# Math Help - tangent plane of level surface

1. ## tangent plane of level surface

Determine the tangent planes of the level surface given by x^2+y^2+x^2 = 1 at the points (x,y,0) and show that they are all parallel to the z-axis.

here is my working thus far:
f(x,y,z) = x^2+y^2+z^2-1

then the tangent plane equation is:

(∂f/dx)|(x,y,0)(x-x) + (∂f/dy)|(x,y,0)(y-y) + (∂f/dz)|(x,y,0)(z-0) = 0

the first two terms go to zero since (x-x = y-y = 0), and the last term leaves 0z=0 (no information!)

Mathematically, where is my problem?

2. Originally Posted by sk1001
Determine the tangent planes of the level surface given by x^2+y^2+x^2 = 1 at the points (x,y,0) and show that they are all parallel to the z-axis.

here is my working thus far:
f(x,y,z) = x^2+y^2+z^2-1

then the tangent plane equation is:

(∂f/dx)|(x,y,0)(x-x) + (∂f/dy)|(x,y,0)(y-y) + (∂f/dz)|(x,y,0)(z-0) = 0

the first two terms go to zero since (x-x = y-y = 0), and the last term leaves 0z=0 (no information!)

Mathematically, where is my problem?
You need to compute the normal to the surface at $(x,y,0)$. The normal is $\nabla f(x,y,0)$. Therefore, the normal is $\bold{n}=(2x,2y,0)$. Notice that $\bold{n}$ has no $z$ coordinate. Therefore, it just consists of $x,y$ coordinates. Thus, it must be parallel with the $z$-axis.

3. at each point (x0,y0,0)

2x0(x-x0) + 2y0(y-y0) = 0

4. Originally Posted by ThePerfectHacker
You need to compute the normal to the surface at $(x,y,0)$. The normal is $\nabla f(x,y,0)$. Therefore, the normal is $\bold{n}=(2x,2y,0)$. Notice that $\bold{n}$ has no $z$ coordinate. Therefore, it just consists of $x,y$ coordinates. Thus, it must be parallel with the $z$-axis.
the question is asking me for the tangent planes though, so there lies my problem.

As shown in my original post, I get the erroneous answer 0z=0! So how do I fix this?

Also, when you say there is no z-coordinate, I cannot see it being parallel to the z-axis, but rather perpendicular to the x-axis. (Imagine x-axis to the right, y-axis out of the page, and z-axis up. Now the xy plane is perpendicular to the z-axis!)

5. Think of

2x0(x-x0) + 2y0(y-y0) = 0

as a line in the x-y plane that is then extended in the pos and neg z directions.

Your original surface is a sphere- the equator is in the x-y plane

The tangent planes at the equator are vertical planes whose equations are given by:

2x0(x-x0) + 2y0(y-y0) = 0

See attachment

6. Originally Posted by sk1001
the question is asking me for the tangent planes though, so there lies my problem.

As shown in my original post, I get the erroneous answer 0z=0! So how do I fix this?

Also, when you say there is no z-coordinate, I cannot see it being parallel to the z-axis, but rather perpendicular to the x-axis. (Imagine x-axis to the right, y-axis out of the page, and z-axis up. Now the xy plane is perpendicular to the z-axis!)
Remember that to find the equation that describes a plane you need to have a point on a plane and the vector to which the plane is normal (perpendicular) to. When we compute the gradient at (x,y,0) we are finding the normal vector to the tangent plane at that point. The normal vector works out to be (2x,2y,0), notice that this vector has no k component, so it must entirely lie in the xy-plane. Therefore, the tangent plane is perpendicular to the xy-plane. But the z-axis is perpendicular to the xy-plane. Now since the tangent plane and the z-axis are both perpendicular to the xy-plane it means they must be parallel with eachother.

7. ok, I think i'm beginning to understand.
before I was confused by the notation (I'm guessing that P should be more explicitely explained as (x0,y0,0)

Let me explain what I understand so far to see if I'm on the right track..

so, to find my tangent planes, I'm using the tangent plane equation,
z-z0 = (∂f/dx)|(x0,y0,0)(x-x0) + (∂f/dy)|(x0,y0,0)(y-y0)
=> z = 2x0(x-x0) + 2y0(y-y0)

Now to prove these are parallel to the z-axis, I'm going to show that the normal to the tangent plane has no z-component, or if I wanted to go a little bit further, that the normal vector n dot produced with the z-axis vector, is 0 (i.e the normal of the tangent plane is perpendicular to the z-axis, thus the tangent plane itself is parallel to the z-axis). Hopefully that makes sense!

So now..
n = <2x 2y 2z>
evaluated at (x0,y0,0), n = <2x0 2y0 0>
no z-component, so parallel to z-axis.

Extra step...
z-axis vector = k = <0 0 1>

n dot k = <0 0 0> = 0
i.e. normal of tangent plane is perp to z-axis, therefore tangent plane is parallel.

8. Originally Posted by sk1001
So now..
n = <2x 2y 2z>
evaluated at (x0,y0,0), n = <2x0 2y0 0>
no z-component, so parallel to z-axis.
Yes, this is what we have been saying.

But you got the equation of the tangent plane wrong. This is because the tangent plane equation that you used only works when $z=f(x,y)$. But $z$ is given implicitly not explicitly. So be careful. You can solve for z then use tangent plane equation but those steps are not necessary.

9. Originally Posted by ThePerfectHacker
Yes, this is what we have been saying.

But you got the equation of the tangent plane wrong. This is because the tangent plane equation that you used only works when $z=f(x,y)$. But $z$ is given implicitly not explicitly. So be careful. You can solve for z then use tangent plane equation but those steps are not necessary.
You do not think it is safe to assume that z=f(x,y)?
I think it is.

If it wern't it complicates things a hell of a lot more.
Thanks for the help

10. Originally Posted by sk1001
You do not think it is safe to assume that z=f(x,y)?
I think it is.

If it wern't it complicates things a hell of a lot more.
Thanks for the help
We have $x^2 + y^2 + z^2 = 1 \implies z = \pm \sqrt{1-x^2-y^2}$.

11. Originally Posted by ThePerfectHacker
We have $x^2 + y^2 + z^2 = 1 \implies z = \pm \sqrt{1-x^2-y^2}$.
right, so then z = f(x,y) !!

12. Originally Posted by sk1001
right, so then z = f(x,y) !!
Now the function is in proper form. Just be careful this is not really a function because it produces two values $\pm$ instead of just one value. What it is, is a set consisting of two parts of functions (the upper part of sphere and lower part of sphere).

13. ok makes sense, but how will that change my tangent plane equation (you mentioned my tangent plane equation was wrong)?

Does this mean I will need chain rule to solve (∂f/∂z)?

14. Originally Posted by sk1001
ok makes sense, but how will that change my tangent plane equation (you mentioned my tangent plane equation was wrong)?

Does this mean I will need chain rule to solve (∂f/∂z)?
You ned to use the chain rule to find $\tfrac{\partial f}{\partial x}, \tfrac{\partial f}{\partial y}$. The tangent plane is given by $z-z_0 = f_x(x_0,y_0)(x-x_0)+f_y(y-y_0)$. However, this formula would only work when $f$ is differenciable at $(x_0,y_0)$. The function $z=\sqrt{1-x^2-y^2}$ it not differenciable on the set $x^2+y^2 = 1$. Think of it this way. You have the function (in a single variable) $y=\sqrt{x}$. You can draw the tangent line at $x_0=0$, that line has equation $x=0$, however you cannot find the line by using the formula $y-y_0 = y'(x_0)(x-x_0)$ because this formula only works when the function is differenciable at the point. Notice that as $x\to 0+$ for $y=\sqrt{x}$ the slope of tangent line shoots away $+\infty$, so the tangent slope does not exist. Likewise, with $\sqrt{1-x^2-y^2}$ on the circle $x^2+y^2=1$. For all these points as you start drawing the tangent plane it becomes vertical and so the function fails to be differenciable. This is way you need to solve this problem by leaving the surface in the form $x^2+y^2+z^2 = 1$ and applying the gradient to this function as you did in the beginning.

15. I'm confused at this point!

Are you saying I should be using the tangent plane equation:
(∂f/dx)|(x,y,0)(x-x0) + (∂f/dy)|(x0,y0,0)(y-y0) + (∂f/dz)|(x0,y0,0)(z-z0) = 0

If I do this though, ∂f/dz = 2z, then when evaluated at (x0,y0,0), this goes to zero and I am left with:

2x0(x-x0) + 2y0(y-y0) = 0

Are you saying this is my tangent plane equation?

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