Gradient and tangent plane/normal line

[fishturtle1]

Homework Statement

Use gradients to find an equation of the tangent plane to the ellipsoid $\frac {x^2}{4} + \frac {y^2}{9} + \frac {z^2}{25} = 3$ at $P = (2, -3, -5)$.

Homework Equations

$\triangledown f$ is a normal vector of f.

The Attempt at a Solution

Let $w = \frac {x^2}{4} + \frac {y^2}{9} + \frac {z^2}{25}$ be the level curve of f at w.

then $w(2, -3, -5) = \frac {4}{4} + \frac {9}{9} + \frac {25}{25} = 3$.

So P is on the level curve w = 3. I can't figure out why we have to do this, I"m just doing it to follow the example. Isn't this arbitrary?

After that step, we solve for $\triangledown f(P)$.

$\triangledown f(P) = <\frac x2, \frac {2y}{9}, \frac {2z}{25} >$
$\triangledown f(2, -3, -5) = <1, -\frac {2}{3}, -\frac {2}{5}>$

So the equation of the tangent plane is
$<1, -\frac {2}{3}, -\frac {2}{5}> \cdot <x - 2, y + 3, z + 5> = 0$
$(x - 2) + -\frac {2}{3}(y + 3) + -\frac {2}{5}(z + 5) = 0$

I'm even more confused to find the normal line at this P. Would it just be $\vec {OP} + \triangledown f(P)$ since $\triangledown f(P)$ is a normal vector to f?

I'm having A LOT of trouble visualizing what is going on here.. please help, thank you.

 

[FactChecker]

Homework Statement

Use gradients to find an equation of the tangent plane to the ellipsoid $\frac {x^2}{4} + \frac {y^2}{9} + \frac {z^2}{25} = 3$ at $P = (2, -3, -5)$.

Homework Equations

$\triangledown f$ is a normal vector of f.

The Attempt at a Solution

Let $w = \frac {x^2}{4} + \frac {y^2}{9} + \frac {z^2}{25}$ be the level curve of f at w.

then $w(2, -3, -5) = \frac {4}{4} + \frac {9}{9} + \frac {25}{25} = 3$.

So P is on the level curve w = 3. I can't figure out why we have to do this, I"m just doing it to follow the example. Isn't this arbitrary?

Good point. That just verifies that P is on the ellipsoid. If you already know that the point is on the surface in question, then this doesn't need to be done.

After that step, we solve for $\triangledown f(P)$.

$\triangledown f(P) = <\frac x2, \frac {2y}{9}, \frac {2z}{25} >$

You shouldn't have 'P' here because P is one particular point and the right hand side is a general formula of (x,y,z)

$\triangledown f(2, -3, -5) = <1, -\frac {2}{3}, -\frac {2}{5}>$

Correct. This is the normal vector at point P

So the equation of the tangent plane is
$<1, -\frac {2}{3}, -\frac {2}{5}> \cdot <x - 2, y + 3, z + 5> = 0$

Correct. This gives you the equation of tangents at point P.

$(x - 2) + -\frac {2}{3}(y + 3) + -\frac {2}{5}(z + 5) = 0$

This simplifies the answer

I'm even more confused to find the normal line at this P. Would it just be $\vec {OP} + \triangledown f(P)$ since $\triangledown f(P)$ is a normal vector to f?

I think that would be a vector from the origin to a point in (X,Y,Z). It's not a normal line. But I don't think the normal line is required for this problem.

 

[fishturtle1]

Good point. That just verifies that P is on the ellipsoid. If you already know that the point is on the surface in question, then this doesn't need to be done.

What really is confusing me is why P has to be on the ellipsoid. I know that since P is on the ellipsoid, the $\triangledown w(P)$ represents the rate of change at P. But if P weren't on the ellipsoid, what would $\triangledown w(P)$ represent?

You shouldn't have 'P' here because P is one particular point and the right hand side is a general formula of (x,y,z)

Got it

I think that would be a vector from the origin to a point in (X,Y,Z). It's not a normal line.

Wouldn't it be though? From the textbook: "$\triangledown w(P)$ is normal to the level curve $w = f(x, y, z) = f(P)$ at $P$ since $\vec r(t) = < x(t), y(t), z(t) >$ is an arbitrary curve through P that lies on the level surface $f(x, y, z) = f(P)$

So $\vec {OP}$ takes us from the origin to P. Then we add $\triangledown w(P)$ which is the normal line. If we didn't have $\vec {OP}$, $<x, y> = \triangledown w(P)$ would represent the normal line at P, but it would start at the origin.
What am I doing wrong here?

 

[FactChecker]

What really is confusing me is why P has to be on the ellipsoid.

You want the tangent plane that goes through a point on the ellipsoid, so you need to specify a point, P, on the ellipsoid.

I know that since P is on the ellipsoid, the $\triangledown w(P)$ represents the rate of change at P. But if P weren't on the ellipsoid, what would $\triangledown w(P)$ represent?

Good question. I don't know. I guess it would be the same thing for an ellipsoid with the same x,y,z formulas on the left hand side but a different constant value on the right.

Wouldn't it be though? From the textbook: "$\triangledown w(P)$ is normal to the level curve $w = f(x, y, z) = f(P)$ at $P$ since $\vec r(t) = < x(t), y(t), z(t) >$ is an arbitrary curve through P that lies on the level surface $f(x, y, z) = f(P)$

So $\vec {OP}$ takes us from the origin to P. Then we add $\triangledown w(P)$ which is the normal line. If we didn't have $\vec {OP}$, $<x, y> = \triangledown w(P)$ would represent the normal line at P,

But it is not an equation of a line. It is a specific fixed point (or vector) with constant values for coordinates (x,y,z). You would need to use the position vector, P, and the direction vector, $\triangledown w(P)$, to make the equation of a line through P in that direction. The way to do that would be to paramaterize the direction vector with t ∈ R to get the line l(t) = P + t $\triangledown w(P)$. From that, you can fill in the (x,y,z) constant values of P and $\triangledown w(P)$ to get the parameterized coordinates of l(t) = (x(t), y(t), z(t)).

 

[Metmann]

I guess it would be the same thing for an ellipsoid with the same x,y,z formulas on the left hand side but a different constant value on the right.

Yes, that's right.

 

[Ray Vickson]

What really is confusing me is why P has to be on the ellipsoid. I know that since P is on the ellipsoid, the $\triangledown w(P)$ represents the rate of change at P. But if P weren't on the ellipsoid, what would $\triangledown w(P)$ represent?Got it
Wouldn't it be though? From the textbook: "$\triangledown w(P)$ is normal to the level curve $w = f(x, y, z) = f(P)$ at $P$ since $\vec r(t) = < x(t), y(t), z(t) >$ is an arbitrary curve through P that lies on the level surface $f(x, y, z) = f(P)$

So $\vec {OP}$ takes us from the origin to P. Then we add $\triangledown w(P)$ which is the normal line. If we didn't have $\vec {OP}$, $<x, y> = \triangledown w(P)$ would represent the normal line at P, but it would start at the origin.
What am I doing wrong here?

You take $w=3$ to get the ellipsoid. But $\nabla w$ is normal to the surface, so points away from the ellipsoid. $\nabla w$ tells you how $w$ would change if you allowed yourself to go to a point near $(2,-3,-5)$, possibly to a new point not on the ellipsoid. If you go from ${\mathbf x}_0 = \langle 2 -3,-5 \rangle$ to ${\mathbf x}_1 = \langle 2+h_x,-3+h_y, -5 + h_z \rangle$ the change in $w$ would be
$$ \Delta w = w({\mathbf x}_1) - w({\mathbf x}_0) = \nabla w({\mathbf x}_0) \cdot \langle h_x,h_y,h_z \rangle$$
to first order in small $h_x,h_y,h_z$. For a given step length $h = \sqrt{h_x^2 + h_y^2 + h_z^2}$, $\Delta w$ is maximized by taking $\langle h_x,h_y,h_z \rangle$ along the direction of the gradient $\nabla w$.

The gradient vector IS the normal vector; it starts at the point $(2,-3,-5)$ and points away from the surface, so, yes indeed, the new point is at $(2,-3, -3) + (h_x,h_y,h_z)$ in the original coordinate system (that is, relative to the origin (0,0,0)).

 

[fishturtle1]

You want the tangent plane that goes through a point on the ellipsoid, so you need to specify a point, P, on the ellipsoid.Good question. I don't know. I guess it would be the same thing for an ellipsoid with the same x,y,z formulas on the left hand side but a different constant value on the right.But it is not an equation of a line. It is a specific fixed point (or vector) with constant values for coordinates (x,y,z). You would need to use the position vector, P, and the direction vector, $\triangledown w(P)$, to make the equation of a line through P in that direction. The way to do that would be to paramaterize the direction vector with t ∈ R to get the line l(t) = P + t $\triangledown w(P)$. From that, you can fill in the (x,y,z) constant values of P and $\triangledown w(P)$ to get the parameterized coordinates of l(t) = (x(t), y(t), z(t)).

Sorry for the late reply, but I think I understand your post now..

I understand now that P + $\triangledown w(P)$ is just a point on the plane and you need the $t$ to make it a line because for every real number t, you get another point on the line..

So my answer to the vector equation for the normal line at P would be:
normal line $l: <x, y> = \vec {OP} + \triangledown f(P)t, t\epsilon\mathbb{R}$
$= <2, -3, -5> + <1, -\frac 23, -\frac 25> t$

Or $l(t) = <2 + t, -3 - \frac 23 t, -5 -\frac 25 t>$

...still thinking on post #6

 

[Mark44]

Sorry for the late reply, but I think I understand your post now..

I understand now that P + $\triangledown w(P)$ is just a point on the plane and you need the $t$ to make it a line because for every real number t, you get another point on the line..

So my answer to the vector equation for the normal line at P would be:
normal line $l: <x, y> = \vec {OP} + \triangledown f(P)t, t\epsilon\mathbb{R}$
$= <2, -3, -5> + <1, -\frac 23, -\frac 25> t$

Or $l(t) = <2 + t, -3 - \frac 23 t, -5 -\frac 25 t>$

...still thinking on post #6

The question asks for the equation of the tangent plane. For this, you need a vector that is normal to the tangent plane and a point on the plane. You don't need a normal line. The vector form of the tangent plane equation is going to need two vectors in that plane, plus a point in the plane.

 

[fishturtle1]

The question asks for the equation of the tangent plane. For this, you need a vector that is normal to the tangent plane and a point on the plane. You don't need a normal line. The vector form of the tangent plane equation is going to need two vectors in that plane, plus a point in the plane.

I probably should have been a clearer and said "Another one of my questions is how to find a normal line on the surface at P" in the OP. It was actually the next question and I was trying to be slick...
My answer to what is the tangent plane is this(I also had this in the OP):

tangent plane: $<1, -\frac 23, -\frac 25> \cdot <x-2, y+3, z+5> = 0$
$(x - 2) -\frac 23 (y + 3) -\frac 25 (z+5) = 0$

 

[FactChecker]

I probably should have been a clearer and said "Another one of my questions is how to find a normal line on the surface at P" in the OP. It was actually the next question and I was trying to be slick...
My answer to what is the tangent plane is this(I also had this in the OP):

tangent plane: $<1, -\frac 23, -\frac 25> \cdot <x-2, y+3, z+5> = 0$
$(x - 2) -\frac 23 (y + 3) -\frac 25 (z+5) = 0$

Good. That's what I thought because it seemed like you already had the answer and then continued with this question.