How do I convert this equation to polar coordinates?
Convert the equation $(x^2 + y^2)^2 = 2xy$ into polar coordinates.
I tried converting the equation in the image to polar coordinates and got $r= sqrt{sin(2theta)}$.
This answer is wrong, but I do not understand why. Can someone please explain what the correct answer is?
trigonometry
edited Nov 24 at 2:48
N. F. Taussig
43.5k93355
asked Nov 23 at 5:09
VRM
1
|
show 4 more comments
Convert the equation $(x^2 + y^2)^2 = 2xy$ into polar coordinates.
I tried converting the equation in the image to polar coordinates and got $r= sqrt{sin(2theta)}$.
This answer is wrong, but I do not understand why. Can someone please explain what the correct answer is?
trigonometry
edited Nov 24 at 2:48
N. F. Taussig
43.5k93355
asked Nov 23 at 5:09
VRM
1
What is the answer given?
– Akash Roy
Nov 23 at 5:14
The answer was not provided.
– VRM
Nov 23 at 5:15
You probably need both the positive and negative square roots.
– John Douma
Nov 23 at 5:16
Yeah that could be the only possibility.
– Akash Roy
Nov 23 at 5:17
There is not an option for that.
– VRM
Nov 23 at 5:17
|
show 4 more comments
Convert the equation $(x^2 + y^2)^2 = 2xy$ into polar coordinates.
I tried converting the equation in the image to polar coordinates and got $r= sqrt{sin(2theta)}$.
This answer is wrong, but I do not understand why. Can someone please explain what the correct answer is?
trigonometry
edited Nov 24 at 2:48
N. F. Taussig
43.5k93355
asked Nov 23 at 5:09
VRM
1
Convert the equation $(x^2 + y^2)^2 = 2xy$ into polar coordinates.
I tried converting the equation in the image to polar coordinates and got $r= sqrt{sin(2theta)}$.
This answer is wrong, but I do not understand why. Can someone please explain what the correct answer is?
trigonometry
trigonometry
edited Nov 24 at 2:48
N. F. Taussig
43.5k93355
asked Nov 23 at 5:09
VRM
1
edited Nov 24 at 2:48
N. F. Taussig
43.5k93355
asked Nov 23 at 5:09
VRM
1
edited Nov 24 at 2:48
N. F. Taussig
43.5k93355
edited Nov 24 at 2:48
N. F. Taussig
43.5k93355
edited Nov 24 at 2:48
N. F. Taussig
43.5k93355
43.5k93355
asked Nov 23 at 5:09
VRM
1
asked Nov 23 at 5:09
VRM
1
asked Nov 23 at 5:09
VRM
1
1
What is the answer given?
– Akash Roy
Nov 23 at 5:14
The answer was not provided.
– VRM
Nov 23 at 5:15
You probably need both the positive and negative square roots.
– John Douma
Nov 23 at 5:16
Yeah that could be the only possibility.
– Akash Roy
Nov 23 at 5:17
There is not an option for that.
– VRM
Nov 23 at 5:17
|
show 4 more comments
What is the answer given?
– Akash Roy
Nov 23 at 5:14
The answer was not provided.
– VRM
Nov 23 at 5:15
You probably need both the positive and negative square roots.
– John Douma
Nov 23 at 5:16
Yeah that could be the only possibility.
– Akash Roy
Nov 23 at 5:17
There is not an option for that.
– VRM
Nov 23 at 5:17
What is the answer given?
– Akash Roy
Nov 23 at 5:14
What is the answer given?
– Akash Roy
Nov 23 at 5:14
The answer was not provided.
– VRM
Nov 23 at 5:15
The answer was not provided.
– VRM
Nov 23 at 5:15
You probably need both the positive and negative square roots.
– John Douma
Nov 23 at 5:16
You probably need both the positive and negative square roots.
– John Douma
Nov 23 at 5:16
Yeah that could be the only possibility.
– Akash Roy
Nov 23 at 5:17
Yeah that could be the only possibility.
– Akash Roy
Nov 23 at 5:17
There is not an option for that.
– VRM
Nov 23 at 5:17
There is not an option for that.
– VRM
Nov 23 at 5:17
|
show 4 more comments
1 Answer
1
active
oldest
votes
Did you get something similar to this?
For the original equation $$(x^2 + y^2)^2 = 2 x y$$
Let $$begin {matrix} x = r cos theta \ y = r sin theta \ r^2 = x^2 + y^2 end {matrix}$$
Plugging all of this in to the original equation we get
$$(r^2)^2 = 2r^2 sin theta cos theta$$
or
$$r^4 = 2 r^2 sin theta cos theta$$
Dividing by $r^2$ we have
$$r^2 = 2 sin theta cos theta$$
However, $2 sin theta cos theta = sin 2 theta$ so we end up with
$$r^2 = sin 2 theta$$
We can leave as-is (I think this may be the answer they're expecting) or take square roots...
$$r = pm sqrt {sin 2 theta}$$
answered Nov 23 at 17:35
bjcolby15
1,1581916
I think it's supposed to be square root of sine double angle.
– VRM
Nov 24 at 2:04
Thanks, VRM - my mistake! Fixed.
– bjcolby15
Nov 24 at 2:43
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
Did you get something similar to this?
For the original equation $$(x^2 + y^2)^2 = 2 x y$$
Let $$begin {matrix} x = r cos theta \ y = r sin theta \ r^2 = x^2 + y^2 end {matrix}$$
Plugging all of this in to the original equation we get
$$(r^2)^2 = 2r^2 sin theta cos theta$$
or
$$r^4 = 2 r^2 sin theta cos theta$$
Dividing by $r^2$ we have
$$r^2 = 2 sin theta cos theta$$
However, $2 sin theta cos theta = sin 2 theta$ so we end up with
$$r^2 = sin 2 theta$$
We can leave as-is (I think this may be the answer they're expecting) or take square roots...
$$r = pm sqrt {sin 2 theta}$$
answered Nov 23 at 17:35
bjcolby15
1,1581916
I think it's supposed to be square root of sine double angle.
– VRM
Nov 24 at 2:04
Thanks, VRM - my mistake! Fixed.
– bjcolby15
Nov 24 at 2:43
add a comment |
Did you get something similar to this?
For the original equation $$(x^2 + y^2)^2 = 2 x y$$
Let $$begin {matrix} x = r cos theta \ y = r sin theta \ r^2 = x^2 + y^2 end {matrix}$$
Plugging all of this in to the original equation we get
$$(r^2)^2 = 2r^2 sin theta cos theta$$
or
$$r^4 = 2 r^2 sin theta cos theta$$
Dividing by $r^2$ we have
$$r^2 = 2 sin theta cos theta$$
However, $2 sin theta cos theta = sin 2 theta$ so we end up with
$$r^2 = sin 2 theta$$
We can leave as-is (I think this may be the answer they're expecting) or take square roots...
$$r = pm sqrt {sin 2 theta}$$
answered Nov 23 at 17:35
bjcolby15
1,1581916
I think it's supposed to be square root of sine double angle.
– VRM
Nov 24 at 2:04
Thanks, VRM - my mistake! Fixed.
– bjcolby15
Nov 24 at 2:43
add a comment |
Did you get something similar to this?
For the original equation $$(x^2 + y^2)^2 = 2 x y$$
Let $$begin {matrix} x = r cos theta \ y = r sin theta \ r^2 = x^2 + y^2 end {matrix}$$
Plugging all of this in to the original equation we get
$$(r^2)^2 = 2r^2 sin theta cos theta$$
or
$$r^4 = 2 r^2 sin theta cos theta$$
Dividing by $r^2$ we have
$$r^2 = 2 sin theta cos theta$$
However, $2 sin theta cos theta = sin 2 theta$ so we end up with
$$r^2 = sin 2 theta$$
We can leave as-is (I think this may be the answer they're expecting) or take square roots...
$$r = pm sqrt {sin 2 theta}$$
answered Nov 23 at 17:35
bjcolby15
1,1581916
Did you get something similar to this?
For the original equation $$(x^2 + y^2)^2 = 2 x y$$
Let $$begin {matrix} x = r cos theta \ y = r sin theta \ r^2 = x^2 + y^2 end {matrix}$$
Plugging all of this in to the original equation we get
$$(r^2)^2 = 2r^2 sin theta cos theta$$
or
$$r^4 = 2 r^2 sin theta cos theta$$
Dividing by $r^2$ we have
$$r^2 = 2 sin theta cos theta$$
However, $2 sin theta cos theta = sin 2 theta$ so we end up with
$$r^2 = sin 2 theta$$
We can leave as-is (I think this may be the answer they're expecting) or take square roots...
$$r = pm sqrt {sin 2 theta}$$
answered Nov 23 at 17:35
bjcolby15
1,1581916
edited Nov 24 at 2:43
answered Nov 23 at 17:35
bjcolby15
1,1581916
answered Nov 23 at 17:35
bjcolby15
1,1581916
answered Nov 23 at 17:35
bjcolby15
1,1581916
1,1581916
I think it's supposed to be square root of sine double angle.
– VRM
Nov 24 at 2:04
Thanks, VRM - my mistake! Fixed.
– bjcolby15
Nov 24 at 2:43
add a comment |
I think it's supposed to be square root of sine double angle.
– VRM
Nov 24 at 2:04
Thanks, VRM - my mistake! Fixed.
– bjcolby15
Nov 24 at 2:43
I think it's supposed to be square root of sine double angle.
– VRM
Nov 24 at 2:04
I think it's supposed to be square root of sine double angle.
– VRM
Nov 24 at 2:04
Thanks, VRM - my mistake! Fixed.
– bjcolby15
Nov 24 at 2:43
Thanks, VRM - my mistake! Fixed.
– bjcolby15
Nov 24 at 2:43
add a comment |
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What is the answer given?
– Akash Roy
Nov 23 at 5:14
The answer was not provided.
– VRM
Nov 23 at 5:15
You probably need both the positive and negative square roots.
– John Douma
Nov 23 at 5:16
Yeah that could be the only possibility.
– Akash Roy
Nov 23 at 5:17
There is not an option for that.
– VRM
Nov 23 at 5:17