Tangent Plane to Level Surfaces Equation Derivation
$begingroup$
I was going through an resource I found online
http://mathonline.wikidot.com/tangent-planes-to-level-surfaces
In this part:
Are they implying that $r'(t_0)$ equals the vector $(x-x_0,y-y_0,z-z_0)$? If so, how did they get that?
They have $r(t)$ equal the vector $(x(t),y(t),z(t))$, are they saying that the derivative of $r(t)$ is $(x-x(t),y-y(t),z-z(t))$ so that $r'(t_0)$ is $(x-x_0,y-y_0,z-z_0)$? I'm just rambling but I don't understand how they were able to transition from $r'(t)$ to $(x-x_0,y-y_0,z-z_0)$ in the picture above, or, just how they got $r'(t)$ in the first place...
calculus multivariable-calculus vectors partial-derivative
asked Jan 3 at 5:24
PurpleBlueJeansPurpleBlueJeans
604
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add a comment |
$begingroup$
I was going through an resource I found online
http://mathonline.wikidot.com/tangent-planes-to-level-surfaces
In this part:
Are they implying that $r'(t_0)$ equals the vector $(x-x_0,y-y_0,z-z_0)$? If so, how did they get that?
They have $r(t)$ equal the vector $(x(t),y(t),z(t))$, are they saying that the derivative of $r(t)$ is $(x-x(t),y-y(t),z-z(t))$ so that $r'(t_0)$ is $(x-x_0,y-y_0,z-z_0)$? I'm just rambling but I don't understand how they were able to transition from $r'(t)$ to $(x-x_0,y-y_0,z-z_0)$ in the picture above, or, just how they got $r'(t)$ in the first place...
calculus multivariable-calculus vectors partial-derivative
asked Jan 3 at 5:24
PurpleBlueJeansPurpleBlueJeans
604
$endgroup$
add a comment |
$begingroup$
I was going through an resource I found online
http://mathonline.wikidot.com/tangent-planes-to-level-surfaces
In this part:
Are they implying that $r'(t_0)$ equals the vector $(x-x_0,y-y_0,z-z_0)$? If so, how did they get that?
They have $r(t)$ equal the vector $(x(t),y(t),z(t))$, are they saying that the derivative of $r(t)$ is $(x-x(t),y-y(t),z-z(t))$ so that $r'(t_0)$ is $(x-x_0,y-y_0,z-z_0)$? I'm just rambling but I don't understand how they were able to transition from $r'(t)$ to $(x-x_0,y-y_0,z-z_0)$ in the picture above, or, just how they got $r'(t)$ in the first place...
calculus multivariable-calculus vectors partial-derivative
asked Jan 3 at 5:24
PurpleBlueJeansPurpleBlueJeans
604
$endgroup$
I was going through an resource I found online
http://mathonline.wikidot.com/tangent-planes-to-level-surfaces
In this part:
Are they implying that $r'(t_0)$ equals the vector $(x-x_0,y-y_0,z-z_0)$? If so, how did they get that?
They have $r(t)$ equal the vector $(x(t),y(t),z(t))$, are they saying that the derivative of $r(t)$ is $(x-x(t),y-y(t),z-z(t))$ so that $r'(t_0)$ is $(x-x_0,y-y_0,z-z_0)$? I'm just rambling but I don't understand how they were able to transition from $r'(t)$ to $(x-x_0,y-y_0,z-z_0)$ in the picture above, or, just how they got $r'(t)$ in the first place...
calculus multivariable-calculus vectors partial-derivative
calculus multivariable-calculus vectors partial-derivative
asked Jan 3 at 5:24
PurpleBlueJeansPurpleBlueJeans
604
asked Jan 3 at 5:24
PurpleBlueJeansPurpleBlueJeans
604
asked Jan 3 at 5:24
PurpleBlueJeansPurpleBlueJeans
604
asked Jan 3 at 5:24
PurpleBlueJeansPurpleBlueJeans
604
asked Jan 3 at 5:24
PurpleBlueJeansPurpleBlueJeans
604
604
add a comment |
add a comment |
1 Answer
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$begingroup$
They are not implying that $r^prime(t_0)$ is equal to $(x-x_0, y-y_0, z-z_0)$.
Instead, their reasoning follows two steps:
First, they prove that for any curve on the surface going through a point $P(x_0, y_0, z_0)$, the gradient $nabla f$ at that point is orthogonal to the curve. That's because $$nabla f (x_0, y_0, z_0).r^prime(t_0)=0$$
Because it's true for any curve on the surface going through $P$, they conclude that $nabla f (x_0, y_0, z_0)$ must be orthogonal to the tangent plane at that point. This, in turns, implies that the equation of the tangent plane is
$$nabla f (x_0, y_0, z_0).(x-x_0, y-y_0, z-z_0)=0$$ That's the definition of a plane given by its normal and an intercept (point $P$).
answered Jan 3 at 6:08
Stefan LafonStefan Lafon
1,38416
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$begingroup$
They are not implying that $r^prime(t_0)$ is equal to $(x-x_0, y-y_0, z-z_0)$.
Instead, their reasoning follows two steps:
First, they prove that for any curve on the surface going through a point $P(x_0, y_0, z_0)$, the gradient $nabla f$ at that point is orthogonal to the curve. That's because $$nabla f (x_0, y_0, z_0).r^prime(t_0)=0$$
Because it's true for any curve on the surface going through $P$, they conclude that $nabla f (x_0, y_0, z_0)$ must be orthogonal to the tangent plane at that point. This, in turns, implies that the equation of the tangent plane is
$$nabla f (x_0, y_0, z_0).(x-x_0, y-y_0, z-z_0)=0$$ That's the definition of a plane given by its normal and an intercept (point $P$).
answered Jan 3 at 6:08
Stefan LafonStefan Lafon
1,38416
$endgroup$
add a comment |
$begingroup$
They are not implying that $r^prime(t_0)$ is equal to $(x-x_0, y-y_0, z-z_0)$.
Instead, their reasoning follows two steps:
First, they prove that for any curve on the surface going through a point $P(x_0, y_0, z_0)$, the gradient $nabla f$ at that point is orthogonal to the curve. That's because $$nabla f (x_0, y_0, z_0).r^prime(t_0)=0$$
Because it's true for any curve on the surface going through $P$, they conclude that $nabla f (x_0, y_0, z_0)$ must be orthogonal to the tangent plane at that point. This, in turns, implies that the equation of the tangent plane is
$$nabla f (x_0, y_0, z_0).(x-x_0, y-y_0, z-z_0)=0$$ That's the definition of a plane given by its normal and an intercept (point $P$).
answered Jan 3 at 6:08
Stefan LafonStefan Lafon
1,38416
$endgroup$
add a comment |
$begingroup$
They are not implying that $r^prime(t_0)$ is equal to $(x-x_0, y-y_0, z-z_0)$.
Instead, their reasoning follows two steps:
First, they prove that for any curve on the surface going through a point $P(x_0, y_0, z_0)$, the gradient $nabla f$ at that point is orthogonal to the curve. That's because $$nabla f (x_0, y_0, z_0).r^prime(t_0)=0$$
Because it's true for any curve on the surface going through $P$, they conclude that $nabla f (x_0, y_0, z_0)$ must be orthogonal to the tangent plane at that point. This, in turns, implies that the equation of the tangent plane is
$$nabla f (x_0, y_0, z_0).(x-x_0, y-y_0, z-z_0)=0$$ That's the definition of a plane given by its normal and an intercept (point $P$).
answered Jan 3 at 6:08
Stefan LafonStefan Lafon
1,38416
$endgroup$
They are not implying that $r^prime(t_0)$ is equal to $(x-x_0, y-y_0, z-z_0)$.
Instead, their reasoning follows two steps:
First, they prove that for any curve on the surface going through a point $P(x_0, y_0, z_0)$, the gradient $nabla f$ at that point is orthogonal to the curve. That's because $$nabla f (x_0, y_0, z_0).r^prime(t_0)=0$$
Because it's true for any curve on the surface going through $P$, they conclude that $nabla f (x_0, y_0, z_0)$ must be orthogonal to the tangent plane at that point. This, in turns, implies that the equation of the tangent plane is
$$nabla f (x_0, y_0, z_0).(x-x_0, y-y_0, z-z_0)=0$$ That's the definition of a plane given by its normal and an intercept (point $P$).
answered Jan 3 at 6:08
Stefan LafonStefan Lafon
1,38416
answered Jan 3 at 6:08
Stefan LafonStefan Lafon
1,38416
answered Jan 3 at 6:08
Stefan LafonStefan Lafon
1,38416
answered Jan 3 at 6:08
Stefan LafonStefan Lafon
1,38416
1,38416
add a comment |
add a comment |
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