Gradient and tangent plane/normal line

  • #1
fishturtle1
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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.
 
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  • #2
fishturtle1 said:

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.
 
  • #3
FactChecker said:
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?

FactChecker said:
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

FactChecker said:
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?
 
  • #4
fishturtle1 said:
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)).
 
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  • #5
FactChecker said:
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.
 
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  • #6
fishturtle1 said:
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)).
 
  • #7
FactChecker said:
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
 
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  • #8
fishturtle1 said:
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.
 
  • #9
Mark44 said:
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##
 
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  • #10
fishturtle1 said:
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.
 
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