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Generalized polar coordinates, how to switch form cartesian to polar
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I was solving a problem: $$left(frac xa + frac ybright)^ 4=4xy,quad a>0 , b>0 $$
Find the are bounded by curve.
It is not hard to see that x and y must have the same sign, that function exist only in first and third quadrant, and that the function is symetrical to ( 0,0).
So i want to transform this to polar coordinates. The regular transformation is $$ x=rcosphi \ y=rsinphi $$
I have noticed that these are only used for circle with centre in (0,0). Whenever the function is more complicated the transformation is different. Why ?
The book solution says that we will moved generalized coordinates use moved generalized coordinates :
$$
x-x_0 = arbeta cos alpha phi\
y-y_0 = brbeta sin alpha phi
$$
Then it says $$ x_0=y_0=0 $$ $$ alpha=2, beta=1 $$ It says we did this to make the curve equation simpler. I see why the first equation is true but the second with $alpha$ and $beta$, I don't understand. Why are $alpha =2, beta = 1$, and where does this generalized polar coordinates equation come from ?
Also, how to think when switching to polar coordinates when dealing with ËÂmore complicated curvesË ( for me they are complicated)
Edit : also i have seen a different transformation used in this problem $$ x=arcos^ 2phi \ y=brsin^ 2phi $$ then this was transformed with double integral identity.
calculus real-analysis multivariable-calculus polar-coordinates multiple-integral
asked Jul 19 at 13:55
Milan Stojanovic
378
add a comment |Â
up vote
2
down vote
favorite
I was solving a problem: $$left(frac xa + frac ybright)^ 4=4xy,quad a>0 , b>0 $$
Find the are bounded by curve.
It is not hard to see that x and y must have the same sign, that function exist only in first and third quadrant, and that the function is symetrical to ( 0,0).
So i want to transform this to polar coordinates. The regular transformation is $$ x=rcosphi \ y=rsinphi $$
I have noticed that these are only used for circle with centre in (0,0). Whenever the function is more complicated the transformation is different. Why ?
The book solution says that we will moved generalized coordinates use moved generalized coordinates :
$$
x-x_0 = arbeta cos alpha phi\
y-y_0 = brbeta sin alpha phi
$$
Then it says $$ x_0=y_0=0 $$ $$ alpha=2, beta=1 $$ It says we did this to make the curve equation simpler. I see why the first equation is true but the second with $alpha$ and $beta$, I don't understand. Why are $alpha =2, beta = 1$, and where does this generalized polar coordinates equation come from ?
Also, how to think when switching to polar coordinates when dealing with ËÂmore complicated curvesË ( for me they are complicated)
Edit : also i have seen a different transformation used in this problem $$ x=arcos^ 2phi \ y=brsin^ 2phi $$ then this was transformed with double integral identity.
calculus real-analysis multivariable-calculus polar-coordinates multiple-integral
asked Jul 19 at 13:55
Milan Stojanovic
378
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was solving a problem: $$left(frac xa + frac ybright)^ 4=4xy,quad a>0 , b>0 $$
Find the are bounded by curve.
It is not hard to see that x and y must have the same sign, that function exist only in first and third quadrant, and that the function is symetrical to ( 0,0).
So i want to transform this to polar coordinates. The regular transformation is $$ x=rcosphi \ y=rsinphi $$
I have noticed that these are only used for circle with centre in (0,0). Whenever the function is more complicated the transformation is different. Why ?
The book solution says that we will moved generalized coordinates use moved generalized coordinates :
$$
x-x_0 = arbeta cos alpha phi\
y-y_0 = brbeta sin alpha phi
$$
Then it says $$ x_0=y_0=0 $$ $$ alpha=2, beta=1 $$ It says we did this to make the curve equation simpler. I see why the first equation is true but the second with $alpha$ and $beta$, I don't understand. Why are $alpha =2, beta = 1$, and where does this generalized polar coordinates equation come from ?
Also, how to think when switching to polar coordinates when dealing with ËÂmore complicated curvesË ( for me they are complicated)
Edit : also i have seen a different transformation used in this problem $$ x=arcos^ 2phi \ y=brsin^ 2phi $$ then this was transformed with double integral identity.
calculus real-analysis multivariable-calculus polar-coordinates multiple-integral
asked Jul 19 at 13:55
Milan Stojanovic
378
I was solving a problem: $$left(frac xa + frac ybright)^ 4=4xy,quad a>0 , b>0 $$
Find the are bounded by curve.
It is not hard to see that x and y must have the same sign, that function exist only in first and third quadrant, and that the function is symetrical to ( 0,0).
So i want to transform this to polar coordinates. The regular transformation is $$ x=rcosphi \ y=rsinphi $$
I have noticed that these are only used for circle with centre in (0,0). Whenever the function is more complicated the transformation is different. Why ?
The book solution says that we will moved generalized coordinates use moved generalized coordinates :
$$
x-x_0 = arbeta cos alpha phi\
y-y_0 = brbeta sin alpha phi
$$
Then it says $$ x_0=y_0=0 $$ $$ alpha=2, beta=1 $$ It says we did this to make the curve equation simpler. I see why the first equation is true but the second with $alpha$ and $beta$, I don't understand. Why are $alpha =2, beta = 1$, and where does this generalized polar coordinates equation come from ?
Also, how to think when switching to polar coordinates when dealing with ËÂmore complicated curvesË ( for me they are complicated)
Edit : also i have seen a different transformation used in this problem $$ x=arcos^ 2phi \ y=brsin^ 2phi $$ then this was transformed with double integral identity.
calculus real-analysis multivariable-calculus polar-coordinates multiple-integral
asked Jul 19 at 13:55
Milan Stojanovic
378
edited Jul 19 at 19:18
asked Jul 19 at 13:55
Milan Stojanovic
378
asked Jul 19 at 13:55
Milan Stojanovic
378
asked Jul 19 at 13:55
Milan Stojanovic
378
378
add a comment |Â
add a comment |Â
1 Answer
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First of all, you can use "ordinary" polar coordinates for lots of curves other than circles centered at the origin. One of my particular favorites is a circle passing through the origin and centered on the $x$-axis. Its polar equation will be $r=2acostheta$ (where $a$ is its radius).
If you have an ellipse $dfracx^2a^2+dfracy^2b^2=1$, then it is natural to think of it as coming from the unit circle by stretching the $x$-axis by a factor of $a$ and the $y$-axis by a factor of $b$. So this would naturally be parametrized by $x=acostheta$, $y=bsintheta$, and you get a modified polar coordinate system by "filling it in" with $x=arcostheta$, $y=arsintheta$. (Note that $theta$ is no longer the angle between the $x$-axis and the ray to the point.)
Suppose we consider a slightly easier version of your question, since we've already covered the effect of stretching with $a$ and $b$. Let's look at the equation
$$(x+y)^4 = 4xy.$$
I'm not sure I get the point of introducing $alpha$. The equation of this curve in usual polar coordinates would be
$$r^2 = frac2sin 2theta(costheta+sintheta)^4,$$
with $0lethetalepi/2$ or $pilethetale 3pi/2$ (to make $sin 2thetage 0$).
If we use $alpha = 2$, with $x=rcos 2theta$ and $y=rsin 2theta$, this equation becomes
$$r^2 = frac2sin 4theta(cos 2theta + sin 2theta)^4,$$
with $0lethetale pi/4$ or $pi/2lethetale 3pi/4$.
Although it's tempting to see that the difference in degree ($4$ on the LHS and $2$ on the RHS) should suggest some $alpha$, I certainly don't see how to simplify the second equation any more easily. (It's tempting because of things like deMoivre's formula, but I don't see where it goes.)
So, how would we try to simplify either expression? We note that
$$costheta+sintheta = sqrt2sinbig(theta+fracpi4big),$$
so we want to substitute $psi = theta+fracpi4$. Then
and then $sin 2theta = sin 2(psi-fracpi4)=sin(2psi - fracpi2) = -cos 2psi$. Then we end up with
$$r^2 = frac-cos 2psi2sin^4psi, quad pi/4lepsile 3pi/4 quadtextorquad 5pi/4lepsile 7pi/4.$$
Now let's try this with the second formulation. Analogously, we have
$$cos 2theta+sin 2theta = sqrt2sinbig(2theta+fracpi4big) = sqrt2sin 2big(theta+fracpi8big).$$
This seems very unhelpful. Perhaps we should try scaling with $alpha = 1/2$?
Then we'd have
$$sinfractheta2 + cosfractheta2 = sqrt2sinbig(fractheta2 +fracpi4big) = sqrt2sinfrac12big(theta+fracpi2big).$$
Does that seem a bit more promising?
At any rate, I am not sure I believe your book is right. A translation of $theta$ is clearly called for before a rescaling of $theta$.
answered Jul 19 at 17:49
Ted Shifrin
59.5k44387
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
First of all, you can use "ordinary" polar coordinates for lots of curves other than circles centered at the origin. One of my particular favorites is a circle passing through the origin and centered on the $x$-axis. Its polar equation will be $r=2acostheta$ (where $a$ is its radius).
If you have an ellipse $dfracx^2a^2+dfracy^2b^2=1$, then it is natural to think of it as coming from the unit circle by stretching the $x$-axis by a factor of $a$ and the $y$-axis by a factor of $b$. So this would naturally be parametrized by $x=acostheta$, $y=bsintheta$, and you get a modified polar coordinate system by "filling it in" with $x=arcostheta$, $y=arsintheta$. (Note that $theta$ is no longer the angle between the $x$-axis and the ray to the point.)
Suppose we consider a slightly easier version of your question, since we've already covered the effect of stretching with $a$ and $b$. Let's look at the equation
$$(x+y)^4 = 4xy.$$
I'm not sure I get the point of introducing $alpha$. The equation of this curve in usual polar coordinates would be
$$r^2 = frac2sin 2theta(costheta+sintheta)^4,$$
with $0lethetalepi/2$ or $pilethetale 3pi/2$ (to make $sin 2thetage 0$).
If we use $alpha = 2$, with $x=rcos 2theta$ and $y=rsin 2theta$, this equation becomes
$$r^2 = frac2sin 4theta(cos 2theta + sin 2theta)^4,$$
with $0lethetale pi/4$ or $pi/2lethetale 3pi/4$.
Although it's tempting to see that the difference in degree ($4$ on the LHS and $2$ on the RHS) should suggest some $alpha$, I certainly don't see how to simplify the second equation any more easily. (It's tempting because of things like deMoivre's formula, but I don't see where it goes.)
So, how would we try to simplify either expression? We note that
$$costheta+sintheta = sqrt2sinbig(theta+fracpi4big),$$
so we want to substitute $psi = theta+fracpi4$. Then
and then $sin 2theta = sin 2(psi-fracpi4)=sin(2psi - fracpi2) = -cos 2psi$. Then we end up with
$$r^2 = frac-cos 2psi2sin^4psi, quad pi/4lepsile 3pi/4 quadtextorquad 5pi/4lepsile 7pi/4.$$
Now let's try this with the second formulation. Analogously, we have
$$cos 2theta+sin 2theta = sqrt2sinbig(2theta+fracpi4big) = sqrt2sin 2big(theta+fracpi8big).$$
This seems very unhelpful. Perhaps we should try scaling with $alpha = 1/2$?
Then we'd have
$$sinfractheta2 + cosfractheta2 = sqrt2sinbig(fractheta2 +fracpi4big) = sqrt2sinfrac12big(theta+fracpi2big).$$
Does that seem a bit more promising?
At any rate, I am not sure I believe your book is right. A translation of $theta$ is clearly called for before a rescaling of $theta$.
answered Jul 19 at 17:49
Ted Shifrin
59.5k44387
add a comment |Â
up vote
1
down vote
First of all, you can use "ordinary" polar coordinates for lots of curves other than circles centered at the origin. One of my particular favorites is a circle passing through the origin and centered on the $x$-axis. Its polar equation will be $r=2acostheta$ (where $a$ is its radius).
If you have an ellipse $dfracx^2a^2+dfracy^2b^2=1$, then it is natural to think of it as coming from the unit circle by stretching the $x$-axis by a factor of $a$ and the $y$-axis by a factor of $b$. So this would naturally be parametrized by $x=acostheta$, $y=bsintheta$, and you get a modified polar coordinate system by "filling it in" with $x=arcostheta$, $y=arsintheta$. (Note that $theta$ is no longer the angle between the $x$-axis and the ray to the point.)
Suppose we consider a slightly easier version of your question, since we've already covered the effect of stretching with $a$ and $b$. Let's look at the equation
$$(x+y)^4 = 4xy.$$
I'm not sure I get the point of introducing $alpha$. The equation of this curve in usual polar coordinates would be
$$r^2 = frac2sin 2theta(costheta+sintheta)^4,$$
with $0lethetalepi/2$ or $pilethetale 3pi/2$ (to make $sin 2thetage 0$).
If we use $alpha = 2$, with $x=rcos 2theta$ and $y=rsin 2theta$, this equation becomes
$$r^2 = frac2sin 4theta(cos 2theta + sin 2theta)^4,$$
with $0lethetale pi/4$ or $pi/2lethetale 3pi/4$.
Although it's tempting to see that the difference in degree ($4$ on the LHS and $2$ on the RHS) should suggest some $alpha$, I certainly don't see how to simplify the second equation any more easily. (It's tempting because of things like deMoivre's formula, but I don't see where it goes.)
So, how would we try to simplify either expression? We note that
$$costheta+sintheta = sqrt2sinbig(theta+fracpi4big),$$
so we want to substitute $psi = theta+fracpi4$. Then
and then $sin 2theta = sin 2(psi-fracpi4)=sin(2psi - fracpi2) = -cos 2psi$. Then we end up with
$$r^2 = frac-cos 2psi2sin^4psi, quad pi/4lepsile 3pi/4 quadtextorquad 5pi/4lepsile 7pi/4.$$
Now let's try this with the second formulation. Analogously, we have
$$cos 2theta+sin 2theta = sqrt2sinbig(2theta+fracpi4big) = sqrt2sin 2big(theta+fracpi8big).$$
This seems very unhelpful. Perhaps we should try scaling with $alpha = 1/2$?
Then we'd have
$$sinfractheta2 + cosfractheta2 = sqrt2sinbig(fractheta2 +fracpi4big) = sqrt2sinfrac12big(theta+fracpi2big).$$
Does that seem a bit more promising?
At any rate, I am not sure I believe your book is right. A translation of $theta$ is clearly called for before a rescaling of $theta$.
answered Jul 19 at 17:49
Ted Shifrin
59.5k44387
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First of all, you can use "ordinary" polar coordinates for lots of curves other than circles centered at the origin. One of my particular favorites is a circle passing through the origin and centered on the $x$-axis. Its polar equation will be $r=2acostheta$ (where $a$ is its radius).
If you have an ellipse $dfracx^2a^2+dfracy^2b^2=1$, then it is natural to think of it as coming from the unit circle by stretching the $x$-axis by a factor of $a$ and the $y$-axis by a factor of $b$. So this would naturally be parametrized by $x=acostheta$, $y=bsintheta$, and you get a modified polar coordinate system by "filling it in" with $x=arcostheta$, $y=arsintheta$. (Note that $theta$ is no longer the angle between the $x$-axis and the ray to the point.)
Suppose we consider a slightly easier version of your question, since we've already covered the effect of stretching with $a$ and $b$. Let's look at the equation
$$(x+y)^4 = 4xy.$$
I'm not sure I get the point of introducing $alpha$. The equation of this curve in usual polar coordinates would be
$$r^2 = frac2sin 2theta(costheta+sintheta)^4,$$
with $0lethetalepi/2$ or $pilethetale 3pi/2$ (to make $sin 2thetage 0$).
If we use $alpha = 2$, with $x=rcos 2theta$ and $y=rsin 2theta$, this equation becomes
$$r^2 = frac2sin 4theta(cos 2theta + sin 2theta)^4,$$
with $0lethetale pi/4$ or $pi/2lethetale 3pi/4$.
Although it's tempting to see that the difference in degree ($4$ on the LHS and $2$ on the RHS) should suggest some $alpha$, I certainly don't see how to simplify the second equation any more easily. (It's tempting because of things like deMoivre's formula, but I don't see where it goes.)
So, how would we try to simplify either expression? We note that
$$costheta+sintheta = sqrt2sinbig(theta+fracpi4big),$$
so we want to substitute $psi = theta+fracpi4$. Then
and then $sin 2theta = sin 2(psi-fracpi4)=sin(2psi - fracpi2) = -cos 2psi$. Then we end up with
$$r^2 = frac-cos 2psi2sin^4psi, quad pi/4lepsile 3pi/4 quadtextorquad 5pi/4lepsile 7pi/4.$$
Now let's try this with the second formulation. Analogously, we have
$$cos 2theta+sin 2theta = sqrt2sinbig(2theta+fracpi4big) = sqrt2sin 2big(theta+fracpi8big).$$
This seems very unhelpful. Perhaps we should try scaling with $alpha = 1/2$?
Then we'd have
$$sinfractheta2 + cosfractheta2 = sqrt2sinbig(fractheta2 +fracpi4big) = sqrt2sinfrac12big(theta+fracpi2big).$$
Does that seem a bit more promising?
At any rate, I am not sure I believe your book is right. A translation of $theta$ is clearly called for before a rescaling of $theta$.
answered Jul 19 at 17:49
Ted Shifrin
59.5k44387
First of all, you can use "ordinary" polar coordinates for lots of curves other than circles centered at the origin. One of my particular favorites is a circle passing through the origin and centered on the $x$-axis. Its polar equation will be $r=2acostheta$ (where $a$ is its radius).
If you have an ellipse $dfracx^2a^2+dfracy^2b^2=1$, then it is natural to think of it as coming from the unit circle by stretching the $x$-axis by a factor of $a$ and the $y$-axis by a factor of $b$. So this would naturally be parametrized by $x=acostheta$, $y=bsintheta$, and you get a modified polar coordinate system by "filling it in" with $x=arcostheta$, $y=arsintheta$. (Note that $theta$ is no longer the angle between the $x$-axis and the ray to the point.)
Suppose we consider a slightly easier version of your question, since we've already covered the effect of stretching with $a$ and $b$. Let's look at the equation
$$(x+y)^4 = 4xy.$$
I'm not sure I get the point of introducing $alpha$. The equation of this curve in usual polar coordinates would be
$$r^2 = frac2sin 2theta(costheta+sintheta)^4,$$
with $0lethetalepi/2$ or $pilethetale 3pi/2$ (to make $sin 2thetage 0$).
If we use $alpha = 2$, with $x=rcos 2theta$ and $y=rsin 2theta$, this equation becomes
$$r^2 = frac2sin 4theta(cos 2theta + sin 2theta)^4,$$
with $0lethetale pi/4$ or $pi/2lethetale 3pi/4$.
Although it's tempting to see that the difference in degree ($4$ on the LHS and $2$ on the RHS) should suggest some $alpha$, I certainly don't see how to simplify the second equation any more easily. (It's tempting because of things like deMoivre's formula, but I don't see where it goes.)
So, how would we try to simplify either expression? We note that
$$costheta+sintheta = sqrt2sinbig(theta+fracpi4big),$$
so we want to substitute $psi = theta+fracpi4$. Then
and then $sin 2theta = sin 2(psi-fracpi4)=sin(2psi - fracpi2) = -cos 2psi$. Then we end up with
$$r^2 = frac-cos 2psi2sin^4psi, quad pi/4lepsile 3pi/4 quadtextorquad 5pi/4lepsile 7pi/4.$$
Now let's try this with the second formulation. Analogously, we have
$$cos 2theta+sin 2theta = sqrt2sinbig(2theta+fracpi4big) = sqrt2sin 2big(theta+fracpi8big).$$
This seems very unhelpful. Perhaps we should try scaling with $alpha = 1/2$?
Then we'd have
$$sinfractheta2 + cosfractheta2 = sqrt2sinbig(fractheta2 +fracpi4big) = sqrt2sinfrac12big(theta+fracpi2big).$$
Does that seem a bit more promising?
At any rate, I am not sure I believe your book is right. A translation of $theta$ is clearly called for before a rescaling of $theta$.
answered Jul 19 at 17:49
Ted Shifrin
59.5k44387
answered Jul 19 at 17:49
Ted Shifrin
59.5k44387
answered Jul 19 at 17:49
Ted Shifrin
59.5k44387
answered Jul 19 at 17:49
Ted Shifrin
59.5k44387
59.5k44387
add a comment |Â
add a comment |Â
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