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Gradient vector $nabla F = (z_x, z_y, -1)$ is normal to the integral surface?
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I have an integral surface $z = z(x, y)$.
Writing this integral surface in implicit form, we get
$$F(x, y, z) = z(x, y) - z = 0$$
I am then told that the gradient vector $nabla F = (z_x, z_y, -1)$ is normal to the integral surface $F(x, y, z) = 0$.
First of all, how was this calculated? I understand how the gradient is calculated, but I don't understand how it was calculated in this case?
And lastly, where did the $-1$ come from and why? Couldn't they also have had $nabla F = (z_x, z_y, 1)$, where this would just be the normal vector in the other direction? Why and how did they pick the $-1$ direction instead?
I apologise. My vector calculus understanding is not particularly strong, and I strive to improve it.
Thank you for any help.
pde vectors vector-analysis surfaces surface-integrals
asked Jul 29 at 9:59
Wyuw
1306
add a comment |Â
up vote
1
down vote
favorite
I have an integral surface $z = z(x, y)$.
Writing this integral surface in implicit form, we get
$$F(x, y, z) = z(x, y) - z = 0$$
I am then told that the gradient vector $nabla F = (z_x, z_y, -1)$ is normal to the integral surface $F(x, y, z) = 0$.
First of all, how was this calculated? I understand how the gradient is calculated, but I don't understand how it was calculated in this case?
And lastly, where did the $-1$ come from and why? Couldn't they also have had $nabla F = (z_x, z_y, 1)$, where this would just be the normal vector in the other direction? Why and how did they pick the $-1$ direction instead?
I apologise. My vector calculus understanding is not particularly strong, and I strive to improve it.
Thank you for any help.
pde vectors vector-analysis surfaces surface-integrals
asked Jul 29 at 9:59
Wyuw
1306
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have an integral surface $z = z(x, y)$.
Writing this integral surface in implicit form, we get
$$F(x, y, z) = z(x, y) - z = 0$$
I am then told that the gradient vector $nabla F = (z_x, z_y, -1)$ is normal to the integral surface $F(x, y, z) = 0$.
First of all, how was this calculated? I understand how the gradient is calculated, but I don't understand how it was calculated in this case?
And lastly, where did the $-1$ come from and why? Couldn't they also have had $nabla F = (z_x, z_y, 1)$, where this would just be the normal vector in the other direction? Why and how did they pick the $-1$ direction instead?
I apologise. My vector calculus understanding is not particularly strong, and I strive to improve it.
Thank you for any help.
pde vectors vector-analysis surfaces surface-integrals
asked Jul 29 at 9:59
Wyuw
1306
I have an integral surface $z = z(x, y)$.
Writing this integral surface in implicit form, we get
$$F(x, y, z) = z(x, y) - z = 0$$
I am then told that the gradient vector $nabla F = (z_x, z_y, -1)$ is normal to the integral surface $F(x, y, z) = 0$.
First of all, how was this calculated? I understand how the gradient is calculated, but I don't understand how it was calculated in this case?
And lastly, where did the $-1$ come from and why? Couldn't they also have had $nabla F = (z_x, z_y, 1)$, where this would just be the normal vector in the other direction? Why and how did they pick the $-1$ direction instead?
I apologise. My vector calculus understanding is not particularly strong, and I strive to improve it.
Thank you for any help.
pde vectors vector-analysis surfaces surface-integrals
asked Jul 29 at 9:59
Wyuw
1306
asked Jul 29 at 9:59
Wyuw
1306
asked Jul 29 at 9:59
Wyuw
1306
asked Jul 29 at 9:59
Wyuw
1306
1306
add a comment |Â
add a comment |Â
2 Answers
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up vote
3
down vote
accepted
maybe its helpful to give the surface equation a different symbol
$$ S = (x,y,z) : z=Z(x,y) $$
then with $F(x,y,z):= Z(x,y) - z$,
$$∇ F (x,y,z) = beginpmatrixpartial_x( Z(x,y) - z)\partial_y( Z(x,y) - z)\partial_z( Z(x,y) - z)endpmatrix= beginpmatrixpartial_x Z(x,y)\partial_yZ(x,y)\ - 1endpmatrix$$
The $-1$ came from the definition of $F$. If you forced $+1$ instead of $-1$ you would be talking about $Z(x,y) + z$ which is not related to $S = F = 0 $.
If you take any curve $mathbf x=mathbf x(t)$ such that $mathbf x(t) ∈ S$ for every $t$, then
$$ F(mathbf x(t)) = 0 $$
taking derivatives,
$$∇ F(mathbf x)· mathbf x' = 0$$
but the collection of all such $mathbf x'$s as you consider different curves $mathbf x$ form the tangent vectors at each point of $S$. Therefore, $∇ F$ is perpendicular to all tangent vectors of $S$ at that point. In dimension 3, there are 2 linearly independent tangent vectors and you get a unique normal vector (up to scaling and sign).
answered Jul 29 at 10:19
Calvin Khor
8,01911132
Ahh, you used different variables $z$ and $Z$, so it makes sense now. But the author used $z$ for both, but treated $z(x, y)$ as a function of 2 variables and $z$ as a constant in the same expression, which makes it confusing, because the author also just said that $z = z(x, y)$. Seems paradoxical? As if $z$ is both a function of $x$ and $y$ and a constant!
– Wyuw
Jul 29 at 10:23
1
@Wyuw yes well, this is the way of things unfortunately and you may need to get used to it
– Calvin Khor
Jul 29 at 10:24
Yes, thank you for your help!
– Wyuw
Jul 29 at 10:25
1
@Wyuw You're welcome. good luck :)
– Calvin Khor
Jul 29 at 10:26
add a comment |Â
up vote
1
down vote
Let us start with an example.
$$ z=x^2+y^2$$
$$ F(x,y,z)=x^2+y^2-z$$
$$nabla F = (z_x, z_y, -1)=< 2x,2y,-1>$$
If a point is given, for example $P(1,2,5)$ Then at that point you have two normal vector to the surface.
Upward normal $$< -2x,-2y,1> = <-2,-4,1>$$
Downward normal $$< 2x,2y,-1> = <2,4,-1>$$
answered Jul 29 at 10:11
Mohammad Riazi-Kermani
27.3k41851
Ahh, thank you for this. I understand your example, but in the case of $F(x, y, z) = z(x, y) - z = 0$, we have that $z = z(x, y)$, so it is not a constant? So how can we say that $nabla F = (z_x, z_y, -1)$? That is why I am confused about where the $-1$ comes from? $z$ is still a function of $x$ and $y$, no? But it is being treated as a constant here, and then it is treated as a variable of $x$ and $y$ in the same expression when they calculate $z_x$ and $z_y$!
– Wyuw
Jul 29 at 10:19
1
$z=z(x,y) $is the explicit equation of the surface surface and $F(x,y,z)=z(x,y,)-z =0 $ is the equation of the same surface implicitly. The vector of partial derivatives of $F(x,y,z)$ gets its $-1$ from the $-z$ in $z(x,y,)-z$
– Mohammad Riazi-Kermani
Jul 29 at 10:27
I understand now. Thank you
– Wyuw
Jul 29 at 10:31
1
Thanks for your attention.
– Mohammad Riazi-Kermani
Jul 29 at 10:38
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
maybe its helpful to give the surface equation a different symbol
$$ S = (x,y,z) : z=Z(x,y) $$
then with $F(x,y,z):= Z(x,y) - z$,
$$∇ F (x,y,z) = beginpmatrixpartial_x( Z(x,y) - z)\partial_y( Z(x,y) - z)\partial_z( Z(x,y) - z)endpmatrix= beginpmatrixpartial_x Z(x,y)\partial_yZ(x,y)\ - 1endpmatrix$$
The $-1$ came from the definition of $F$. If you forced $+1$ instead of $-1$ you would be talking about $Z(x,y) + z$ which is not related to $S = F = 0 $.
If you take any curve $mathbf x=mathbf x(t)$ such that $mathbf x(t) ∈ S$ for every $t$, then
$$ F(mathbf x(t)) = 0 $$
taking derivatives,
$$∇ F(mathbf x)· mathbf x' = 0$$
but the collection of all such $mathbf x'$s as you consider different curves $mathbf x$ form the tangent vectors at each point of $S$. Therefore, $∇ F$ is perpendicular to all tangent vectors of $S$ at that point. In dimension 3, there are 2 linearly independent tangent vectors and you get a unique normal vector (up to scaling and sign).
answered Jul 29 at 10:19
Calvin Khor
8,01911132
Ahh, you used different variables $z$ and $Z$, so it makes sense now. But the author used $z$ for both, but treated $z(x, y)$ as a function of 2 variables and $z$ as a constant in the same expression, which makes it confusing, because the author also just said that $z = z(x, y)$. Seems paradoxical? As if $z$ is both a function of $x$ and $y$ and a constant!
– Wyuw
Jul 29 at 10:23
1
@Wyuw yes well, this is the way of things unfortunately and you may need to get used to it
– Calvin Khor
Jul 29 at 10:24
Yes, thank you for your help!
– Wyuw
Jul 29 at 10:25
1
@Wyuw You're welcome. good luck :)
– Calvin Khor
Jul 29 at 10:26
add a comment |Â
up vote
3
down vote
accepted
maybe its helpful to give the surface equation a different symbol
$$ S = (x,y,z) : z=Z(x,y) $$
then with $F(x,y,z):= Z(x,y) - z$,
$$∇ F (x,y,z) = beginpmatrixpartial_x( Z(x,y) - z)\partial_y( Z(x,y) - z)\partial_z( Z(x,y) - z)endpmatrix= beginpmatrixpartial_x Z(x,y)\partial_yZ(x,y)\ - 1endpmatrix$$
The $-1$ came from the definition of $F$. If you forced $+1$ instead of $-1$ you would be talking about $Z(x,y) + z$ which is not related to $S = F = 0 $.
If you take any curve $mathbf x=mathbf x(t)$ such that $mathbf x(t) ∈ S$ for every $t$, then
$$ F(mathbf x(t)) = 0 $$
taking derivatives,
$$∇ F(mathbf x)· mathbf x' = 0$$
but the collection of all such $mathbf x'$s as you consider different curves $mathbf x$ form the tangent vectors at each point of $S$. Therefore, $∇ F$ is perpendicular to all tangent vectors of $S$ at that point. In dimension 3, there are 2 linearly independent tangent vectors and you get a unique normal vector (up to scaling and sign).
answered Jul 29 at 10:19
Calvin Khor
8,01911132
Ahh, you used different variables $z$ and $Z$, so it makes sense now. But the author used $z$ for both, but treated $z(x, y)$ as a function of 2 variables and $z$ as a constant in the same expression, which makes it confusing, because the author also just said that $z = z(x, y)$. Seems paradoxical? As if $z$ is both a function of $x$ and $y$ and a constant!
– Wyuw
Jul 29 at 10:23
1
@Wyuw yes well, this is the way of things unfortunately and you may need to get used to it
– Calvin Khor
Jul 29 at 10:24
Yes, thank you for your help!
– Wyuw
Jul 29 at 10:25
1
@Wyuw You're welcome. good luck :)
– Calvin Khor
Jul 29 at 10:26
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
maybe its helpful to give the surface equation a different symbol
$$ S = (x,y,z) : z=Z(x,y) $$
then with $F(x,y,z):= Z(x,y) - z$,
$$∇ F (x,y,z) = beginpmatrixpartial_x( Z(x,y) - z)\partial_y( Z(x,y) - z)\partial_z( Z(x,y) - z)endpmatrix= beginpmatrixpartial_x Z(x,y)\partial_yZ(x,y)\ - 1endpmatrix$$
The $-1$ came from the definition of $F$. If you forced $+1$ instead of $-1$ you would be talking about $Z(x,y) + z$ which is not related to $S = F = 0 $.
If you take any curve $mathbf x=mathbf x(t)$ such that $mathbf x(t) ∈ S$ for every $t$, then
$$ F(mathbf x(t)) = 0 $$
taking derivatives,
$$∇ F(mathbf x)· mathbf x' = 0$$
but the collection of all such $mathbf x'$s as you consider different curves $mathbf x$ form the tangent vectors at each point of $S$. Therefore, $∇ F$ is perpendicular to all tangent vectors of $S$ at that point. In dimension 3, there are 2 linearly independent tangent vectors and you get a unique normal vector (up to scaling and sign).
answered Jul 29 at 10:19
Calvin Khor
8,01911132
maybe its helpful to give the surface equation a different symbol
$$ S = (x,y,z) : z=Z(x,y) $$
then with $F(x,y,z):= Z(x,y) - z$,
$$∇ F (x,y,z) = beginpmatrixpartial_x( Z(x,y) - z)\partial_y( Z(x,y) - z)\partial_z( Z(x,y) - z)endpmatrix= beginpmatrixpartial_x Z(x,y)\partial_yZ(x,y)\ - 1endpmatrix$$
The $-1$ came from the definition of $F$. If you forced $+1$ instead of $-1$ you would be talking about $Z(x,y) + z$ which is not related to $S = F = 0 $.
If you take any curve $mathbf x=mathbf x(t)$ such that $mathbf x(t) ∈ S$ for every $t$, then
$$ F(mathbf x(t)) = 0 $$
taking derivatives,
$$∇ F(mathbf x)· mathbf x' = 0$$
but the collection of all such $mathbf x'$s as you consider different curves $mathbf x$ form the tangent vectors at each point of $S$. Therefore, $∇ F$ is perpendicular to all tangent vectors of $S$ at that point. In dimension 3, there are 2 linearly independent tangent vectors and you get a unique normal vector (up to scaling and sign).
answered Jul 29 at 10:19
Calvin Khor
8,01911132
edited Jul 29 at 10:27
answered Jul 29 at 10:19
Calvin Khor
8,01911132
answered Jul 29 at 10:19
Calvin Khor
8,01911132
answered Jul 29 at 10:19
Calvin Khor
8,01911132
8,01911132
Ahh, you used different variables $z$ and $Z$, so it makes sense now. But the author used $z$ for both, but treated $z(x, y)$ as a function of 2 variables and $z$ as a constant in the same expression, which makes it confusing, because the author also just said that $z = z(x, y)$. Seems paradoxical? As if $z$ is both a function of $x$ and $y$ and a constant!
– Wyuw
Jul 29 at 10:23
1
@Wyuw yes well, this is the way of things unfortunately and you may need to get used to it
– Calvin Khor
Jul 29 at 10:24
Yes, thank you for your help!
– Wyuw
Jul 29 at 10:25
1
@Wyuw You're welcome. good luck :)
– Calvin Khor
Jul 29 at 10:26
add a comment |Â
Ahh, you used different variables $z$ and $Z$, so it makes sense now. But the author used $z$ for both, but treated $z(x, y)$ as a function of 2 variables and $z$ as a constant in the same expression, which makes it confusing, because the author also just said that $z = z(x, y)$. Seems paradoxical? As if $z$ is both a function of $x$ and $y$ and a constant!
– Wyuw
Jul 29 at 10:23
1
@Wyuw yes well, this is the way of things unfortunately and you may need to get used to it
– Calvin Khor
Jul 29 at 10:24
Yes, thank you for your help!
– Wyuw
Jul 29 at 10:25
1
@Wyuw You're welcome. good luck :)
– Calvin Khor
Jul 29 at 10:26
Ahh, you used different variables $z$ and $Z$, so it makes sense now. But the author used $z$ for both, but treated $z(x, y)$ as a function of 2 variables and $z$ as a constant in the same expression, which makes it confusing, because the author also just said that $z = z(x, y)$. Seems paradoxical? As if $z$ is both a function of $x$ and $y$ and a constant!
– Wyuw
Jul 29 at 10:23
Ahh, you used different variables $z$ and $Z$, so it makes sense now. But the author used $z$ for both, but treated $z(x, y)$ as a function of 2 variables and $z$ as a constant in the same expression, which makes it confusing, because the author also just said that $z = z(x, y)$. Seems paradoxical? As if $z$ is both a function of $x$ and $y$ and a constant!
– Wyuw
Jul 29 at 10:23
1
1
@Wyuw yes well, this is the way of things unfortunately and you may need to get used to it
– Calvin Khor
Jul 29 at 10:24
@Wyuw yes well, this is the way of things unfortunately and you may need to get used to it
– Calvin Khor
Jul 29 at 10:24
Yes, thank you for your help!
– Wyuw
Jul 29 at 10:25
Yes, thank you for your help!
– Wyuw
Jul 29 at 10:25
1
1
@Wyuw You're welcome. good luck :)
– Calvin Khor
Jul 29 at 10:26
@Wyuw You're welcome. good luck :)
– Calvin Khor
Jul 29 at 10:26
add a comment |Â
up vote
1
down vote
Let us start with an example.
$$ z=x^2+y^2$$
$$ F(x,y,z)=x^2+y^2-z$$
$$nabla F = (z_x, z_y, -1)=< 2x,2y,-1>$$
If a point is given, for example $P(1,2,5)$ Then at that point you have two normal vector to the surface.
Upward normal $$< -2x,-2y,1> = <-2,-4,1>$$
Downward normal $$< 2x,2y,-1> = <2,4,-1>$$
answered Jul 29 at 10:11
Mohammad Riazi-Kermani
27.3k41851
Ahh, thank you for this. I understand your example, but in the case of $F(x, y, z) = z(x, y) - z = 0$, we have that $z = z(x, y)$, so it is not a constant? So how can we say that $nabla F = (z_x, z_y, -1)$? That is why I am confused about where the $-1$ comes from? $z$ is still a function of $x$ and $y$, no? But it is being treated as a constant here, and then it is treated as a variable of $x$ and $y$ in the same expression when they calculate $z_x$ and $z_y$!
– Wyuw
Jul 29 at 10:19
1
$z=z(x,y) $is the explicit equation of the surface surface and $F(x,y,z)=z(x,y,)-z =0 $ is the equation of the same surface implicitly. The vector of partial derivatives of $F(x,y,z)$ gets its $-1$ from the $-z$ in $z(x,y,)-z$
– Mohammad Riazi-Kermani
Jul 29 at 10:27
I understand now. Thank you
– Wyuw
Jul 29 at 10:31
1
Thanks for your attention.
– Mohammad Riazi-Kermani
Jul 29 at 10:38
add a comment |Â
up vote
1
down vote
Let us start with an example.
$$ z=x^2+y^2$$
$$ F(x,y,z)=x^2+y^2-z$$
$$nabla F = (z_x, z_y, -1)=< 2x,2y,-1>$$
If a point is given, for example $P(1,2,5)$ Then at that point you have two normal vector to the surface.
Upward normal $$< -2x,-2y,1> = <-2,-4,1>$$
Downward normal $$< 2x,2y,-1> = <2,4,-1>$$
answered Jul 29 at 10:11
Mohammad Riazi-Kermani
27.3k41851
Ahh, thank you for this. I understand your example, but in the case of $F(x, y, z) = z(x, y) - z = 0$, we have that $z = z(x, y)$, so it is not a constant? So how can we say that $nabla F = (z_x, z_y, -1)$? That is why I am confused about where the $-1$ comes from? $z$ is still a function of $x$ and $y$, no? But it is being treated as a constant here, and then it is treated as a variable of $x$ and $y$ in the same expression when they calculate $z_x$ and $z_y$!
– Wyuw
Jul 29 at 10:19
1
$z=z(x,y) $is the explicit equation of the surface surface and $F(x,y,z)=z(x,y,)-z =0 $ is the equation of the same surface implicitly. The vector of partial derivatives of $F(x,y,z)$ gets its $-1$ from the $-z$ in $z(x,y,)-z$
– Mohammad Riazi-Kermani
Jul 29 at 10:27
I understand now. Thank you
– Wyuw
Jul 29 at 10:31
1
Thanks for your attention.
– Mohammad Riazi-Kermani
Jul 29 at 10:38
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let us start with an example.
$$ z=x^2+y^2$$
$$ F(x,y,z)=x^2+y^2-z$$
$$nabla F = (z_x, z_y, -1)=< 2x,2y,-1>$$
If a point is given, for example $P(1,2,5)$ Then at that point you have two normal vector to the surface.
Upward normal $$< -2x,-2y,1> = <-2,-4,1>$$
Downward normal $$< 2x,2y,-1> = <2,4,-1>$$
answered Jul 29 at 10:11
Mohammad Riazi-Kermani
27.3k41851
Let us start with an example.
$$ z=x^2+y^2$$
$$ F(x,y,z)=x^2+y^2-z$$
$$nabla F = (z_x, z_y, -1)=< 2x,2y,-1>$$
If a point is given, for example $P(1,2,5)$ Then at that point you have two normal vector to the surface.
Upward normal $$< -2x,-2y,1> = <-2,-4,1>$$
Downward normal $$< 2x,2y,-1> = <2,4,-1>$$
answered Jul 29 at 10:11
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 10:11
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 10:11
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 10:11
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
Ahh, thank you for this. I understand your example, but in the case of $F(x, y, z) = z(x, y) - z = 0$, we have that $z = z(x, y)$, so it is not a constant? So how can we say that $nabla F = (z_x, z_y, -1)$? That is why I am confused about where the $-1$ comes from? $z$ is still a function of $x$ and $y$, no? But it is being treated as a constant here, and then it is treated as a variable of $x$ and $y$ in the same expression when they calculate $z_x$ and $z_y$!
– Wyuw
Jul 29 at 10:19
1
$z=z(x,y) $is the explicit equation of the surface surface and $F(x,y,z)=z(x,y,)-z =0 $ is the equation of the same surface implicitly. The vector of partial derivatives of $F(x,y,z)$ gets its $-1$ from the $-z$ in $z(x,y,)-z$
– Mohammad Riazi-Kermani
Jul 29 at 10:27
I understand now. Thank you
– Wyuw
Jul 29 at 10:31
1
Thanks for your attention.
– Mohammad Riazi-Kermani
Jul 29 at 10:38
add a comment |Â
Ahh, thank you for this. I understand your example, but in the case of $F(x, y, z) = z(x, y) - z = 0$, we have that $z = z(x, y)$, so it is not a constant? So how can we say that $nabla F = (z_x, z_y, -1)$? That is why I am confused about where the $-1$ comes from? $z$ is still a function of $x$ and $y$, no? But it is being treated as a constant here, and then it is treated as a variable of $x$ and $y$ in the same expression when they calculate $z_x$ and $z_y$!
– Wyuw
Jul 29 at 10:19
1
$z=z(x,y) $is the explicit equation of the surface surface and $F(x,y,z)=z(x,y,)-z =0 $ is the equation of the same surface implicitly. The vector of partial derivatives of $F(x,y,z)$ gets its $-1$ from the $-z$ in $z(x,y,)-z$
– Mohammad Riazi-Kermani
Jul 29 at 10:27
I understand now. Thank you
– Wyuw
Jul 29 at 10:31
1
Thanks for your attention.
– Mohammad Riazi-Kermani
Jul 29 at 10:38
Ahh, thank you for this. I understand your example, but in the case of $F(x, y, z) = z(x, y) - z = 0$, we have that $z = z(x, y)$, so it is not a constant? So how can we say that $nabla F = (z_x, z_y, -1)$? That is why I am confused about where the $-1$ comes from? $z$ is still a function of $x$ and $y$, no? But it is being treated as a constant here, and then it is treated as a variable of $x$ and $y$ in the same expression when they calculate $z_x$ and $z_y$!
– Wyuw
Jul 29 at 10:19
Ahh, thank you for this. I understand your example, but in the case of $F(x, y, z) = z(x, y) - z = 0$, we have that $z = z(x, y)$, so it is not a constant? So how can we say that $nabla F = (z_x, z_y, -1)$? That is why I am confused about where the $-1$ comes from? $z$ is still a function of $x$ and $y$, no? But it is being treated as a constant here, and then it is treated as a variable of $x$ and $y$ in the same expression when they calculate $z_x$ and $z_y$!
– Wyuw
Jul 29 at 10:19
1
1
$z=z(x,y) $is the explicit equation of the surface surface and $F(x,y,z)=z(x,y,)-z =0 $ is the equation of the same surface implicitly. The vector of partial derivatives of $F(x,y,z)$ gets its $-1$ from the $-z$ in $z(x,y,)-z$
– Mohammad Riazi-Kermani
Jul 29 at 10:27
$z=z(x,y) $is the explicit equation of the surface surface and $F(x,y,z)=z(x,y,)-z =0 $ is the equation of the same surface implicitly. The vector of partial derivatives of $F(x,y,z)$ gets its $-1$ from the $-z$ in $z(x,y,)-z$
– Mohammad Riazi-Kermani
Jul 29 at 10:27
I understand now. Thank you
– Wyuw
Jul 29 at 10:31
I understand now. Thank you
– Wyuw
Jul 29 at 10:31
1
1
Thanks for your attention.
– Mohammad Riazi-Kermani
Jul 29 at 10:38
Thanks for your attention.
– Mohammad Riazi-Kermani
Jul 29 at 10:38
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