Find the orthogonal trajectory to the family of curves $x^2+y^2=4c^2$ where $c$ is a constant.
-1
$begingroup$
Differentiate w.r.t $x$:
$$x+yfrac{dx}{dy}=4c^2$$
hyperbolic-functions
edited Dec 6 '18 at 12:39
amWhy
1
asked Dec 6 '18 at 12:14
Best 4 YouBest 4 You
1
$endgroup$
add a comment |
-1
$begingroup$
Differentiate w.r.t $x$:
$$x+yfrac{dx}{dy}=4c^2$$
hyperbolic-functions
edited Dec 6 '18 at 12:39
amWhy
1
asked Dec 6 '18 at 12:14
Best 4 YouBest 4 You
1
$endgroup$
$begingroup$
Your differentiation is wrong. The derivative of a constant is 0 and you want $frac{dy}{dx}$, not $frac{dx}{dy}$. You should have $x+ yfrac{dy}{dx}= 0$. From that $frac{dy}{dx}= -frac{x}{y}$ and the orthogonal complement will satisfy $frac{dy}{dx}= frac{y}{x}$.
$endgroup$
– user247327
Dec 6 '18 at 12:33
$begingroup$
I would argue the derivative w.r.t $x$ equals $2x + y frac{delta y}{delta x} = 0$, thus $frac{delta y}{delta x}=frac{-2x}{y}$. Hope this helps.
$endgroup$
– Mathbeginner
Dec 6 '18 at 12:44
add a comment |
-1
-1
-1
$begingroup$
Differentiate w.r.t $x$:
$$x+yfrac{dx}{dy}=4c^2$$
hyperbolic-functions
edited Dec 6 '18 at 12:39
amWhy
1
asked Dec 6 '18 at 12:14
Best 4 YouBest 4 You
1
$endgroup$
Differentiate w.r.t $x$:
$$x+yfrac{dx}{dy}=4c^2$$
hyperbolic-functions
hyperbolic-functions
edited Dec 6 '18 at 12:39
amWhy
1
asked Dec 6 '18 at 12:14
Best 4 YouBest 4 You
1
edited Dec 6 '18 at 12:39
amWhy
1
asked Dec 6 '18 at 12:14
Best 4 YouBest 4 You
1
edited Dec 6 '18 at 12:39
amWhy
1
edited Dec 6 '18 at 12:39
amWhy
1
edited Dec 6 '18 at 12:39
amWhy
1
1
asked Dec 6 '18 at 12:14
Best 4 YouBest 4 You
1
asked Dec 6 '18 at 12:14
Best 4 YouBest 4 You
1
asked Dec 6 '18 at 12:14
Best 4 YouBest 4 You
1
1
$begingroup$
Your differentiation is wrong. The derivative of a constant is 0 and you want $frac{dy}{dx}$, not $frac{dx}{dy}$. You should have $x+ yfrac{dy}{dx}= 0$. From that $frac{dy}{dx}= -frac{x}{y}$ and the orthogonal complement will satisfy $frac{dy}{dx}= frac{y}{x}$.
$endgroup$
– user247327
Dec 6 '18 at 12:33
$begingroup$
I would argue the derivative w.r.t $x$ equals $2x + y frac{delta y}{delta x} = 0$, thus $frac{delta y}{delta x}=frac{-2x}{y}$. Hope this helps.
$endgroup$
– Mathbeginner
Dec 6 '18 at 12:44
add a comment |
$begingroup$
Your differentiation is wrong. The derivative of a constant is 0 and you want $frac{dy}{dx}$, not $frac{dx}{dy}$. You should have $x+ yfrac{dy}{dx}= 0$. From that $frac{dy}{dx}= -frac{x}{y}$ and the orthogonal complement will satisfy $frac{dy}{dx}= frac{y}{x}$.
$endgroup$
– user247327
Dec 6 '18 at 12:33
$begingroup$
I would argue the derivative w.r.t $x$ equals $2x + y frac{delta y}{delta x} = 0$, thus $frac{delta y}{delta x}=frac{-2x}{y}$. Hope this helps.
$endgroup$
– Mathbeginner
Dec 6 '18 at 12:44
$begingroup$
Your differentiation is wrong. The derivative of a constant is 0 and you want $frac{dy}{dx}$, not $frac{dx}{dy}$. You should have $x+ yfrac{dy}{dx}= 0$. From that $frac{dy}{dx}= -frac{x}{y}$ and the orthogonal complement will satisfy $frac{dy}{dx}= frac{y}{x}$.
$endgroup$
– user247327
Dec 6 '18 at 12:33
$begingroup$
Your differentiation is wrong. The derivative of a constant is 0 and you want $frac{dy}{dx}$, not $frac{dx}{dy}$. You should have $x+ yfrac{dy}{dx}= 0$. From that $frac{dy}{dx}= -frac{x}{y}$ and the orthogonal complement will satisfy $frac{dy}{dx}= frac{y}{x}$.
$endgroup$
– user247327
Dec 6 '18 at 12:33
$begingroup$
I would argue the derivative w.r.t $x$ equals $2x + y frac{delta y}{delta x} = 0$, thus $frac{delta y}{delta x}=frac{-2x}{y}$. Hope this helps.
$endgroup$
– Mathbeginner
Dec 6 '18 at 12:44
$begingroup$
I would argue the derivative w.r.t $x$ equals $2x + y frac{delta y}{delta x} = 0$, thus $frac{delta y}{delta x}=frac{-2x}{y}$. Hope this helps.
$endgroup$
– Mathbeginner
Dec 6 '18 at 12:44
add a comment |
1 Answer
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$begingroup$
$$x+yy'=0$$ turns to
$$x-frac y{y'}=0$$ which integrates as $$y=cx.$$
This was expected as the given family is made of circles centered at the origin.
answered Dec 6 '18 at 12:48
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
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1 Answer
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0
$begingroup$
$$x+yy'=0$$ turns to
$$x-frac y{y'}=0$$ which integrates as $$y=cx.$$
This was expected as the given family is made of circles centered at the origin.
answered Dec 6 '18 at 12:48
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
0
$begingroup$
$$x+yy'=0$$ turns to
$$x-frac y{y'}=0$$ which integrates as $$y=cx.$$
This was expected as the given family is made of circles centered at the origin.
answered Dec 6 '18 at 12:48
Yves DaoustYves Daoust
126k672226
$endgroup$
add a comment |
0
0
0
$begingroup$
$$x+yy'=0$$ turns to
$$x-frac y{y'}=0$$ which integrates as $$y=cx.$$
This was expected as the given family is made of circles centered at the origin.
answered Dec 6 '18 at 12:48
Yves DaoustYves Daoust
126k672226
$endgroup$
$$x+yy'=0$$ turns to
$$x-frac y{y'}=0$$ which integrates as $$y=cx.$$
This was expected as the given family is made of circles centered at the origin.
answered Dec 6 '18 at 12:48
Yves DaoustYves Daoust
126k672226
answered Dec 6 '18 at 12:48
Yves DaoustYves Daoust
126k672226
answered Dec 6 '18 at 12:48
Yves DaoustYves Daoust
126k672226
answered Dec 6 '18 at 12:48
Yves DaoustYves Daoust
126k672226
126k672226
add a comment |
add a comment |
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$begingroup$
Your differentiation is wrong. The derivative of a constant is 0 and you want $frac{dy}{dx}$, not $frac{dx}{dy}$. You should have $x+ yfrac{dy}{dx}= 0$. From that $frac{dy}{dx}= -frac{x}{y}$ and the orthogonal complement will satisfy $frac{dy}{dx}= frac{y}{x}$.
$endgroup$
– user247327
Dec 6 '18 at 12:33
$begingroup$
I would argue the derivative w.r.t $x$ equals $2x + y frac{delta y}{delta x} = 0$, thus $frac{delta y}{delta x}=frac{-2x}{y}$. Hope this helps.
$endgroup$
– Mathbeginner
Dec 6 '18 at 12:44