Mixing
Find the unit normal to an ellipse given by an equation
$begingroup$
The equation of the ellipse is given as being: $$x^2 -xy + y^2 = 7$$We're instructed to find a unit normal to the curve at a general point $P(x,y)$, and also at point $(-1,2)$ in particular.
My intuition is that I need to parameterize this equation, but I have no idea how to go about doing that. I thought that once I had the parametric equations, I could take the derivative of that set of equations with respect to $t$, and get a general equation for the slope at any point. With that I could probably use the dot product to find the perpendicular normal.
That's my reasoning, but of course, I'm really unsure if any of that is correct.
vectors conic-sections parametrization
asked Jan 30 at 23:37
WalkerGainWalkerGain
112
$endgroup$
add a comment |
$begingroup$
The equation of the ellipse is given as being: $$x^2 -xy + y^2 = 7$$We're instructed to find a unit normal to the curve at a general point $P(x,y)$, and also at point $(-1,2)$ in particular.
My intuition is that I need to parameterize this equation, but I have no idea how to go about doing that. I thought that once I had the parametric equations, I could take the derivative of that set of equations with respect to $t$, and get a general equation for the slope at any point. With that I could probably use the dot product to find the perpendicular normal.
That's my reasoning, but of course, I'm really unsure if any of that is correct.
vectors conic-sections parametrization
asked Jan 30 at 23:37
WalkerGainWalkerGain
112
$endgroup$
$begingroup$
You could get $dy/dx$ using implicit differentiation, then take negative reciprocal for slope of normal.
$endgroup$
– coffeemath
Jan 30 at 23:41
add a comment |
$begingroup$
The equation of the ellipse is given as being: $$x^2 -xy + y^2 = 7$$We're instructed to find a unit normal to the curve at a general point $P(x,y)$, and also at point $(-1,2)$ in particular.
My intuition is that I need to parameterize this equation, but I have no idea how to go about doing that. I thought that once I had the parametric equations, I could take the derivative of that set of equations with respect to $t$, and get a general equation for the slope at any point. With that I could probably use the dot product to find the perpendicular normal.
That's my reasoning, but of course, I'm really unsure if any of that is correct.
vectors conic-sections parametrization
asked Jan 30 at 23:37
WalkerGainWalkerGain
112
$endgroup$
The equation of the ellipse is given as being: $$x^2 -xy + y^2 = 7$$We're instructed to find a unit normal to the curve at a general point $P(x,y)$, and also at point $(-1,2)$ in particular.
My intuition is that I need to parameterize this equation, but I have no idea how to go about doing that. I thought that once I had the parametric equations, I could take the derivative of that set of equations with respect to $t$, and get a general equation for the slope at any point. With that I could probably use the dot product to find the perpendicular normal.
That's my reasoning, but of course, I'm really unsure if any of that is correct.
vectors conic-sections parametrization
vectors conic-sections parametrization
asked Jan 30 at 23:37
WalkerGainWalkerGain
112
asked Jan 30 at 23:37
WalkerGainWalkerGain
112
asked Jan 30 at 23:37
WalkerGainWalkerGain
112
asked Jan 30 at 23:37
WalkerGainWalkerGain
112
asked Jan 30 at 23:37
WalkerGainWalkerGain
112
112
$begingroup$
You could get $dy/dx$ using implicit differentiation, then take negative reciprocal for slope of normal.
$endgroup$
– coffeemath
Jan 30 at 23:41
add a comment |
$begingroup$
You could get $dy/dx$ using implicit differentiation, then take negative reciprocal for slope of normal.
$endgroup$
– coffeemath
Jan 30 at 23:41
$begingroup$
You could get $dy/dx$ using implicit differentiation, then take negative reciprocal for slope of normal.
$endgroup$
– coffeemath
Jan 30 at 23:41
$begingroup$
You could get $dy/dx$ using implicit differentiation, then take negative reciprocal for slope of normal.
$endgroup$
– coffeemath
Jan 30 at 23:41
add a comment |
2 Answers
2
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oldest
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$begingroup$
We have $2x-y-xy'+2yy'=0$ so $y' =frac {y-2x} {2y-x}$. At any point $(x,y)$ on the ellipse the slope of the normal line is $frac {2y-x} {2x-y}$. At $(-1,2)$ it is $-5/4$. (A unit vector along the normal is $(frac 1 {sqrt {1+m^{2}}},frac m {sqrt {1+m^{2}}})$ where $m$ is the slope).
answered Jan 30 at 23:52
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
add a comment |
$begingroup$
The general parametrization of an axis aligned ellipse is $$pmatrix{x = a cos(t) \ y = b sin(t) }$$
But this is oblique ellipse, which can be made parametric, just as above, but with a rotation.
$$pmatrix{x = a cos(t) cos(theta) - b sin(t) sin(theta) \ y = a cos(t) sin(theta) + b sin(t) cos(theta) }$$
Plug into the ellipse equation, and force all coefficients of $sin(t)$ or $cos(t)$ to be zero. With some effort, you will get to
$$ pmatrix{ x = sqrt{7} cos(t) - frac{sqrt{7}}{sqrt{3}} sin(t) \
y = sqrt{7} cos(t) + frac{sqrt{7}}{sqrt{3}} sin(t) }$$
which is a result of $theta = frac{pi}{4}$, $a=sqrt{2}sqrt{7}$ and $b=sqrt{2}sqrt{7}/sqrt{3}$.
Now apply the tricks of differential geometry to find the tangent, normal, curvature etc.
Another approach is to construct the implicit derivative of $f(x,y) = x^2 - x y + y^2 -7$
$$ {rm d}f = frac{partial f}{partial x} {rm d}x + frac{partial f}{partial y}{rm d}y $$
$$ {rm d}f = (2x-y) {rm d}x + (2y-x){rm d}y = 0$$
Note that the tangent and the normal are
$$begin{aligned} vec{e} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{{rm d}x \ {rm d}y} & vec{n} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{-{rm d}y \ {rm d}x} end{aligned}$$
and from the implicit derivate, we can arbitrarily make ${rm d}x = frac{x-2 y}{2x -y} {rm d}y$ which is used above to evaluate
$$begin{aligned} vec{e} & = pmatrix{frac{x-2y}{sqrt{5x^2-8xy+5y^2}} \ frac{2x-y}{sqrt{5x^2-8xy+5y^2}}} & vec{n} & = pmatrix{ frac{y-2x}{sqrt{5x^2-8xy+5y^2}} \ frac{x-2y}{sqrt{5x^2-8xy+5y^2}} } end{aligned}$$
Now use different values of $(x,y)$ find a solution.
answered Jan 31 at 6:14
ja72ja72
7,54212044
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
We have $2x-y-xy'+2yy'=0$ so $y' =frac {y-2x} {2y-x}$. At any point $(x,y)$ on the ellipse the slope of the normal line is $frac {2y-x} {2x-y}$. At $(-1,2)$ it is $-5/4$. (A unit vector along the normal is $(frac 1 {sqrt {1+m^{2}}},frac m {sqrt {1+m^{2}}})$ where $m$ is the slope).
answered Jan 30 at 23:52
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
add a comment |
$begingroup$
We have $2x-y-xy'+2yy'=0$ so $y' =frac {y-2x} {2y-x}$. At any point $(x,y)$ on the ellipse the slope of the normal line is $frac {2y-x} {2x-y}$. At $(-1,2)$ it is $-5/4$. (A unit vector along the normal is $(frac 1 {sqrt {1+m^{2}}},frac m {sqrt {1+m^{2}}})$ where $m$ is the slope).
answered Jan 30 at 23:52
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
add a comment |
$begingroup$
We have $2x-y-xy'+2yy'=0$ so $y' =frac {y-2x} {2y-x}$. At any point $(x,y)$ on the ellipse the slope of the normal line is $frac {2y-x} {2x-y}$. At $(-1,2)$ it is $-5/4$. (A unit vector along the normal is $(frac 1 {sqrt {1+m^{2}}},frac m {sqrt {1+m^{2}}})$ where $m$ is the slope).
answered Jan 30 at 23:52
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
$endgroup$
We have $2x-y-xy'+2yy'=0$ so $y' =frac {y-2x} {2y-x}$. At any point $(x,y)$ on the ellipse the slope of the normal line is $frac {2y-x} {2x-y}$. At $(-1,2)$ it is $-5/4$. (A unit vector along the normal is $(frac 1 {sqrt {1+m^{2}}},frac m {sqrt {1+m^{2}}})$ where $m$ is the slope).
answered Jan 30 at 23:52
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 30 at 23:52
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 30 at 23:52
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
answered Jan 30 at 23:52
Kavi Rama MurthyKavi Rama Murthy
72.6k53170
72.6k53170
add a comment |
add a comment |
$begingroup$
The general parametrization of an axis aligned ellipse is $$pmatrix{x = a cos(t) \ y = b sin(t) }$$
But this is oblique ellipse, which can be made parametric, just as above, but with a rotation.
$$pmatrix{x = a cos(t) cos(theta) - b sin(t) sin(theta) \ y = a cos(t) sin(theta) + b sin(t) cos(theta) }$$
Plug into the ellipse equation, and force all coefficients of $sin(t)$ or $cos(t)$ to be zero. With some effort, you will get to
$$ pmatrix{ x = sqrt{7} cos(t) - frac{sqrt{7}}{sqrt{3}} sin(t) \
y = sqrt{7} cos(t) + frac{sqrt{7}}{sqrt{3}} sin(t) }$$
which is a result of $theta = frac{pi}{4}$, $a=sqrt{2}sqrt{7}$ and $b=sqrt{2}sqrt{7}/sqrt{3}$.
Now apply the tricks of differential geometry to find the tangent, normal, curvature etc.
Another approach is to construct the implicit derivative of $f(x,y) = x^2 - x y + y^2 -7$
$$ {rm d}f = frac{partial f}{partial x} {rm d}x + frac{partial f}{partial y}{rm d}y $$
$$ {rm d}f = (2x-y) {rm d}x + (2y-x){rm d}y = 0$$
Note that the tangent and the normal are
$$begin{aligned} vec{e} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{{rm d}x \ {rm d}y} & vec{n} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{-{rm d}y \ {rm d}x} end{aligned}$$
and from the implicit derivate, we can arbitrarily make ${rm d}x = frac{x-2 y}{2x -y} {rm d}y$ which is used above to evaluate
$$begin{aligned} vec{e} & = pmatrix{frac{x-2y}{sqrt{5x^2-8xy+5y^2}} \ frac{2x-y}{sqrt{5x^2-8xy+5y^2}}} & vec{n} & = pmatrix{ frac{y-2x}{sqrt{5x^2-8xy+5y^2}} \ frac{x-2y}{sqrt{5x^2-8xy+5y^2}} } end{aligned}$$
Now use different values of $(x,y)$ find a solution.
answered Jan 31 at 6:14
ja72ja72
7,54212044
$endgroup$
add a comment |
$begingroup$
The general parametrization of an axis aligned ellipse is $$pmatrix{x = a cos(t) \ y = b sin(t) }$$
But this is oblique ellipse, which can be made parametric, just as above, but with a rotation.
$$pmatrix{x = a cos(t) cos(theta) - b sin(t) sin(theta) \ y = a cos(t) sin(theta) + b sin(t) cos(theta) }$$
Plug into the ellipse equation, and force all coefficients of $sin(t)$ or $cos(t)$ to be zero. With some effort, you will get to
$$ pmatrix{ x = sqrt{7} cos(t) - frac{sqrt{7}}{sqrt{3}} sin(t) \
y = sqrt{7} cos(t) + frac{sqrt{7}}{sqrt{3}} sin(t) }$$
which is a result of $theta = frac{pi}{4}$, $a=sqrt{2}sqrt{7}$ and $b=sqrt{2}sqrt{7}/sqrt{3}$.
Now apply the tricks of differential geometry to find the tangent, normal, curvature etc.
Another approach is to construct the implicit derivative of $f(x,y) = x^2 - x y + y^2 -7$
$$ {rm d}f = frac{partial f}{partial x} {rm d}x + frac{partial f}{partial y}{rm d}y $$
$$ {rm d}f = (2x-y) {rm d}x + (2y-x){rm d}y = 0$$
Note that the tangent and the normal are
$$begin{aligned} vec{e} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{{rm d}x \ {rm d}y} & vec{n} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{-{rm d}y \ {rm d}x} end{aligned}$$
and from the implicit derivate, we can arbitrarily make ${rm d}x = frac{x-2 y}{2x -y} {rm d}y$ which is used above to evaluate
$$begin{aligned} vec{e} & = pmatrix{frac{x-2y}{sqrt{5x^2-8xy+5y^2}} \ frac{2x-y}{sqrt{5x^2-8xy+5y^2}}} & vec{n} & = pmatrix{ frac{y-2x}{sqrt{5x^2-8xy+5y^2}} \ frac{x-2y}{sqrt{5x^2-8xy+5y^2}} } end{aligned}$$
Now use different values of $(x,y)$ find a solution.
answered Jan 31 at 6:14
ja72ja72
7,54212044
$endgroup$
add a comment |
$begingroup$
The general parametrization of an axis aligned ellipse is $$pmatrix{x = a cos(t) \ y = b sin(t) }$$
But this is oblique ellipse, which can be made parametric, just as above, but with a rotation.
$$pmatrix{x = a cos(t) cos(theta) - b sin(t) sin(theta) \ y = a cos(t) sin(theta) + b sin(t) cos(theta) }$$
Plug into the ellipse equation, and force all coefficients of $sin(t)$ or $cos(t)$ to be zero. With some effort, you will get to
$$ pmatrix{ x = sqrt{7} cos(t) - frac{sqrt{7}}{sqrt{3}} sin(t) \
y = sqrt{7} cos(t) + frac{sqrt{7}}{sqrt{3}} sin(t) }$$
which is a result of $theta = frac{pi}{4}$, $a=sqrt{2}sqrt{7}$ and $b=sqrt{2}sqrt{7}/sqrt{3}$.
Now apply the tricks of differential geometry to find the tangent, normal, curvature etc.
Another approach is to construct the implicit derivative of $f(x,y) = x^2 - x y + y^2 -7$
$$ {rm d}f = frac{partial f}{partial x} {rm d}x + frac{partial f}{partial y}{rm d}y $$
$$ {rm d}f = (2x-y) {rm d}x + (2y-x){rm d}y = 0$$
Note that the tangent and the normal are
$$begin{aligned} vec{e} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{{rm d}x \ {rm d}y} & vec{n} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{-{rm d}y \ {rm d}x} end{aligned}$$
and from the implicit derivate, we can arbitrarily make ${rm d}x = frac{x-2 y}{2x -y} {rm d}y$ which is used above to evaluate
$$begin{aligned} vec{e} & = pmatrix{frac{x-2y}{sqrt{5x^2-8xy+5y^2}} \ frac{2x-y}{sqrt{5x^2-8xy+5y^2}}} & vec{n} & = pmatrix{ frac{y-2x}{sqrt{5x^2-8xy+5y^2}} \ frac{x-2y}{sqrt{5x^2-8xy+5y^2}} } end{aligned}$$
Now use different values of $(x,y)$ find a solution.
answered Jan 31 at 6:14
ja72ja72
7,54212044
$endgroup$
The general parametrization of an axis aligned ellipse is $$pmatrix{x = a cos(t) \ y = b sin(t) }$$
But this is oblique ellipse, which can be made parametric, just as above, but with a rotation.
$$pmatrix{x = a cos(t) cos(theta) - b sin(t) sin(theta) \ y = a cos(t) sin(theta) + b sin(t) cos(theta) }$$
Plug into the ellipse equation, and force all coefficients of $sin(t)$ or $cos(t)$ to be zero. With some effort, you will get to
$$ pmatrix{ x = sqrt{7} cos(t) - frac{sqrt{7}}{sqrt{3}} sin(t) \
y = sqrt{7} cos(t) + frac{sqrt{7}}{sqrt{3}} sin(t) }$$
which is a result of $theta = frac{pi}{4}$, $a=sqrt{2}sqrt{7}$ and $b=sqrt{2}sqrt{7}/sqrt{3}$.
Now apply the tricks of differential geometry to find the tangent, normal, curvature etc.
Another approach is to construct the implicit derivative of $f(x,y) = x^2 - x y + y^2 -7$
$$ {rm d}f = frac{partial f}{partial x} {rm d}x + frac{partial f}{partial y}{rm d}y $$
$$ {rm d}f = (2x-y) {rm d}x + (2y-x){rm d}y = 0$$
Note that the tangent and the normal are
$$begin{aligned} vec{e} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{{rm d}x \ {rm d}y} & vec{n} & = frac{1}{sqrt{{rm d}x^2 + {rm d}y^2}} pmatrix{-{rm d}y \ {rm d}x} end{aligned}$$
and from the implicit derivate, we can arbitrarily make ${rm d}x = frac{x-2 y}{2x -y} {rm d}y$ which is used above to evaluate
$$begin{aligned} vec{e} & = pmatrix{frac{x-2y}{sqrt{5x^2-8xy+5y^2}} \ frac{2x-y}{sqrt{5x^2-8xy+5y^2}}} & vec{n} & = pmatrix{ frac{y-2x}{sqrt{5x^2-8xy+5y^2}} \ frac{x-2y}{sqrt{5x^2-8xy+5y^2}} } end{aligned}$$
Now use different values of $(x,y)$ find a solution.
answered Jan 31 at 6:14
ja72ja72
7,54212044
edited Jan 31 at 6:31
answered Jan 31 at 6:14
ja72ja72
7,54212044
answered Jan 31 at 6:14
ja72ja72
7,54212044
answered Jan 31 at 6:14
ja72ja72
7,54212044
7,54212044
add a comment |
add a comment |
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$begingroup$
You could get $dy/dx$ using implicit differentiation, then take negative reciprocal for slope of normal.
$endgroup$
– coffeemath
Jan 30 at 23:41