theta + pi)$).
answered Nov 29 '18 at 13:54
ArthurArthur
111k7107189
answered Nov 29 '18 at 13:54
ArthurArthur
111k7107189
answered Nov 29 '18 at 13:54
ArthurArthur
111k7107189
answered Nov 29 '18 at 13:54
ArthurArthur
111k7107189
111k7107189
The question I did satisfies the first condition you mentioned, but can you explain the reason?
– Nikhil Krishna
Nov 29 '18 at 14:00
1
@NikhilKrishna Because an angle of $theta$ and an angle of $theta + 2pi$ gives the same ray out from the origin. So if the value of $r$ is the same, then that means that the graph from the two (technically different) angles will overlap exactly. The second condition is basically the same thing, except an angle of $theta$ is exactly in the opposite direction of an angle of $theta + pi$, so if the value of $r$ is completely opposite as well, that means the graph once again overlaps.
– Arthur
Nov 29 '18 at 14:02
add a comment |
The question I did satisfies the first condition you mentioned, but can you explain the reason?
– Nikhil Krishna
Nov 29 '18 at 14:00
1
@NikhilKrishna Because an angle of $theta$ and an angle of $theta + 2pi$ gives the same ray out from the origin. So if the value of $r$ is the same, then that means that the graph from the two (technically different) angles will overlap exactly. The second condition is basically the same thing, except an angle of $theta$ is exactly in the opposite direction of an angle of $theta + pi$, so if the value of $r$ is completely opposite as well, that means the graph once again overlaps.
– Arthur
Nov 29 '18 at 14:02
The question I did satisfies the first condition you mentioned, but can you explain the reason?
– Nikhil Krishna
Nov 29 '18 at 14:00
The question I did satisfies the first condition you mentioned, but can you explain the reason?
– Nikhil Krishna
Nov 29 '18 at 14:00
1
1
@NikhilKrishna Because an angle of $theta$ and an angle of $theta + 2pi$ gives the same ray out from the origin. So if the value of $r$ is the same, then that means that the graph from the two (technically different) angles will overlap exactly. The second condition is basically the same thing, except an angle of $theta$ is exactly in the opposite direction of an angle of $theta + pi$, so if the value of $r$ is completely opposite as well, that means the graph once again overlaps.
– Arthur
Nov 29 '18 at 14:02
@NikhilKrishna Because an angle of $theta$ and an angle of $theta + 2pi$ gives the same ray out from the origin. So if the value of $r$ is the same, then that means that the graph from the two (technically different) angles will overlap exactly. The second condition is basically the same thing, except an angle of $theta$ is exactly in the opposite direction of an angle of $theta + pi$, so if the value of $r$ is completely opposite as well, that means the graph once again overlaps.
– Arthur
Nov 29 '18 at 14:02
add a comment |
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