How to show that $($grad $F)(a,b,c)$ is normal to $r^2=a^2+b^2+c^2$ at $(a,b,c)$? The 2019 Stack Overflow Developer Survey Results Are InEquation of A Tangent Line To a SphereVector analysis: understanding formulas for normal and tangentSteepest part of a surface.Show curve is tangent to a surface using gradientGradient of a real valued function defined on a sphereShow tangent of evolute is normal to original curveProve: $lim_varepsilonto 0int_S(x_0,varepsilon)u(x)operatornamegrad(varphi(x))cdot N,ds=u(x_0)$Angle between two intersecting lines that appear on a cut faceExpressing gradient F $nabla F$ in the form of symmetric equation.Show that the curvature of simple arc C is given by $kappa(q)=fracVert f'(a)times f''(a)VertVert f''(a)Vert^3$
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How to show that $($grad $F)(a,b,c)$ is normal to $r^2=a^2+b^2+c^2$ at $(a,b,c)$?
The 2019 Stack Overflow Developer Survey Results Are InEquation of A Tangent Line To a SphereVector analysis: understanding formulas for normal and tangentSteepest part of a surface.Show curve is tangent to a surface using gradientGradient of a real valued function defined on a sphereShow tangent of evolute is normal to original curveProve: $lim_varepsilonto 0int_S(x_0,varepsilon)u(x)operatornamegrad(varphi(x))cdot N,ds=u(x_0)$Angle between two intersecting lines that appear on a cut faceExpressing gradient F $nabla F$ in the form of symmetric equation.Show that the curvature of simple arc C is given by $kappa(q)=fracVert f'(a)times f''(a)VertVert f''(a)Vert^3$
$begingroup$
I need help with this problem:
Let $F:mathbbR^3$
$mathbf0rightarrowmathbbR$ be defined by $$F(mathbfr)=frac1r, quad mathbfrneqmathbf0$$ where $mathbfr=(x,y,z)$ and $r=VertmathbfrVert=sqrt(x^2+y^2+x^2)$. Prove that $$(grad F)(mathbfr)=-frac1r^3mathbfr$$ Show that $($grad $F)(a,b,c)$ is normal to the sphere $r^2=a^2+b^2+c^2$ at the point $(a,b,c)$ and pints in the direction of increasing values of F. Find the equation of the tangent plane to the sphere $(a,b,c)$.
I've already proved that $(grad F)(mathbfr)=-frac1r^3mathbfr$, but I don't know hot to show that the $(grad F)(a,b,c)$ is normal to the sphere. Should I use dot product? Also how do I find the equation of the tangent plane? Hope you can help me.
vector-analysis
asked Mar 24 at 4:12
davidllerenavdavidllerenav
3228
$endgroup$
add a comment |
$begingroup$
I need help with this problem:
Let $F:mathbbR^3$
$mathbf0rightarrowmathbbR$ be defined by $$F(mathbfr)=frac1r, quad mathbfrneqmathbf0$$ where $mathbfr=(x,y,z)$ and $r=VertmathbfrVert=sqrt(x^2+y^2+x^2)$. Prove that $$(grad F)(mathbfr)=-frac1r^3mathbfr$$ Show that $($grad $F)(a,b,c)$ is normal to the sphere $r^2=a^2+b^2+c^2$ at the point $(a,b,c)$ and pints in the direction of increasing values of F. Find the equation of the tangent plane to the sphere $(a,b,c)$.
I've already proved that $(grad F)(mathbfr)=-frac1r^3mathbfr$, but I don't know hot to show that the $(grad F)(a,b,c)$ is normal to the sphere. Should I use dot product? Also how do I find the equation of the tangent plane? Hope you can help me.
vector-analysis
asked Mar 24 at 4:12
davidllerenavdavidllerenav
3228
$endgroup$
add a comment |
$begingroup$
I need help with this problem:
Let $F:mathbbR^3$
$mathbf0rightarrowmathbbR$ be defined by $$F(mathbfr)=frac1r, quad mathbfrneqmathbf0$$ where $mathbfr=(x,y,z)$ and $r=VertmathbfrVert=sqrt(x^2+y^2+x^2)$. Prove that $$(grad F)(mathbfr)=-frac1r^3mathbfr$$ Show that $($grad $F)(a,b,c)$ is normal to the sphere $r^2=a^2+b^2+c^2$ at the point $(a,b,c)$ and pints in the direction of increasing values of F. Find the equation of the tangent plane to the sphere $(a,b,c)$.
I've already proved that $(grad F)(mathbfr)=-frac1r^3mathbfr$, but I don't know hot to show that the $(grad F)(a,b,c)$ is normal to the sphere. Should I use dot product? Also how do I find the equation of the tangent plane? Hope you can help me.
vector-analysis
asked Mar 24 at 4:12
davidllerenavdavidllerenav
3228
$endgroup$
I need help with this problem:
Let $F:mathbbR^3$
$mathbf0rightarrowmathbbR$ be defined by $$F(mathbfr)=frac1r, quad mathbfrneqmathbf0$$ where $mathbfr=(x,y,z)$ and $r=VertmathbfrVert=sqrt(x^2+y^2+x^2)$. Prove that $$(grad F)(mathbfr)=-frac1r^3mathbfr$$ Show that $($grad $F)(a,b,c)$ is normal to the sphere $r^2=a^2+b^2+c^2$ at the point $(a,b,c)$ and pints in the direction of increasing values of F. Find the equation of the tangent plane to the sphere $(a,b,c)$.
I've already proved that $(grad F)(mathbfr)=-frac1r^3mathbfr$, but I don't know hot to show that the $(grad F)(a,b,c)$ is normal to the sphere. Should I use dot product? Also how do I find the equation of the tangent plane? Hope you can help me.
vector-analysis
vector-analysis
asked Mar 24 at 4:12
davidllerenavdavidllerenav
3228
asked Mar 24 at 4:12
davidllerenavdavidllerenav
3228
asked Mar 24 at 4:12
davidllerenavdavidllerenav
3228
asked Mar 24 at 4:12
davidllerenavdavidllerenav
3228
asked Mar 24 at 4:12
davidllerenavdavidllerenav
3228
3228
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are two facts about the gradient in this problem that you have to use. The first one, is that the gradient vector $nabla F$ always points in the direction of maximum increase of $F$. The second one, is that the gradient vector is normal to the level surface of $F(x,y,z)$ at a point $(x,y,z)$. There are really good proofs here to the bottom of the page, under each blue note.
So, we have already solved half of the problem. Now to show that $nabla F$ is perpendicular to the sphere $r^2 = a^2 + b^2 + c^2$ at a point $(a,b,c)$, we have to show that the sphere is the level surface of $F$ at the point $(a,b,c)$. The level surface of $F$ at a point $(a,b,c)$ is the surface given by the equation $F(x,y,z) = k$ where $k = F(a,b,c)$.
So we have
$$ F(x,y,z) = F(a,b,c) \ frac1sqrtx^2 + y^2 + z^2 = frac1sqrta^2 + b^2 + c^2 \ sqrtr^2 = sqrta^2 + b^2 + c^2 \ r^2 = a^2 + b^2 + c^2$$
So the sphere $r^2 = a^2 + b^2 + c^2$ (which is a sphere centered at the origin that has the point $(a,b,c)$ on it's surface) is indeed the level surface of $F$ at the point $(a,b,c)$ and so $nabla F(a,b,c)$ is perpendicular to it.
To find the equation of the tangent plane to the sphere at the point $(a,b,c)$, you need a normal vector to the sphere in that point. We already have it: it's the gradient. The equation of a tangent plane to a surface at a point $P$ is given by:
$$ vec n cdot left(vec r - vec Pright) = 0 $$
where $vec P$ is the position vector of the point $P$ and $vec n$ is the normal vector. So in our case, the equation for the plane is:
$$ nabla F cdot left(leftlangle x, y, z rightrangle - leftlangle a, b, c rightrangleright) = 0 $$ and then just carry the calculations on.
answered Mar 24 at 8:53
Stefan OctavianStefan Octavian
1987
$endgroup$
$begingroup$
The level set is always $F(x,y,z)=F(a,b,c)$ where $(a,b,c)$ is a point? For example, in the level set of $f(x,y)=x^2-y^2$ is $f(x,y)=k$ where $k=f(a,b)$?
$endgroup$
– davidllerenav
Mar 25 at 20:19
$begingroup$
The level curves/surfaces/higher-dimension-equivalent of a function f of multiple variables is where $f(vec P) = k$ and $vec P$ is in a sense a notation for describing $f(x,y,...)$. Now, if we have a point $(a,b)$, we know that it might be on a level curve of the function $f(x,y)$. But, if $(a,b)$ is on a level curve, $f(x,y) = k$, it must be that $f(a,b) = k$.
$endgroup$
– Stefan Octavian
Mar 26 at 18:14
$begingroup$
You can read more about level curves to the bottom of this page.
$endgroup$
– Stefan Octavian
Mar 26 at 18:20
$begingroup$
Ok, I'll definitely read more about them. So, if I have $f(x,y)=x^2-y^2$, then the level set $x^2-y^2=c$ at point $(a,b)$ means that $f(a,b)=c$, thus the level set is $f(x,y)=f(a,b)Rightarrow x^2-y^2=a^2-b^2$?
$endgroup$
– davidllerenav
Mar 26 at 18:43
$begingroup$
Yes, briefly said. It's because $(a,b)$ has to be on that level set and so $f(a,b) = c$
$endgroup$
– Stefan Octavian
Mar 27 at 5:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
There are two facts about the gradient in this problem that you have to use. The first one, is that the gradient vector $nabla F$ always points in the direction of maximum increase of $F$. The second one, is that the gradient vector is normal to the level surface of $F(x,y,z)$ at a point $(x,y,z)$. There are really good proofs here to the bottom of the page, under each blue note.
So, we have already solved half of the problem. Now to show that $nabla F$ is perpendicular to the sphere $r^2 = a^2 + b^2 + c^2$ at a point $(a,b,c)$, we have to show that the sphere is the level surface of $F$ at the point $(a,b,c)$. The level surface of $F$ at a point $(a,b,c)$ is the surface given by the equation $F(x,y,z) = k$ where $k = F(a,b,c)$.
So we have
$$ F(x,y,z) = F(a,b,c) \ frac1sqrtx^2 + y^2 + z^2 = frac1sqrta^2 + b^2 + c^2 \ sqrtr^2 = sqrta^2 + b^2 + c^2 \ r^2 = a^2 + b^2 + c^2$$
So the sphere $r^2 = a^2 + b^2 + c^2$ (which is a sphere centered at the origin that has the point $(a,b,c)$ on it's surface) is indeed the level surface of $F$ at the point $(a,b,c)$ and so $nabla F(a,b,c)$ is perpendicular to it.
To find the equation of the tangent plane to the sphere at the point $(a,b,c)$, you need a normal vector to the sphere in that point. We already have it: it's the gradient. The equation of a tangent plane to a surface at a point $P$ is given by:
$$ vec n cdot left(vec r - vec Pright) = 0 $$
where $vec P$ is the position vector of the point $P$ and $vec n$ is the normal vector. So in our case, the equation for the plane is:
$$ nabla F cdot left(leftlangle x, y, z rightrangle - leftlangle a, b, c rightrangleright) = 0 $$ and then just carry the calculations on.
answered Mar 24 at 8:53
Stefan OctavianStefan Octavian
1987
$endgroup$
$begingroup$
The level set is always $F(x,y,z)=F(a,b,c)$ where $(a,b,c)$ is a point? For example, in the level set of $f(x,y)=x^2-y^2$ is $f(x,y)=k$ where $k=f(a,b)$?
$endgroup$
– davidllerenav
Mar 25 at 20:19
$begingroup$
The level curves/surfaces/higher-dimension-equivalent of a function f of multiple variables is where $f(vec P) = k$ and $vec P$ is in a sense a notation for describing $f(x,y,...)$. Now, if we have a point $(a,b)$, we know that it might be on a level curve of the function $f(x,y)$. But, if $(a,b)$ is on a level curve, $f(x,y) = k$, it must be that $f(a,b) = k$.
$endgroup$
– Stefan Octavian
Mar 26 at 18:14
$begingroup$
You can read more about level curves to the bottom of this page.
$endgroup$
– Stefan Octavian
Mar 26 at 18:20
$begingroup$
Ok, I'll definitely read more about them. So, if I have $f(x,y)=x^2-y^2$, then the level set $x^2-y^2=c$ at point $(a,b)$ means that $f(a,b)=c$, thus the level set is $f(x,y)=f(a,b)Rightarrow x^2-y^2=a^2-b^2$?
$endgroup$
– davidllerenav
Mar 26 at 18:43
$begingroup$
Yes, briefly said. It's because $(a,b)$ has to be on that level set and so $f(a,b) = c$
$endgroup$
– Stefan Octavian
Mar 27 at 5:12
add a comment |
$begingroup$
There are two facts about the gradient in this problem that you have to use. The first one, is that the gradient vector $nabla F$ always points in the direction of maximum increase of $F$. The second one, is that the gradient vector is normal to the level surface of $F(x,y,z)$ at a point $(x,y,z)$. There are really good proofs here to the bottom of the page, under each blue note.
So, we have already solved half of the problem. Now to show that $nabla F$ is perpendicular to the sphere $r^2 = a^2 + b^2 + c^2$ at a point $(a,b,c)$, we have to show that the sphere is the level surface of $F$ at the point $(a,b,c)$. The level surface of $F$ at a point $(a,b,c)$ is the surface given by the equation $F(x,y,z) = k$ where $k = F(a,b,c)$.
So we have
$$ F(x,y,z) = F(a,b,c) \ frac1sqrtx^2 + y^2 + z^2 = frac1sqrta^2 + b^2 + c^2 \ sqrtr^2 = sqrta^2 + b^2 + c^2 \ r^2 = a^2 + b^2 + c^2$$
So the sphere $r^2 = a^2 + b^2 + c^2$ (which is a sphere centered at the origin that has the point $(a,b,c)$ on it's surface) is indeed the level surface of $F$ at the point $(a,b,c)$ and so $nabla F(a,b,c)$ is perpendicular to it.
To find the equation of the tangent plane to the sphere at the point $(a,b,c)$, you need a normal vector to the sphere in that point. We already have it: it's the gradient. The equation of a tangent plane to a surface at a point $P$ is given by:
$$ vec n cdot left(vec r - vec Pright) = 0 $$
where $vec P$ is the position vector of the point $P$ and $vec n$ is the normal vector. So in our case, the equation for the plane is:
$$ nabla F cdot left(leftlangle x, y, z rightrangle - leftlangle a, b, c rightrangleright) = 0 $$ and then just carry the calculations on.
answered Mar 24 at 8:53
Stefan OctavianStefan Octavian
1987
$endgroup$
$begingroup$
The level set is always $F(x,y,z)=F(a,b,c)$ where $(a,b,c)$ is a point? For example, in the level set of $f(x,y)=x^2-y^2$ is $f(x,y)=k$ where $k=f(a,b)$?
$endgroup$
– davidllerenav
Mar 25 at 20:19
$begingroup$
The level curves/surfaces/higher-dimension-equivalent of a function f of multiple variables is where $f(vec P) = k$ and $vec P$ is in a sense a notation for describing $f(x,y,...)$. Now, if we have a point $(a,b)$, we know that it might be on a level curve of the function $f(x,y)$. But, if $(a,b)$ is on a level curve, $f(x,y) = k$, it must be that $f(a,b) = k$.
$endgroup$
– Stefan Octavian
Mar 26 at 18:14
$begingroup$
You can read more about level curves to the bottom of this page.
$endgroup$
– Stefan Octavian
Mar 26 at 18:20
$begingroup$
Ok, I'll definitely read more about them. So, if I have $f(x,y)=x^2-y^2$, then the level set $x^2-y^2=c$ at point $(a,b)$ means that $f(a,b)=c$, thus the level set is $f(x,y)=f(a,b)Rightarrow x^2-y^2=a^2-b^2$?
$endgroup$
– davidllerenav
Mar 26 at 18:43
$begingroup$
Yes, briefly said. It's because $(a,b)$ has to be on that level set and so $f(a,b) = c$
$endgroup$
– Stefan Octavian
Mar 27 at 5:12
add a comment |
$begingroup$
There are two facts about the gradient in this problem that you have to use. The first one, is that the gradient vector $nabla F$ always points in the direction of maximum increase of $F$. The second one, is that the gradient vector is normal to the level surface of $F(x,y,z)$ at a point $(x,y,z)$. There are really good proofs here to the bottom of the page, under each blue note.
So, we have already solved half of the problem. Now to show that $nabla F$ is perpendicular to the sphere $r^2 = a^2 + b^2 + c^2$ at a point $(a,b,c)$, we have to show that the sphere is the level surface of $F$ at the point $(a,b,c)$. The level surface of $F$ at a point $(a,b,c)$ is the surface given by the equation $F(x,y,z) = k$ where $k = F(a,b,c)$.
So we have
$$ F(x,y,z) = F(a,b,c) \ frac1sqrtx^2 + y^2 + z^2 = frac1sqrta^2 + b^2 + c^2 \ sqrtr^2 = sqrta^2 + b^2 + c^2 \ r^2 = a^2 + b^2 + c^2$$
So the sphere $r^2 = a^2 + b^2 + c^2$ (which is a sphere centered at the origin that has the point $(a,b,c)$ on it's surface) is indeed the level surface of $F$ at the point $(a,b,c)$ and so $nabla F(a,b,c)$ is perpendicular to it.
To find the equation of the tangent plane to the sphere at the point $(a,b,c)$, you need a normal vector to the sphere in that point. We already have it: it's the gradient. The equation of a tangent plane to a surface at a point $P$ is given by:
$$ vec n cdot left(vec r - vec Pright) = 0 $$
where $vec P$ is the position vector of the point $P$ and $vec n$ is the normal vector. So in our case, the equation for the plane is:
$$ nabla F cdot left(leftlangle x, y, z rightrangle - leftlangle a, b, c rightrangleright) = 0 $$ and then just carry the calculations on.
answered Mar 24 at 8:53
Stefan OctavianStefan Octavian
1987
$endgroup$
There are two facts about the gradient in this problem that you have to use. The first one, is that the gradient vector $nabla F$ always points in the direction of maximum increase of $F$. The second one, is that the gradient vector is normal to the level surface of $F(x,y,z)$ at a point $(x,y,z)$. There are really good proofs here to the bottom of the page, under each blue note.
So, we have already solved half of the problem. Now to show that $nabla F$ is perpendicular to the sphere $r^2 = a^2 + b^2 + c^2$ at a point $(a,b,c)$, we have to show that the sphere is the level surface of $F$ at the point $(a,b,c)$. The level surface of $F$ at a point $(a,b,c)$ is the surface given by the equation $F(x,y,z) = k$ where $k = F(a,b,c)$.
So we have
$$ F(x,y,z) = F(a,b,c) \ frac1sqrtx^2 + y^2 + z^2 = frac1sqrta^2 + b^2 + c^2 \ sqrtr^2 = sqrta^2 + b^2 + c^2 \ r^2 = a^2 + b^2 + c^2$$
So the sphere $r^2 = a^2 + b^2 + c^2$ (which is a sphere centered at the origin that has the point $(a,b,c)$ on it's surface) is indeed the level surface of $F$ at the point $(a,b,c)$ and so $nabla F(a,b,c)$ is perpendicular to it.
To find the equation of the tangent plane to the sphere at the point $(a,b,c)$, you need a normal vector to the sphere in that point. We already have it: it's the gradient. The equation of a tangent plane to a surface at a point $P$ is given by:
$$ vec n cdot left(vec r - vec Pright) = 0 $$
where $vec P$ is the position vector of the point $P$ and $vec n$ is the normal vector. So in our case, the equation for the plane is:
$$ nabla F cdot left(leftlangle x, y, z rightrangle - leftlangle a, b, c rightrangleright) = 0 $$ and then just carry the calculations on.
answered Mar 24 at 8:53
Stefan OctavianStefan Octavian
1987
edited Mar 24 at 8:59
answered Mar 24 at 8:53
Stefan OctavianStefan Octavian
1987
answered Mar 24 at 8:53
Stefan OctavianStefan Octavian
1987
answered Mar 24 at 8:53
Stefan OctavianStefan Octavian
1987
1987
$begingroup$
The level set is always $F(x,y,z)=F(a,b,c)$ where $(a,b,c)$ is a point? For example, in the level set of $f(x,y)=x^2-y^2$ is $f(x,y)=k$ where $k=f(a,b)$?
$endgroup$
– davidllerenav
Mar 25 at 20:19
$begingroup$
The level curves/surfaces/higher-dimension-equivalent of a function f of multiple variables is where $f(vec P) = k$ and $vec P$ is in a sense a notation for describing $f(x,y,...)$. Now, if we have a point $(a,b)$, we know that it might be on a level curve of the function $f(x,y)$. But, if $(a,b)$ is on a level curve, $f(x,y) = k$, it must be that $f(a,b) = k$.
$endgroup$
– Stefan Octavian
Mar 26 at 18:14
$begingroup$
You can read more about level curves to the bottom of this page.
$endgroup$
– Stefan Octavian
Mar 26 at 18:20
$begingroup$
Ok, I'll definitely read more about them. So, if I have $f(x,y)=x^2-y^2$, then the level set $x^2-y^2=c$ at point $(a,b)$ means that $f(a,b)=c$, thus the level set is $f(x,y)=f(a,b)Rightarrow x^2-y^2=a^2-b^2$?
$endgroup$
– davidllerenav
Mar 26 at 18:43
$begingroup$
Yes, briefly said. It's because $(a,b)$ has to be on that level set and so $f(a,b) = c$
$endgroup$
– Stefan Octavian
Mar 27 at 5:12
add a comment |
$begingroup$
The level set is always $F(x,y,z)=F(a,b,c)$ where $(a,b,c)$ is a point? For example, in the level set of $f(x,y)=x^2-y^2$ is $f(x,y)=k$ where $k=f(a,b)$?
$endgroup$
– davidllerenav
Mar 25 at 20:19
$begingroup$
The level curves/surfaces/higher-dimension-equivalent of a function f of multiple variables is where $f(vec P) = k$ and $vec P$ is in a sense a notation for describing $f(x,y,...)$. Now, if we have a point $(a,b)$, we know that it might be on a level curve of the function $f(x,y)$. But, if $(a,b)$ is on a level curve, $f(x,y) = k$, it must be that $f(a,b) = k$.
$endgroup$
– Stefan Octavian
Mar 26 at 18:14
$begingroup$
You can read more about level curves to the bottom of this page.
$endgroup$
– Stefan Octavian
Mar 26 at 18:20
$begingroup$
Ok, I'll definitely read more about them. So, if I have $f(x,y)=x^2-y^2$, then the level set $x^2-y^2=c$ at point $(a,b)$ means that $f(a,b)=c$, thus the level set is $f(x,y)=f(a,b)Rightarrow x^2-y^2=a^2-b^2$?
$endgroup$
– davidllerenav
Mar 26 at 18:43
$begingroup$
Yes, briefly said. It's because $(a,b)$ has to be on that level set and so $f(a,b) = c$
$endgroup$
– Stefan Octavian
Mar 27 at 5:12
$begingroup$
The level set is always $F(x,y,z)=F(a,b,c)$ where $(a,b,c)$ is a point? For example, in the level set of $f(x,y)=x^2-y^2$ is $f(x,y)=k$ where $k=f(a,b)$?
$endgroup$
– davidllerenav
Mar 25 at 20:19
$begingroup$
The level set is always $F(x,y,z)=F(a,b,c)$ where $(a,b,c)$ is a point? For example, in the level set of $f(x,y)=x^2-y^2$ is $f(x,y)=k$ where $k=f(a,b)$?
$endgroup$
– davidllerenav
Mar 25 at 20:19
$begingroup$
The level curves/surfaces/higher-dimension-equivalent of a function f of multiple variables is where $f(vec P) = k$ and $vec P$ is in a sense a notation for describing $f(x,y,...)$. Now, if we have a point $(a,b)$, we know that it might be on a level curve of the function $f(x,y)$. But, if $(a,b)$ is on a level curve, $f(x,y) = k$, it must be that $f(a,b) = k$.
$endgroup$
– Stefan Octavian
Mar 26 at 18:14
$begingroup$
The level curves/surfaces/higher-dimension-equivalent of a function f of multiple variables is where $f(vec P) = k$ and $vec P$ is in a sense a notation for describing $f(x,y,...)$. Now, if we have a point $(a,b)$, we know that it might be on a level curve of the function $f(x,y)$. But, if $(a,b)$ is on a level curve, $f(x,y) = k$, it must be that $f(a,b) = k$.
$endgroup$
– Stefan Octavian
Mar 26 at 18:14
$begingroup$
You can read more about level curves to the bottom of this page.
$endgroup$
– Stefan Octavian
Mar 26 at 18:20
$begingroup$
You can read more about level curves to the bottom of this page.
$endgroup$
– Stefan Octavian
Mar 26 at 18:20
$begingroup$
Ok, I'll definitely read more about them. So, if I have $f(x,y)=x^2-y^2$, then the level set $x^2-y^2=c$ at point $(a,b)$ means that $f(a,b)=c$, thus the level set is $f(x,y)=f(a,b)Rightarrow x^2-y^2=a^2-b^2$?
$endgroup$
– davidllerenav
Mar 26 at 18:43
$begingroup$
Ok, I'll definitely read more about them. So, if I have $f(x,y)=x^2-y^2$, then the level set $x^2-y^2=c$ at point $(a,b)$ means that $f(a,b)=c$, thus the level set is $f(x,y)=f(a,b)Rightarrow x^2-y^2=a^2-b^2$?
$endgroup$
– davidllerenav
Mar 26 at 18:43
$begingroup$
Yes, briefly said. It's because $(a,b)$ has to be on that level set and so $f(a,b) = c$
$endgroup$
– Stefan Octavian
Mar 27 at 5:12
$begingroup$
Yes, briefly said. It's because $(a,b)$ has to be on that level set and so $f(a,b) = c$
$endgroup$
– Stefan Octavian
Mar 27 at 5:12
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