Is there an intuitive way to understand $frac{xspace dy-yspace dx}{x^2+y^2}=d(arctanfrac yx)$
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1
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My book says-
$$frac{xspace dy-yspace dx}{x^2+y^2}=d(arctanfrac yx)$$
Specifically I am solving differential equations where I may have to transform the LHS into the RHS. Is there an intuitive way so that one can easily understand that the LHS corresponds to the RHS without explicitly evaluating the RHS?
Thanks for any help!
calculus differential-equations
asked Nov 16 at 18:22
tatan
5,52462555
add a comment |
up vote
1
down vote
favorite
My book says-
$$frac{xspace dy-yspace dx}{x^2+y^2}=d(arctanfrac yx)$$
Specifically I am solving differential equations where I may have to transform the LHS into the RHS. Is there an intuitive way so that one can easily understand that the LHS corresponds to the RHS without explicitly evaluating the RHS?
Thanks for any help!
calculus differential-equations
asked Nov 16 at 18:22
tatan
5,52462555
Think about polar coordinates.
– Lord Shark the Unknown
Nov 16 at 18:30
4
$$frac{xspace dy-yspace dx}{x^2+y^2}=frac{frac{xspace dy-yspace dx}{x^2}}{1+frac{y^2}{x^2}}=frac{d(frac{y}{x})}{1+frac{y^2}{x^2}}=d(arctanfrac yx)$$
– Nosrati
Nov 16 at 18:30
@LordSharktheUnknown I don't know much polar coordinates
– tatan
Nov 16 at 18:38
@Nosrati Thanks. It helped
– tatan
Nov 16 at 18:39
@Nosrati how do you get a space between $x$ and $dy$? I can't in my answer :-( . Thanks!
– Oscar Lanzi
Nov 16 at 19:15
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
My book says-
$$frac{xspace dy-yspace dx}{x^2+y^2}=d(arctanfrac yx)$$
Specifically I am solving differential equations where I may have to transform the LHS into the RHS. Is there an intuitive way so that one can easily understand that the LHS corresponds to the RHS without explicitly evaluating the RHS?
Thanks for any help!
calculus differential-equations
asked Nov 16 at 18:22
tatan
5,52462555
My book says-
$$frac{xspace dy-yspace dx}{x^2+y^2}=d(arctanfrac yx)$$
Specifically I am solving differential equations where I may have to transform the LHS into the RHS. Is there an intuitive way so that one can easily understand that the LHS corresponds to the RHS without explicitly evaluating the RHS?
Thanks for any help!
calculus differential-equations
calculus differential-equations
asked Nov 16 at 18:22
tatan
5,52462555
asked Nov 16 at 18:22
tatan
5,52462555
asked Nov 16 at 18:22
tatan
5,52462555
asked Nov 16 at 18:22
tatan
5,52462555
asked Nov 16 at 18:22
tatan
5,52462555
5,52462555
Think about polar coordinates.
– Lord Shark the Unknown
Nov 16 at 18:30
4
$$frac{xspace dy-yspace dx}{x^2+y^2}=frac{frac{xspace dy-yspace dx}{x^2}}{1+frac{y^2}{x^2}}=frac{d(frac{y}{x})}{1+frac{y^2}{x^2}}=d(arctanfrac yx)$$
– Nosrati
Nov 16 at 18:30
@LordSharktheUnknown I don't know much polar coordinates
– tatan
Nov 16 at 18:38
@Nosrati Thanks. It helped
– tatan
Nov 16 at 18:39
@Nosrati how do you get a space between $x$ and $dy$? I can't in my answer :-( . Thanks!
– Oscar Lanzi
Nov 16 at 19:15
add a comment |
Think about polar coordinates.
– Lord Shark the Unknown
Nov 16 at 18:30
4
$$frac{xspace dy-yspace dx}{x^2+y^2}=frac{frac{xspace dy-yspace dx}{x^2}}{1+frac{y^2}{x^2}}=frac{d(frac{y}{x})}{1+frac{y^2}{x^2}}=d(arctanfrac yx)$$
– Nosrati
Nov 16 at 18:30
@LordSharktheUnknown I don't know much polar coordinates
– tatan
Nov 16 at 18:38
@Nosrati Thanks. It helped
– tatan
Nov 16 at 18:39
@Nosrati how do you get a space between $x$ and $dy$? I can't in my answer :-( . Thanks!
– Oscar Lanzi
Nov 16 at 19:15
Think about polar coordinates.
– Lord Shark the Unknown
Nov 16 at 18:30
Think about polar coordinates.
– Lord Shark the Unknown
Nov 16 at 18:30
4
4
$$frac{xspace dy-yspace dx}{x^2+y^2}=frac{frac{xspace dy-yspace dx}{x^2}}{1+frac{y^2}{x^2}}=frac{d(frac{y}{x})}{1+frac{y^2}{x^2}}=d(arctanfrac yx)$$
– Nosrati
Nov 16 at 18:30
$$frac{xspace dy-yspace dx}{x^2+y^2}=frac{frac{xspace dy-yspace dx}{x^2}}{1+frac{y^2}{x^2}}=frac{d(frac{y}{x})}{1+frac{y^2}{x^2}}=d(arctanfrac yx)$$
– Nosrati
Nov 16 at 18:30
@LordSharktheUnknown I don't know much polar coordinates
– tatan
Nov 16 at 18:38
@LordSharktheUnknown I don't know much polar coordinates
– tatan
Nov 16 at 18:38
@Nosrati Thanks. It helped
– tatan
Nov 16 at 18:39
@Nosrati Thanks. It helped
– tatan
Nov 16 at 18:39
@Nosrati how do you get a space between $x$ and $dy$? I can't in my answer :-( . Thanks!
– Oscar Lanzi
Nov 16 at 19:15
@Nosrati how do you get a space between $x$ and $dy$? I can't in my answer :-( . Thanks!
– Oscar Lanzi
Nov 16 at 19:15
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
Draw a right triangle with vertices at $(0,0),(x,0),(x,y)$. From SOH CAH TOA render $tantheta=(y/x)$, where $theta$ is the angle at $(0,0)$. From the derivative of the tangent function on the left side, and the quotient rule on the right, obtain the differential relation
$sec^2theta dtheta=dfrac{xdy-ydx}{x^2}$.
Return to the right triangle and from SOH CAH TOA again, render $sectheta=1/costheta=(sqrt{x^2+y^2})/x$. Substitute into the equation above and solve for $dtheta$.
answered Nov 16 at 19:03
Oscar Lanzi
11.6k11935
add a comment |
up vote
0
down vote
hint
use
$$d(f(x,y))=xfrac{partial f}{partial y}(x,y)+yfrac{partial f}{partial y}(x,y)$$
and
$$frac{partial}{partial x}arctan(g(x,y))=$$
$$frac{partial}{partial x} g(x,y).frac{1}{1+g(x,y)^2}$$
answered Nov 16 at 18:29
hamam_Abdallah
36.8k21533
1
This helps me to go from RHS to LHS, not from LHS to RHS, right?
– tatan
Nov 16 at 18:35
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
Draw a right triangle with vertices at $(0,0),(x,0),(x,y)$. From SOH CAH TOA render $tantheta=(y/x)$, where $theta$ is the angle at $(0,0)$. From the derivative of the tangent function on the left side, and the quotient rule on the right, obtain the differential relation
$sec^2theta dtheta=dfrac{xdy-ydx}{x^2}$.
Return to the right triangle and from SOH CAH TOA again, render $sectheta=1/costheta=(sqrt{x^2+y^2})/x$. Substitute into the equation above and solve for $dtheta$.
answered Nov 16 at 19:03
Oscar Lanzi
11.6k11935
add a comment |
up vote
2
down vote
Draw a right triangle with vertices at $(0,0),(x,0),(x,y)$. From SOH CAH TOA render $tantheta=(y/x)$, where $theta$ is the angle at $(0,0)$. From the derivative of the tangent function on the left side, and the quotient rule on the right, obtain the differential relation
$sec^2theta dtheta=dfrac{xdy-ydx}{x^2}$.
Return to the right triangle and from SOH CAH TOA again, render $sectheta=1/costheta=(sqrt{x^2+y^2})/x$. Substitute into the equation above and solve for $dtheta$.
answered Nov 16 at 19:03
Oscar Lanzi
11.6k11935
add a comment |
up vote
2
down vote
up vote
2
down vote
Draw a right triangle with vertices at $(0,0),(x,0),(x,y)$. From SOH CAH TOA render $tantheta=(y/x)$, where $theta$ is the angle at $(0,0)$. From the derivative of the tangent function on the left side, and the quotient rule on the right, obtain the differential relation
$sec^2theta dtheta=dfrac{xdy-ydx}{x^2}$.
Return to the right triangle and from SOH CAH TOA again, render $sectheta=1/costheta=(sqrt{x^2+y^2})/x$. Substitute into the equation above and solve for $dtheta$.
answered Nov 16 at 19:03
Oscar Lanzi
11.6k11935
Draw a right triangle with vertices at $(0,0),(x,0),(x,y)$. From SOH CAH TOA render $tantheta=(y/x)$, where $theta$ is the angle at $(0,0)$. From the derivative of the tangent function on the left side, and the quotient rule on the right, obtain the differential relation
$sec^2theta dtheta=dfrac{xdy-ydx}{x^2}$.
Return to the right triangle and from SOH CAH TOA again, render $sectheta=1/costheta=(sqrt{x^2+y^2})/x$. Substitute into the equation above and solve for $dtheta$.
answered Nov 16 at 19:03
Oscar Lanzi
11.6k11935
edited Nov 16 at 19:09
answered Nov 16 at 19:03
Oscar Lanzi
11.6k11935
answered Nov 16 at 19:03
Oscar Lanzi
11.6k11935
answered Nov 16 at 19:03
Oscar Lanzi
11.6k11935
11.6k11935
add a comment |
add a comment |
up vote
0
down vote
hint
use
$$d(f(x,y))=xfrac{partial f}{partial y}(x,y)+yfrac{partial f}{partial y}(x,y)$$
and
$$frac{partial}{partial x}arctan(g(x,y))=$$
$$frac{partial}{partial x} g(x,y).frac{1}{1+g(x,y)^2}$$
answered Nov 16 at 18:29
hamam_Abdallah
36.8k21533
1
This helps me to go from RHS to LHS, not from LHS to RHS, right?
– tatan
Nov 16 at 18:35
add a comment |
up vote
0
down vote
hint
use
$$d(f(x,y))=xfrac{partial f}{partial y}(x,y)+yfrac{partial f}{partial y}(x,y)$$
and
$$frac{partial}{partial x}arctan(g(x,y))=$$
$$frac{partial}{partial x} g(x,y).frac{1}{1+g(x,y)^2}$$
answered Nov 16 at 18:29
hamam_Abdallah
36.8k21533
1
This helps me to go from RHS to LHS, not from LHS to RHS, right?
– tatan
Nov 16 at 18:35
add a comment |
up vote
0
down vote
up vote
0
down vote
hint
use
$$d(f(x,y))=xfrac{partial f}{partial y}(x,y)+yfrac{partial f}{partial y}(x,y)$$
and
$$frac{partial}{partial x}arctan(g(x,y))=$$
$$frac{partial}{partial x} g(x,y).frac{1}{1+g(x,y)^2}$$
answered Nov 16 at 18:29
hamam_Abdallah
36.8k21533
hint
use
$$d(f(x,y))=xfrac{partial f}{partial y}(x,y)+yfrac{partial f}{partial y}(x,y)$$
and
$$frac{partial}{partial x}arctan(g(x,y))=$$
$$frac{partial}{partial x} g(x,y).frac{1}{1+g(x,y)^2}$$
answered Nov 16 at 18:29
hamam_Abdallah
36.8k21533
answered Nov 16 at 18:29
hamam_Abdallah
36.8k21533
answered Nov 16 at 18:29
hamam_Abdallah
36.8k21533
answered Nov 16 at 18:29
hamam_Abdallah
36.8k21533
36.8k21533
1
This helps me to go from RHS to LHS, not from LHS to RHS, right?
– tatan
Nov 16 at 18:35
add a comment |
1
This helps me to go from RHS to LHS, not from LHS to RHS, right?
– tatan
Nov 16 at 18:35
1
1
This helps me to go from RHS to LHS, not from LHS to RHS, right?
– tatan
Nov 16 at 18:35
This helps me to go from RHS to LHS, not from LHS to RHS, right?
– tatan
Nov 16 at 18:35
add a comment |
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Think about polar coordinates.
– Lord Shark the Unknown
Nov 16 at 18:30
4
$$frac{xspace dy-yspace dx}{x^2+y^2}=frac{frac{xspace dy-yspace dx}{x^2}}{1+frac{y^2}{x^2}}=frac{d(frac{y}{x})}{1+frac{y^2}{x^2}}=d(arctanfrac yx)$$
– Nosrati
Nov 16 at 18:30
@LordSharktheUnknown I don't know much polar coordinates
– tatan
Nov 16 at 18:38
@Nosrati Thanks. It helped
– tatan
Nov 16 at 18:39
@Nosrati how do you get a space between $x$ and $dy$? I can't in my answer :-( . Thanks!
– Oscar Lanzi
Nov 16 at 19:15