Two polar curves intersect problem .
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Suppose we have two curves given by: $$r=20sin2theta $$ $$r= 20 cos2theta$$
Now by solving the equations, we get the solution as $theta = frac{pi}{8}$.
However, on graphing the equations, I can find 8 points of intersection . Where could I have done a mistake?
geometry euclidean-geometry coordinate-systems polar-coordinates
edited Nov 21 at 6:20
Tusky
632618
asked Nov 21 at 5:25
user187604
2579
add a comment |
up vote
4
down vote
favorite
Suppose we have two curves given by: $$r=20sin2theta $$ $$r= 20 cos2theta$$
Now by solving the equations, we get the solution as $theta = frac{pi}{8}$.
However, on graphing the equations, I can find 8 points of intersection . Where could I have done a mistake?
geometry euclidean-geometry coordinate-systems polar-coordinates
edited Nov 21 at 6:20
Tusky
632618
asked Nov 21 at 5:25
user187604
2579
NB both parameterizations are periodic with period $pi$, so if $theta$ is a solution, so are $theta pm pi, theta pm 2 pi, ldots$---but this need not account for all solutions, and in this case it does not.
– Travis
Nov 21 at 6:29
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Suppose we have two curves given by: $$r=20sin2theta $$ $$r= 20 cos2theta$$
Now by solving the equations, we get the solution as $theta = frac{pi}{8}$.
However, on graphing the equations, I can find 8 points of intersection . Where could I have done a mistake?
geometry euclidean-geometry coordinate-systems polar-coordinates
edited Nov 21 at 6:20
Tusky
632618
asked Nov 21 at 5:25
user187604
2579
Suppose we have two curves given by: $$r=20sin2theta $$ $$r= 20 cos2theta$$
Now by solving the equations, we get the solution as $theta = frac{pi}{8}$.
However, on graphing the equations, I can find 8 points of intersection . Where could I have done a mistake?
geometry euclidean-geometry coordinate-systems polar-coordinates
geometry euclidean-geometry coordinate-systems polar-coordinates
edited Nov 21 at 6:20
Tusky
632618
asked Nov 21 at 5:25
user187604
2579
edited Nov 21 at 6:20
Tusky
632618
asked Nov 21 at 5:25
user187604
2579
edited Nov 21 at 6:20
Tusky
632618
edited Nov 21 at 6:20
Tusky
632618
edited Nov 21 at 6:20
Tusky
632618
632618
asked Nov 21 at 5:25
user187604
2579
asked Nov 21 at 5:25
user187604
2579
asked Nov 21 at 5:25
user187604
2579
2579
NB both parameterizations are periodic with period $pi$, so if $theta$ is a solution, so are $theta pm pi, theta pm 2 pi, ldots$---but this need not account for all solutions, and in this case it does not.
– Travis
Nov 21 at 6:29
add a comment |
NB both parameterizations are periodic with period $pi$, so if $theta$ is a solution, so are $theta pm pi, theta pm 2 pi, ldots$---but this need not account for all solutions, and in this case it does not.
– Travis
Nov 21 at 6:29
NB both parameterizations are periodic with period $pi$, so if $theta$ is a solution, so are $theta pm pi, theta pm 2 pi, ldots$---but this need not account for all solutions, and in this case it does not.
– Travis
Nov 21 at 6:29
NB both parameterizations are periodic with period $pi$, so if $theta$ is a solution, so are $theta pm pi, theta pm 2 pi, ldots$---but this need not account for all solutions, and in this case it does not.
– Travis
Nov 21 at 6:29
add a comment |
2 Answers
2
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oldest
votes
up vote
2
down vote
accepted
You have to take into account two issues. One of them is the periodicity of the $tan$ function. $tan(2theta)$ has a periodicity of $pi/2$, so you get $pi/8, 5pi/8, 9pi/8, 13pi/8$ as possible solutions in the $[0,2pi)$ interval. The other issue to consider is that the $r$ in the two equations is not necessarily the same. You also get solution if $r_1=-r_2$ and $sin(2theta)=-cos(2theta)$. From the definition of $arctan$ you get $theta=-pi/8$, which yields $3pi/8, 7pi/8, 11pi/8, 15pi/8$ in the above mentioned interval.
answered Nov 21 at 6:42
Andrei
10.4k21025
add a comment |
up vote
1
down vote
You're missing the point that $$ frac{pi}{8} $$ is just one of the solutions to this set of polar equations.
Let's find the general solution by setting both values of $r$ as equal to each other:
$$ 20 sin{2theta} = 20cos{2theta} implies sin{2theta} = cos{2theta} $$
Solving this trigonometric equation, we get our general solutions for $theta$ as:
$$ theta = frac{1}{8}(4pi * n + pi ) ; n in{Z} $$
Setting $n$ = $0$, you get your first solution: $frac{pi}{8}$.
With multiple values of $n$, we get multiple values of $theta$ and hence you can find $4$ values of $theta$ in the interval $[0,2pi]$ and hence you have your multiple points of intersection.
answered Nov 21 at 5:52
Tusky
632618
This will give you only 4 solutions in the $[0,2pi)$ interval
– Andrei
Nov 21 at 6:36
@Andrei Oops my bad, edited my answer!
– Tusky
Nov 22 at 7:30
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have to take into account two issues. One of them is the periodicity of the $tan$ function. $tan(2theta)$ has a periodicity of $pi/2$, so you get $pi/8, 5pi/8, 9pi/8, 13pi/8$ as possible solutions in the $[0,2pi)$ interval. The other issue to consider is that the $r$ in the two equations is not necessarily the same. You also get solution if $r_1=-r_2$ and $sin(2theta)=-cos(2theta)$. From the definition of $arctan$ you get $theta=-pi/8$, which yields $3pi/8, 7pi/8, 11pi/8, 15pi/8$ in the above mentioned interval.
answered Nov 21 at 6:42
Andrei
10.4k21025
add a comment |
up vote
2
down vote
accepted
You have to take into account two issues. One of them is the periodicity of the $tan$ function. $tan(2theta)$ has a periodicity of $pi/2$, so you get $pi/8, 5pi/8, 9pi/8, 13pi/8$ as possible solutions in the $[0,2pi)$ interval. The other issue to consider is that the $r$ in the two equations is not necessarily the same. You also get solution if $r_1=-r_2$ and $sin(2theta)=-cos(2theta)$. From the definition of $arctan$ you get $theta=-pi/8$, which yields $3pi/8, 7pi/8, 11pi/8, 15pi/8$ in the above mentioned interval.
answered Nov 21 at 6:42
Andrei
10.4k21025
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have to take into account two issues. One of them is the periodicity of the $tan$ function. $tan(2theta)$ has a periodicity of $pi/2$, so you get $pi/8, 5pi/8, 9pi/8, 13pi/8$ as possible solutions in the $[0,2pi)$ interval. The other issue to consider is that the $r$ in the two equations is not necessarily the same. You also get solution if $r_1=-r_2$ and $sin(2theta)=-cos(2theta)$. From the definition of $arctan$ you get $theta=-pi/8$, which yields $3pi/8, 7pi/8, 11pi/8, 15pi/8$ in the above mentioned interval.
answered Nov 21 at 6:42
Andrei
10.4k21025
You have to take into account two issues. One of them is the periodicity of the $tan$ function. $tan(2theta)$ has a periodicity of $pi/2$, so you get $pi/8, 5pi/8, 9pi/8, 13pi/8$ as possible solutions in the $[0,2pi)$ interval. The other issue to consider is that the $r$ in the two equations is not necessarily the same. You also get solution if $r_1=-r_2$ and $sin(2theta)=-cos(2theta)$. From the definition of $arctan$ you get $theta=-pi/8$, which yields $3pi/8, 7pi/8, 11pi/8, 15pi/8$ in the above mentioned interval.
answered Nov 21 at 6:42
Andrei
10.4k21025
answered Nov 21 at 6:42
Andrei
10.4k21025
answered Nov 21 at 6:42
Andrei
10.4k21025
answered Nov 21 at 6:42
Andrei
10.4k21025
10.4k21025
add a comment |
add a comment |
up vote
1
down vote
You're missing the point that $$ frac{pi}{8} $$ is just one of the solutions to this set of polar equations.
Let's find the general solution by setting both values of $r$ as equal to each other:
$$ 20 sin{2theta} = 20cos{2theta} implies sin{2theta} = cos{2theta} $$
Solving this trigonometric equation, we get our general solutions for $theta$ as:
$$ theta = frac{1}{8}(4pi * n + pi ) ; n in{Z} $$
Setting $n$ = $0$, you get your first solution: $frac{pi}{8}$.
With multiple values of $n$, we get multiple values of $theta$ and hence you can find $4$ values of $theta$ in the interval $[0,2pi]$ and hence you have your multiple points of intersection.
answered Nov 21 at 5:52
Tusky
632618
This will give you only 4 solutions in the $[0,2pi)$ interval
– Andrei
Nov 21 at 6:36
@Andrei Oops my bad, edited my answer!
– Tusky
Nov 22 at 7:30
add a comment |
up vote
1
down vote
You're missing the point that $$ frac{pi}{8} $$ is just one of the solutions to this set of polar equations.
Let's find the general solution by setting both values of $r$ as equal to each other:
$$ 20 sin{2theta} = 20cos{2theta} implies sin{2theta} = cos{2theta} $$
Solving this trigonometric equation, we get our general solutions for $theta$ as:
$$ theta = frac{1}{8}(4pi * n + pi ) ; n in{Z} $$
Setting $n$ = $0$, you get your first solution: $frac{pi}{8}$.
With multiple values of $n$, we get multiple values of $theta$ and hence you can find $4$ values of $theta$ in the interval $[0,2pi]$ and hence you have your multiple points of intersection.
answered Nov 21 at 5:52
Tusky
632618
This will give you only 4 solutions in the $[0,2pi)$ interval
– Andrei
Nov 21 at 6:36
@Andrei Oops my bad, edited my answer!
– Tusky
Nov 22 at 7:30
add a comment |
up vote
1
down vote
up vote
1
down vote
You're missing the point that $$ frac{pi}{8} $$ is just one of the solutions to this set of polar equations.
Let's find the general solution by setting both values of $r$ as equal to each other:
$$ 20 sin{2theta} = 20cos{2theta} implies sin{2theta} = cos{2theta} $$
Solving this trigonometric equation, we get our general solutions for $theta$ as:
$$ theta = frac{1}{8}(4pi * n + pi ) ; n in{Z} $$
Setting $n$ = $0$, you get your first solution: $frac{pi}{8}$.
With multiple values of $n$, we get multiple values of $theta$ and hence you can find $4$ values of $theta$ in the interval $[0,2pi]$ and hence you have your multiple points of intersection.
answered Nov 21 at 5:52
Tusky
632618
You're missing the point that $$ frac{pi}{8} $$ is just one of the solutions to this set of polar equations.
Let's find the general solution by setting both values of $r$ as equal to each other:
$$ 20 sin{2theta} = 20cos{2theta} implies sin{2theta} = cos{2theta} $$
Solving this trigonometric equation, we get our general solutions for $theta$ as:
$$ theta = frac{1}{8}(4pi * n + pi ) ; n in{Z} $$
Setting $n$ = $0$, you get your first solution: $frac{pi}{8}$.
With multiple values of $n$, we get multiple values of $theta$ and hence you can find $4$ values of $theta$ in the interval $[0,2pi]$ and hence you have your multiple points of intersection.
answered Nov 21 at 5:52
Tusky
632618
edited Nov 22 at 7:31
answered Nov 21 at 5:52
Tusky
632618
answered Nov 21 at 5:52
Tusky
632618
answered Nov 21 at 5:52
Tusky
632618
632618
This will give you only 4 solutions in the $[0,2pi)$ interval
– Andrei
Nov 21 at 6:36
@Andrei Oops my bad, edited my answer!
– Tusky
Nov 22 at 7:30
add a comment |
This will give you only 4 solutions in the $[0,2pi)$ interval
– Andrei
Nov 21 at 6:36
@Andrei Oops my bad, edited my answer!
– Tusky
Nov 22 at 7:30
This will give you only 4 solutions in the $[0,2pi)$ interval
– Andrei
Nov 21 at 6:36
This will give you only 4 solutions in the $[0,2pi)$ interval
– Andrei
Nov 21 at 6:36
@Andrei Oops my bad, edited my answer!
– Tusky
Nov 22 at 7:30
@Andrei Oops my bad, edited my answer!
– Tusky
Nov 22 at 7:30
add a comment |
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NB both parameterizations are periodic with period $pi$, so if $theta$ is a solution, so are $theta pm pi, theta pm 2 pi, ldots$---but this need not account for all solutions, and in this case it does not.
– Travis
Nov 21 at 6:29