How do we really get the angle of a vector from the components?












4












$begingroup$


Usually when people discuss getting the polar form of a vector $v$, they present the following two formulas:



$$text{Magnitude}(v) = sqrt{x^2 + y^2}$$



$$text{Angle}(v) = arctan left(frac{y}{x} right)$$



$$ text{ Where } space v = begin{bmatrix} x \ y end{bmatrix}$$



I believe that this formula for the angle is only partially true. I think a better and more complete formula for the angle should be:



$$ text{Angle}(v) = begin{cases}
arctan left(frac{y}{x} right) &; space x gt 0 \
pi +arctan left(frac{y}{x} right) &; space x lt 0 \
{begin{cases}
operatorname{sign}(y) frac{pi}{2} &; y neq 0 \
text{undefined} &; space y = 0
end{cases}} &; x = 0
end{cases} $$



Is there some sort of way to simplify this or to better express this, or is this it?






trigonometry vectors







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    The "atan2" function was devised to help with this.
    $endgroup$
    – Blue
    Jan 20 at 14:59






  • 2




    $begingroup$
    BTW: You can see atan2 being used in the "Converting between polar and Cartesian coordinates" section of Wikipedia's "polar coordinate system" entry. Who are these people you've been talking-to who "usually" just use arctan?
    $endgroup$
    – Blue
    Jan 20 at 15:06






  • 1




    $begingroup$
    That's unfortunate. :/
    $endgroup$
    – Blue
    Jan 20 at 15:13






  • 1




    $begingroup$
    @Blue Thank you for your sympathy. :)
    $endgroup$
    – Gustav
    Jan 20 at 15:16






  • 3




    $begingroup$
    I'll note that the computer algebra Mathematica uses a "smart" ArcTan function: when fed just one argument, it gives the regular inverse-tangent; when given both $x$ and $y$ arguments, it gives the quadrant-savvy version of the angle (equivalent to atan2).
    $endgroup$
    – Blue
    Jan 20 at 15:20


















4












$begingroup$


Usually when people discuss getting the polar form of a vector $v$, they present the following two formulas:



$$text{Magnitude}(v) = sqrt{x^2 + y^2}$$



$$text{Angle}(v) = arctan left(frac{y}{x} right)$$



$$ text{ Where } space v = begin{bmatrix} x \ y end{bmatrix}$$



I believe that this formula for the angle is only partially true. I think a better and more complete formula for the angle should be:



$$ text{Angle}(v) = begin{cases}
arctan left(frac{y}{x} right) &; space x gt 0 \
pi +arctan left(frac{y}{x} right) &; space x lt 0 \
{begin{cases}
operatorname{sign}(y) frac{pi}{2} &; y neq 0 \
text{undefined} &; space y = 0
end{cases}} &; x = 0
end{cases} $$



Is there some sort of way to simplify this or to better express this, or is this it?






trigonometry vectors







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    The "atan2" function was devised to help with this.
    $endgroup$
    – Blue
    Jan 20 at 14:59






  • 2




    $begingroup$
    BTW: You can see atan2 being used in the "Converting between polar and Cartesian coordinates" section of Wikipedia's "polar coordinate system" entry. Who are these people you've been talking-to who "usually" just use arctan?
    $endgroup$
    – Blue
    Jan 20 at 15:06






  • 1




    $begingroup$
    That's unfortunate. :/
    $endgroup$
    – Blue
    Jan 20 at 15:13






  • 1




    $begingroup$
    @Blue Thank you for your sympathy. :)
    $endgroup$
    – Gustav
    Jan 20 at 15:16






  • 3




    $begingroup$
    I'll note that the computer algebra Mathematica uses a "smart" ArcTan function: when fed just one argument, it gives the regular inverse-tangent; when given both $x$ and $y$ arguments, it gives the quadrant-savvy version of the angle (equivalent to atan2).
    $endgroup$
    – Blue
    Jan 20 at 15:20
















4












4








4


1



$begingroup$


Usually when people discuss getting the polar form of a vector $v$, they present the following two formulas:



$$text{Magnitude}(v) = sqrt{x^2 + y^2}$$



$$text{Angle}(v) = arctan left(frac{y}{x} right)$$



$$ text{ Where } space v = begin{bmatrix} x \ y end{bmatrix}$$



I believe that this formula for the angle is only partially true. I think a better and more complete formula for the angle should be:



$$ text{Angle}(v) = begin{cases}
arctan left(frac{y}{x} right) &; space x gt 0 \
pi +arctan left(frac{y}{x} right) &; space x lt 0 \
{begin{cases}
operatorname{sign}(y) frac{pi}{2} &; y neq 0 \
text{undefined} &; space y = 0
end{cases}} &; x = 0
end{cases} $$



Is there some sort of way to simplify this or to better express this, or is this it?






trigonometry vectors







share|cite|improve this question











$endgroup$




Usually when people discuss getting the polar form of a vector $v$, they present the following two formulas:



$$text{Magnitude}(v) = sqrt{x^2 + y^2}$$



$$text{Angle}(v) = arctan left(frac{y}{x} right)$$



$$ text{ Where } space v = begin{bmatrix} x \ y end{bmatrix}$$



I believe that this formula for the angle is only partially true. I think a better and more complete formula for the angle should be:



$$ text{Angle}(v) = begin{cases}
arctan left(frac{y}{x} right) &; space x gt 0 \
pi +arctan left(frac{y}{x} right) &; space x lt 0 \
{begin{cases}
operatorname{sign}(y) frac{pi}{2} &; y neq 0 \
text{undefined} &; space y = 0
end{cases}} &; x = 0
end{cases} $$



Is there some sort of way to simplify this or to better express this, or is this it?






trigonometry vectors




trigonometry vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 15:12









Bernard

121k740116




121k740116










asked Jan 20 at 14:56









GustavGustav

1469




1469








  • 4




    $begingroup$
    The "atan2" function was devised to help with this.
    $endgroup$
    – Blue
    Jan 20 at 14:59






  • 2




    $begingroup$
    BTW: You can see atan2 being used in the "Converting between polar and Cartesian coordinates" section of Wikipedia's "polar coordinate system" entry. Who are these people you've been talking-to who "usually" just use arctan?
    $endgroup$
    – Blue
    Jan 20 at 15:06






  • 1




    $begingroup$
    That's unfortunate. :/
    $endgroup$
    – Blue
    Jan 20 at 15:13






  • 1




    $begingroup$
    @Blue Thank you for your sympathy. :)
    $endgroup$
    – Gustav
    Jan 20 at 15:16






  • 3




    $begingroup$
    I'll note that the computer algebra Mathematica uses a "smart" ArcTan function: when fed just one argument, it gives the regular inverse-tangent; when given both $x$ and $y$ arguments, it gives the quadrant-savvy version of the angle (equivalent to atan2).
    $endgroup$
    – Blue
    Jan 20 at 15:20
















  • 4




    $begingroup$
    The "atan2" function was devised to help with this.
    $endgroup$
    – Blue
    Jan 20 at 14:59






  • 2




    $begingroup$
    BTW: You can see atan2 being used in the "Converting between polar and Cartesian coordinates" section of Wikipedia's "polar coordinate system" entry. Who are these people you've been talking-to who "usually" just use arctan?
    $endgroup$
    – Blue
    Jan 20 at 15:06






  • 1




    $begingroup$
    That's unfortunate. :/
    $endgroup$
    – Blue
    Jan 20 at 15:13






  • 1




    $begingroup$
    @Blue Thank you for your sympathy. :)
    $endgroup$
    – Gustav
    Jan 20 at 15:16






  • 3




    $begingroup$
    I'll note that the computer algebra Mathematica uses a "smart" ArcTan function: when fed just one argument, it gives the regular inverse-tangent; when given both $x$ and $y$ arguments, it gives the quadrant-savvy version of the angle (equivalent to atan2).
    $endgroup$
    – Blue
    Jan 20 at 15:20










4




4




$begingroup$
The "atan2" function was devised to help with this.
$endgroup$
– Blue
Jan 20 at 14:59




$begingroup$
The "atan2" function was devised to help with this.
$endgroup$
– Blue
Jan 20 at 14:59




2




2




$begingroup$
BTW: You can see atan2 being used in the "Converting between polar and Cartesian coordinates" section of Wikipedia's "polar coordinate system" entry. Who are these people you've been talking-to who "usually" just use arctan?
$endgroup$
– Blue
Jan 20 at 15:06




$begingroup$
BTW: You can see atan2 being used in the "Converting between polar and Cartesian coordinates" section of Wikipedia's "polar coordinate system" entry. Who are these people you've been talking-to who "usually" just use arctan?
$endgroup$
– Blue
Jan 20 at 15:06




1




1




$begingroup$
That's unfortunate. :/
$endgroup$
– Blue
Jan 20 at 15:13




$begingroup$
That's unfortunate. :/
$endgroup$
– Blue
Jan 20 at 15:13




1




1




$begingroup$
@Blue Thank you for your sympathy. :)
$endgroup$
– Gustav
Jan 20 at 15:16




$begingroup$
@Blue Thank you for your sympathy. :)
$endgroup$
– Gustav
Jan 20 at 15:16




3




3




$begingroup$
I'll note that the computer algebra Mathematica uses a "smart" ArcTan function: when fed just one argument, it gives the regular inverse-tangent; when given both $x$ and $y$ arguments, it gives the quadrant-savvy version of the angle (equivalent to atan2).
$endgroup$
– Blue
Jan 20 at 15:20






$begingroup$
I'll note that the computer algebra Mathematica uses a "smart" ArcTan function: when fed just one argument, it gives the regular inverse-tangent; when given both $x$ and $y$ arguments, it gives the quadrant-savvy version of the angle (equivalent to atan2).
$endgroup$
– Blue
Jan 20 at 15:20












2 Answers
2






active

oldest

votes


















7












$begingroup$

If you want an answer in form of a mathematical function definition, using only functions that were in common use in undergraduate instruction fifty years ago, I do not think you can do much better than the excellent answer by Rhys Hughes.
As noted in the answer, this is how mathematicians often define a function equivalent to yours in textbooks.
The only detail you might want to add is something to deal with the case $x = y = 0.$



Note that the formulas in that answer do not tell you how to find $theta,$
but they do uniquely identify the output of the function for any possible input.
There is always some value of $theta$ that will satisfy both equations when
$x^2 + y^2 neq 0,$ and there will never be more than one value of $theta$ that satisfies both equations.



If you want a formula to compute the angle using only functions that were in common use in undergraduate instruction fifty years ago, I think the formula you wrote is close to the best you can get, though I would handle one or two cases a bit differently.



If you want a nice way to represent your function in other formulas, you can borrow the two-parameter arc tangent function that is defined in many software packages.
That is, define a function $operatorname{atan2}(y, x)$
whose value is the angle of the vector $begin{bmatrix} x \ y end{bmatrix}.$
You can define $operatorname{atan2}(y, x)$ either in the style of the complex analysis textbooks described in the other answer, or you can define it in your style:



$$
operatorname{atan2}(y, x) = begin{cases}
arctanleft(frac yxright) & x gt 0 \
arctanleft(frac yxright) + pi quad & x lt 0, y geq 0 \
arctanleft(frac yxright) - pi quad & x lt 0, y < 0 \
fracpi2 & x = 0, y > 0 \
-fracpi2 & x = 0, y < 0 \
text{undefined} & x = 0, y = 0.
end{cases}
$$



When this is implemented in software I think the "undefined" case usually returns $0.$



If you have to write formulas involving the direction angles of several two-dimensional vectors in terms of their components, then you might find the notation
$operatorname{atan2}(y, x)$ convenient.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Gustav Wow! I mean wow squared. You complain that "they" leave a little bit for you to figure out instead of giving you exactly what you want on a silver platter, then you speculate this is because they are lazy? Wow. (This is an example of what they mean by "irony". If you visit the_internet.com you can apply for a refund of the monry you paid...)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:29










  • $begingroup$
    @Gustav Not frustrated at all, more amused.
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:36










  • $begingroup$
    @Gustav I understood the question. And I have no problem with anything you said about the answer - my comment had to do with you complaining about him being lazy. (Do you know what the word "irony" means?)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:45








  • 1




    $begingroup$
    No problem; I've deleted earlier comments.
    $endgroup$
    – David K
    Jan 20 at 17:12






  • 1




    $begingroup$
    @David C. Ullrich. Could you also remove your previous comments as they no longer serve any purpose?
    $endgroup$
    – Gustav
    Jan 20 at 17:20





















4












$begingroup$

This looks very similar to complex numbers, where $text{Magnitude}(v)$ is displayed by $|v|$ and $text{Angle}(v)$ is written $text{arg}(v)$. We have that $arg(v)$ is the unique angle $in (-pi, pi]$ where:



$$costheta =frac{x}{|v|}=frac{x}{sqrt{x^2+y^2}}$$
and
$$sintheta=frac{y}{|v|}=frac{y}{sqrt{x^2+y^2}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    When working with only arctan, if we look at its graph, we can clearly see that it only gives values in the interval [$frac{-pi}{2}$ , $frac{pi}{2}$]. This only covers the first and fourth quadrants. So, in order to access the second and third quadrants, we need another definition. On top of that we need a definition for when the denomination in the usual definition would be zero, etc.
    $endgroup$
    – Gustav
    Jan 20 at 15:22








  • 1




    $begingroup$
    When the denominator equals to zero, the 2-norm or magnitude of the vector is 0 anyway. In other words, the 0 vector has no angle (Linear Algebra tells us that 0 is in the null space of any system fully described by the $R^2$).
    $endgroup$
    – Joel Biffin
    Jan 20 at 15:52










  • $begingroup$
    @Joel When the denominator is zero, only one of the components is zero, not both. So no, the magnitude is not zero.
    $endgroup$
    – Gustav
    Jan 20 at 15:58












  • $begingroup$
    @Gustav you literally called $sqrt{x^2+y^2}$ the magnitude in your question!
    $endgroup$
    – Rhys Hughes
    Jan 20 at 16:01










  • $begingroup$
    2 comments back, I accidentally wrote denomination instead of denominator. With this I mean the value of x. Sorry about that.
    $endgroup$
    – Gustav
    Jan 20 at 16:03











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

If you want an answer in form of a mathematical function definition, using only functions that were in common use in undergraduate instruction fifty years ago, I do not think you can do much better than the excellent answer by Rhys Hughes.
As noted in the answer, this is how mathematicians often define a function equivalent to yours in textbooks.
The only detail you might want to add is something to deal with the case $x = y = 0.$



Note that the formulas in that answer do not tell you how to find $theta,$
but they do uniquely identify the output of the function for any possible input.
There is always some value of $theta$ that will satisfy both equations when
$x^2 + y^2 neq 0,$ and there will never be more than one value of $theta$ that satisfies both equations.



If you want a formula to compute the angle using only functions that were in common use in undergraduate instruction fifty years ago, I think the formula you wrote is close to the best you can get, though I would handle one or two cases a bit differently.



If you want a nice way to represent your function in other formulas, you can borrow the two-parameter arc tangent function that is defined in many software packages.
That is, define a function $operatorname{atan2}(y, x)$
whose value is the angle of the vector $begin{bmatrix} x \ y end{bmatrix}.$
You can define $operatorname{atan2}(y, x)$ either in the style of the complex analysis textbooks described in the other answer, or you can define it in your style:



$$
operatorname{atan2}(y, x) = begin{cases}
arctanleft(frac yxright) & x gt 0 \
arctanleft(frac yxright) + pi quad & x lt 0, y geq 0 \
arctanleft(frac yxright) - pi quad & x lt 0, y < 0 \
fracpi2 & x = 0, y > 0 \
-fracpi2 & x = 0, y < 0 \
text{undefined} & x = 0, y = 0.
end{cases}
$$



When this is implemented in software I think the "undefined" case usually returns $0.$



If you have to write formulas involving the direction angles of several two-dimensional vectors in terms of their components, then you might find the notation
$operatorname{atan2}(y, x)$ convenient.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Gustav Wow! I mean wow squared. You complain that "they" leave a little bit for you to figure out instead of giving you exactly what you want on a silver platter, then you speculate this is because they are lazy? Wow. (This is an example of what they mean by "irony". If you visit the_internet.com you can apply for a refund of the monry you paid...)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:29










  • $begingroup$
    @Gustav Not frustrated at all, more amused.
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:36










  • $begingroup$
    @Gustav I understood the question. And I have no problem with anything you said about the answer - my comment had to do with you complaining about him being lazy. (Do you know what the word "irony" means?)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:45








  • 1




    $begingroup$
    No problem; I've deleted earlier comments.
    $endgroup$
    – David K
    Jan 20 at 17:12






  • 1




    $begingroup$
    @David C. Ullrich. Could you also remove your previous comments as they no longer serve any purpose?
    $endgroup$
    – Gustav
    Jan 20 at 17:20


















7












$begingroup$

If you want an answer in form of a mathematical function definition, using only functions that were in common use in undergraduate instruction fifty years ago, I do not think you can do much better than the excellent answer by Rhys Hughes.
As noted in the answer, this is how mathematicians often define a function equivalent to yours in textbooks.
The only detail you might want to add is something to deal with the case $x = y = 0.$



Note that the formulas in that answer do not tell you how to find $theta,$
but they do uniquely identify the output of the function for any possible input.
There is always some value of $theta$ that will satisfy both equations when
$x^2 + y^2 neq 0,$ and there will never be more than one value of $theta$ that satisfies both equations.



If you want a formula to compute the angle using only functions that were in common use in undergraduate instruction fifty years ago, I think the formula you wrote is close to the best you can get, though I would handle one or two cases a bit differently.



If you want a nice way to represent your function in other formulas, you can borrow the two-parameter arc tangent function that is defined in many software packages.
That is, define a function $operatorname{atan2}(y, x)$
whose value is the angle of the vector $begin{bmatrix} x \ y end{bmatrix}.$
You can define $operatorname{atan2}(y, x)$ either in the style of the complex analysis textbooks described in the other answer, or you can define it in your style:



$$
operatorname{atan2}(y, x) = begin{cases}
arctanleft(frac yxright) & x gt 0 \
arctanleft(frac yxright) + pi quad & x lt 0, y geq 0 \
arctanleft(frac yxright) - pi quad & x lt 0, y < 0 \
fracpi2 & x = 0, y > 0 \
-fracpi2 & x = 0, y < 0 \
text{undefined} & x = 0, y = 0.
end{cases}
$$



When this is implemented in software I think the "undefined" case usually returns $0.$



If you have to write formulas involving the direction angles of several two-dimensional vectors in terms of their components, then you might find the notation
$operatorname{atan2}(y, x)$ convenient.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Gustav Wow! I mean wow squared. You complain that "they" leave a little bit for you to figure out instead of giving you exactly what you want on a silver platter, then you speculate this is because they are lazy? Wow. (This is an example of what they mean by "irony". If you visit the_internet.com you can apply for a refund of the monry you paid...)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:29










  • $begingroup$
    @Gustav Not frustrated at all, more amused.
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:36










  • $begingroup$
    @Gustav I understood the question. And I have no problem with anything you said about the answer - my comment had to do with you complaining about him being lazy. (Do you know what the word "irony" means?)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:45








  • 1




    $begingroup$
    No problem; I've deleted earlier comments.
    $endgroup$
    – David K
    Jan 20 at 17:12






  • 1




    $begingroup$
    @David C. Ullrich. Could you also remove your previous comments as they no longer serve any purpose?
    $endgroup$
    – Gustav
    Jan 20 at 17:20
















7












7








7





$begingroup$

If you want an answer in form of a mathematical function definition, using only functions that were in common use in undergraduate instruction fifty years ago, I do not think you can do much better than the excellent answer by Rhys Hughes.
As noted in the answer, this is how mathematicians often define a function equivalent to yours in textbooks.
The only detail you might want to add is something to deal with the case $x = y = 0.$



Note that the formulas in that answer do not tell you how to find $theta,$
but they do uniquely identify the output of the function for any possible input.
There is always some value of $theta$ that will satisfy both equations when
$x^2 + y^2 neq 0,$ and there will never be more than one value of $theta$ that satisfies both equations.



If you want a formula to compute the angle using only functions that were in common use in undergraduate instruction fifty years ago, I think the formula you wrote is close to the best you can get, though I would handle one or two cases a bit differently.



If you want a nice way to represent your function in other formulas, you can borrow the two-parameter arc tangent function that is defined in many software packages.
That is, define a function $operatorname{atan2}(y, x)$
whose value is the angle of the vector $begin{bmatrix} x \ y end{bmatrix}.$
You can define $operatorname{atan2}(y, x)$ either in the style of the complex analysis textbooks described in the other answer, or you can define it in your style:



$$
operatorname{atan2}(y, x) = begin{cases}
arctanleft(frac yxright) & x gt 0 \
arctanleft(frac yxright) + pi quad & x lt 0, y geq 0 \
arctanleft(frac yxright) - pi quad & x lt 0, y < 0 \
fracpi2 & x = 0, y > 0 \
-fracpi2 & x = 0, y < 0 \
text{undefined} & x = 0, y = 0.
end{cases}
$$



When this is implemented in software I think the "undefined" case usually returns $0.$



If you have to write formulas involving the direction angles of several two-dimensional vectors in terms of their components, then you might find the notation
$operatorname{atan2}(y, x)$ convenient.






share|cite|improve this answer









$endgroup$



If you want an answer in form of a mathematical function definition, using only functions that were in common use in undergraduate instruction fifty years ago, I do not think you can do much better than the excellent answer by Rhys Hughes.
As noted in the answer, this is how mathematicians often define a function equivalent to yours in textbooks.
The only detail you might want to add is something to deal with the case $x = y = 0.$



Note that the formulas in that answer do not tell you how to find $theta,$
but they do uniquely identify the output of the function for any possible input.
There is always some value of $theta$ that will satisfy both equations when
$x^2 + y^2 neq 0,$ and there will never be more than one value of $theta$ that satisfies both equations.



If you want a formula to compute the angle using only functions that were in common use in undergraduate instruction fifty years ago, I think the formula you wrote is close to the best you can get, though I would handle one or two cases a bit differently.



If you want a nice way to represent your function in other formulas, you can borrow the two-parameter arc tangent function that is defined in many software packages.
That is, define a function $operatorname{atan2}(y, x)$
whose value is the angle of the vector $begin{bmatrix} x \ y end{bmatrix}.$
You can define $operatorname{atan2}(y, x)$ either in the style of the complex analysis textbooks described in the other answer, or you can define it in your style:



$$
operatorname{atan2}(y, x) = begin{cases}
arctanleft(frac yxright) & x gt 0 \
arctanleft(frac yxright) + pi quad & x lt 0, y geq 0 \
arctanleft(frac yxright) - pi quad & x lt 0, y < 0 \
fracpi2 & x = 0, y > 0 \
-fracpi2 & x = 0, y < 0 \
text{undefined} & x = 0, y = 0.
end{cases}
$$



When this is implemented in software I think the "undefined" case usually returns $0.$



If you have to write formulas involving the direction angles of several two-dimensional vectors in terms of their components, then you might find the notation
$operatorname{atan2}(y, x)$ convenient.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 20 at 16:01









David KDavid K

54.5k343120




54.5k343120












  • $begingroup$
    @Gustav Wow! I mean wow squared. You complain that "they" leave a little bit for you to figure out instead of giving you exactly what you want on a silver platter, then you speculate this is because they are lazy? Wow. (This is an example of what they mean by "irony". If you visit the_internet.com you can apply for a refund of the monry you paid...)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:29










  • $begingroup$
    @Gustav Not frustrated at all, more amused.
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:36










  • $begingroup$
    @Gustav I understood the question. And I have no problem with anything you said about the answer - my comment had to do with you complaining about him being lazy. (Do you know what the word "irony" means?)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:45








  • 1




    $begingroup$
    No problem; I've deleted earlier comments.
    $endgroup$
    – David K
    Jan 20 at 17:12






  • 1




    $begingroup$
    @David C. Ullrich. Could you also remove your previous comments as they no longer serve any purpose?
    $endgroup$
    – Gustav
    Jan 20 at 17:20




















  • $begingroup$
    @Gustav Wow! I mean wow squared. You complain that "they" leave a little bit for you to figure out instead of giving you exactly what you want on a silver platter, then you speculate this is because they are lazy? Wow. (This is an example of what they mean by "irony". If you visit the_internet.com you can apply for a refund of the monry you paid...)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:29










  • $begingroup$
    @Gustav Not frustrated at all, more amused.
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:36










  • $begingroup$
    @Gustav I understood the question. And I have no problem with anything you said about the answer - my comment had to do with you complaining about him being lazy. (Do you know what the word "irony" means?)
    $endgroup$
    – David C. Ullrich
    Jan 20 at 16:45








  • 1




    $begingroup$
    No problem; I've deleted earlier comments.
    $endgroup$
    – David K
    Jan 20 at 17:12






  • 1




    $begingroup$
    @David C. Ullrich. Could you also remove your previous comments as they no longer serve any purpose?
    $endgroup$
    – Gustav
    Jan 20 at 17:20


















$begingroup$
@Gustav Wow! I mean wow squared. You complain that "they" leave a little bit for you to figure out instead of giving you exactly what you want on a silver platter, then you speculate this is because they are lazy? Wow. (This is an example of what they mean by "irony". If you visit the_internet.com you can apply for a refund of the monry you paid...)
$endgroup$
– David C. Ullrich
Jan 20 at 16:29




$begingroup$
@Gustav Wow! I mean wow squared. You complain that "they" leave a little bit for you to figure out instead of giving you exactly what you want on a silver platter, then you speculate this is because they are lazy? Wow. (This is an example of what they mean by "irony". If you visit the_internet.com you can apply for a refund of the monry you paid...)
$endgroup$
– David C. Ullrich
Jan 20 at 16:29












$begingroup$
@Gustav Not frustrated at all, more amused.
$endgroup$
– David C. Ullrich
Jan 20 at 16:36




$begingroup$
@Gustav Not frustrated at all, more amused.
$endgroup$
– David C. Ullrich
Jan 20 at 16:36












$begingroup$
@Gustav I understood the question. And I have no problem with anything you said about the answer - my comment had to do with you complaining about him being lazy. (Do you know what the word "irony" means?)
$endgroup$
– David C. Ullrich
Jan 20 at 16:45






$begingroup$
@Gustav I understood the question. And I have no problem with anything you said about the answer - my comment had to do with you complaining about him being lazy. (Do you know what the word "irony" means?)
$endgroup$
– David C. Ullrich
Jan 20 at 16:45






1




1




$begingroup$
No problem; I've deleted earlier comments.
$endgroup$
– David K
Jan 20 at 17:12




$begingroup$
No problem; I've deleted earlier comments.
$endgroup$
– David K
Jan 20 at 17:12




1




1




$begingroup$
@David C. Ullrich. Could you also remove your previous comments as they no longer serve any purpose?
$endgroup$
– Gustav
Jan 20 at 17:20






$begingroup$
@David C. Ullrich. Could you also remove your previous comments as they no longer serve any purpose?
$endgroup$
– Gustav
Jan 20 at 17:20













4












$begingroup$

This looks very similar to complex numbers, where $text{Magnitude}(v)$ is displayed by $|v|$ and $text{Angle}(v)$ is written $text{arg}(v)$. We have that $arg(v)$ is the unique angle $in (-pi, pi]$ where:



$$costheta =frac{x}{|v|}=frac{x}{sqrt{x^2+y^2}}$$
and
$$sintheta=frac{y}{|v|}=frac{y}{sqrt{x^2+y^2}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    When working with only arctan, if we look at its graph, we can clearly see that it only gives values in the interval [$frac{-pi}{2}$ , $frac{pi}{2}$]. This only covers the first and fourth quadrants. So, in order to access the second and third quadrants, we need another definition. On top of that we need a definition for when the denomination in the usual definition would be zero, etc.
    $endgroup$
    – Gustav
    Jan 20 at 15:22








  • 1




    $begingroup$
    When the denominator equals to zero, the 2-norm or magnitude of the vector is 0 anyway. In other words, the 0 vector has no angle (Linear Algebra tells us that 0 is in the null space of any system fully described by the $R^2$).
    $endgroup$
    – Joel Biffin
    Jan 20 at 15:52










  • $begingroup$
    @Joel When the denominator is zero, only one of the components is zero, not both. So no, the magnitude is not zero.
    $endgroup$
    – Gustav
    Jan 20 at 15:58












  • $begingroup$
    @Gustav you literally called $sqrt{x^2+y^2}$ the magnitude in your question!
    $endgroup$
    – Rhys Hughes
    Jan 20 at 16:01










  • $begingroup$
    2 comments back, I accidentally wrote denomination instead of denominator. With this I mean the value of x. Sorry about that.
    $endgroup$
    – Gustav
    Jan 20 at 16:03
















4












$begingroup$

This looks very similar to complex numbers, where $text{Magnitude}(v)$ is displayed by $|v|$ and $text{Angle}(v)$ is written $text{arg}(v)$. We have that $arg(v)$ is the unique angle $in (-pi, pi]$ where:



$$costheta =frac{x}{|v|}=frac{x}{sqrt{x^2+y^2}}$$
and
$$sintheta=frac{y}{|v|}=frac{y}{sqrt{x^2+y^2}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    When working with only arctan, if we look at its graph, we can clearly see that it only gives values in the interval [$frac{-pi}{2}$ , $frac{pi}{2}$]. This only covers the first and fourth quadrants. So, in order to access the second and third quadrants, we need another definition. On top of that we need a definition for when the denomination in the usual definition would be zero, etc.
    $endgroup$
    – Gustav
    Jan 20 at 15:22








  • 1




    $begingroup$
    When the denominator equals to zero, the 2-norm or magnitude of the vector is 0 anyway. In other words, the 0 vector has no angle (Linear Algebra tells us that 0 is in the null space of any system fully described by the $R^2$).
    $endgroup$
    – Joel Biffin
    Jan 20 at 15:52










  • $begingroup$
    @Joel When the denominator is zero, only one of the components is zero, not both. So no, the magnitude is not zero.
    $endgroup$
    – Gustav
    Jan 20 at 15:58












  • $begingroup$
    @Gustav you literally called $sqrt{x^2+y^2}$ the magnitude in your question!
    $endgroup$
    – Rhys Hughes
    Jan 20 at 16:01










  • $begingroup$
    2 comments back, I accidentally wrote denomination instead of denominator. With this I mean the value of x. Sorry about that.
    $endgroup$
    – Gustav
    Jan 20 at 16:03














4












4








4





$begingroup$

This looks very similar to complex numbers, where $text{Magnitude}(v)$ is displayed by $|v|$ and $text{Angle}(v)$ is written $text{arg}(v)$. We have that $arg(v)$ is the unique angle $in (-pi, pi]$ where:



$$costheta =frac{x}{|v|}=frac{x}{sqrt{x^2+y^2}}$$
and
$$sintheta=frac{y}{|v|}=frac{y}{sqrt{x^2+y^2}}$$






share|cite|improve this answer









$endgroup$



This looks very similar to complex numbers, where $text{Magnitude}(v)$ is displayed by $|v|$ and $text{Angle}(v)$ is written $text{arg}(v)$. We have that $arg(v)$ is the unique angle $in (-pi, pi]$ where:



$$costheta =frac{x}{|v|}=frac{x}{sqrt{x^2+y^2}}$$
and
$$sintheta=frac{y}{|v|}=frac{y}{sqrt{x^2+y^2}}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 20 at 15:06









Rhys HughesRhys Hughes

6,7801530




6,7801530












  • $begingroup$
    When working with only arctan, if we look at its graph, we can clearly see that it only gives values in the interval [$frac{-pi}{2}$ , $frac{pi}{2}$]. This only covers the first and fourth quadrants. So, in order to access the second and third quadrants, we need another definition. On top of that we need a definition for when the denomination in the usual definition would be zero, etc.
    $endgroup$
    – Gustav
    Jan 20 at 15:22








  • 1




    $begingroup$
    When the denominator equals to zero, the 2-norm or magnitude of the vector is 0 anyway. In other words, the 0 vector has no angle (Linear Algebra tells us that 0 is in the null space of any system fully described by the $R^2$).
    $endgroup$
    – Joel Biffin
    Jan 20 at 15:52










  • $begingroup$
    @Joel When the denominator is zero, only one of the components is zero, not both. So no, the magnitude is not zero.
    $endgroup$
    – Gustav
    Jan 20 at 15:58












  • $begingroup$
    @Gustav you literally called $sqrt{x^2+y^2}$ the magnitude in your question!
    $endgroup$
    – Rhys Hughes
    Jan 20 at 16:01










  • $begingroup$
    2 comments back, I accidentally wrote denomination instead of denominator. With this I mean the value of x. Sorry about that.
    $endgroup$
    – Gustav
    Jan 20 at 16:03


















  • $begingroup$
    When working with only arctan, if we look at its graph, we can clearly see that it only gives values in the interval [$frac{-pi}{2}$ , $frac{pi}{2}$]. This only covers the first and fourth quadrants. So, in order to access the second and third quadrants, we need another definition. On top of that we need a definition for when the denomination in the usual definition would be zero, etc.
    $endgroup$
    – Gustav
    Jan 20 at 15:22








  • 1




    $begingroup$
    When the denominator equals to zero, the 2-norm or magnitude of the vector is 0 anyway. In other words, the 0 vector has no angle (Linear Algebra tells us that 0 is in the null space of any system fully described by the $R^2$).
    $endgroup$
    – Joel Biffin
    Jan 20 at 15:52










  • $begingroup$
    @Joel When the denominator is zero, only one of the components is zero, not both. So no, the magnitude is not zero.
    $endgroup$
    – Gustav
    Jan 20 at 15:58












  • $begingroup$
    @Gustav you literally called $sqrt{x^2+y^2}$ the magnitude in your question!
    $endgroup$
    – Rhys Hughes
    Jan 20 at 16:01










  • $begingroup$
    2 comments back, I accidentally wrote denomination instead of denominator. With this I mean the value of x. Sorry about that.
    $endgroup$
    – Gustav
    Jan 20 at 16:03
















$begingroup$
When working with only arctan, if we look at its graph, we can clearly see that it only gives values in the interval [$frac{-pi}{2}$ , $frac{pi}{2}$]. This only covers the first and fourth quadrants. So, in order to access the second and third quadrants, we need another definition. On top of that we need a definition for when the denomination in the usual definition would be zero, etc.
$endgroup$
– Gustav
Jan 20 at 15:22






$begingroup$
When working with only arctan, if we look at its graph, we can clearly see that it only gives values in the interval [$frac{-pi}{2}$ , $frac{pi}{2}$]. This only covers the first and fourth quadrants. So, in order to access the second and third quadrants, we need another definition. On top of that we need a definition for when the denomination in the usual definition would be zero, etc.
$endgroup$
– Gustav
Jan 20 at 15:22






1




1




$begingroup$
When the denominator equals to zero, the 2-norm or magnitude of the vector is 0 anyway. In other words, the 0 vector has no angle (Linear Algebra tells us that 0 is in the null space of any system fully described by the $R^2$).
$endgroup$
– Joel Biffin
Jan 20 at 15:52




$begingroup$
When the denominator equals to zero, the 2-norm or magnitude of the vector is 0 anyway. In other words, the 0 vector has no angle (Linear Algebra tells us that 0 is in the null space of any system fully described by the $R^2$).
$endgroup$
– Joel Biffin
Jan 20 at 15:52












$begingroup$
@Joel When the denominator is zero, only one of the components is zero, not both. So no, the magnitude is not zero.
$endgroup$
– Gustav
Jan 20 at 15:58






$begingroup$
@Joel When the denominator is zero, only one of the components is zero, not both. So no, the magnitude is not zero.
$endgroup$
– Gustav
Jan 20 at 15:58














$begingroup$
@Gustav you literally called $sqrt{x^2+y^2}$ the magnitude in your question!
$endgroup$
– Rhys Hughes
Jan 20 at 16:01




$begingroup$
@Gustav you literally called $sqrt{x^2+y^2}$ the magnitude in your question!
$endgroup$
– Rhys Hughes
Jan 20 at 16:01












$begingroup$
2 comments back, I accidentally wrote denomination instead of denominator. With this I mean the value of x. Sorry about that.
$endgroup$
– Gustav
Jan 20 at 16:03




$begingroup$
2 comments back, I accidentally wrote denomination instead of denominator. With this I mean the value of x. Sorry about that.
$endgroup$
– Gustav
Jan 20 at 16:03


















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