Calculus IV: points on the ellipsoid x^2 + 2y^2 + 2z^2 = 5

[relle]

The problem I'm having trouble with is as follows:

Find the points on the ellipsoid x^2 + 2y^2 + 2z^2 = 5 where the tangent plane is parallel to the plane x + 2y + 2z = 10.

I know that I have to find the normal to the given plane, then find the generic normals to the ellipsoid, but I am lost from here. I am only used to finding the normal when given a point.

Any direction on this problem will be greatly, GREATLY appreciated!

 

[pka]

The gradient \(\displaystyle \nabla f = 2xi + 4yj + 4zk\) is parallel to the vector \(\displaystyle 2i + 4j + 4k\) at the point \(\displaystyle \left( {1,1,1} \right).\)

 

[relle]

Calculus IV

I am unsure as how you obtained the vector 2i+4j+4k at (1,1,1).

Is the gradient of f, 2xi+4yj+4zk, the normal vector of th ellipsoid since the normal vector is equal to the gradient?

I found the normal vector of the plane x+2y+2z=10 to be <1,2,2>.

Am I missing something?

Thanks

 

[galactus]

If (x,y,z) is on the ellipsoid, then

\(\displaystyle {\nabla}f(x,y,z)=2(xi+2yj+2zk)\) is normal there and hence so is

\(\displaystyle n_{1}=xi+2yj+2zk;\)

\(\displaystyle n_{1}\) must be parallel to \(\displaystyle i+2j+2k\) which is normal

to the given plane so \(\displaystyle n_{1}=cn_{2}\) for some constant c.

Equate corresponding components, then sub into the ellipsoid equation to find c and the points on the ellipsoid.

 

[kris]

What happened to the s in front of n1, and isn't there another possible point where the two are parallel? just thought i'd ask.

 

[stapel]

What happened to the s in front of n1...?

Where are you seeing an "s"?

isn't there another possible point where the two are parallel?

Let's not leave the student "hanging": Please reply showing your steps and your results (the steps and results that raised the question you asked). Thank you.

Eliz.

 

[kris]

sorry, I meant for that to be a 2. I was just following along with those who posted before. What I was thinking was that wouldn't there be two planes parallel to the given plane? One plane at one end of the ellipsoid and one on the other end?

 

[pka]

Yes kris you are correct.
The other point would (-1,-1,-1).
But the students should do some things for themselves.

 

[jay]

I like to read these so I can broaden my knowledge of calculus (and I never know when it could be useful).

In this case, I can follow up to the point where there are two points. I understand why there would be two, but why -1(1,1,1,)? Is it because this point is the exact opposite of (1,1,1) which signifies the end of the ellipsoid?

Pray explain :lol:

 

[pka]

The vector \(\displaystyle i + 2j + 2k\) is the normal to the given plane.
The gradient to the ellipse is \(\displaystyle \nabla f(x,y,z) = 2xi + 4yj + 4zk.\)
Both points (1,1,1) & (-1,-1,-1) are on the ellipse!
Moreover, \(\displaystyle \nabla f(1,1,1) = - \nabla f( - 1, - 1, - 1).\)
That is, at both of the points the gradient is parallel to the normal of the given plane.
That means that the tangent planes to the ellipse at these points will be parallel to the given plane.

 

[kris]

Can anyone explain why the two just disappears? Everything seems kind of straight forward except for this!

 

[pka]

Can anyone explain why the two just disappears? Everything seems kind of straight forward except for this!

Can you tell us what you know about vector anaylsis?
Do you understand what it means to say that two vectors are parallel?
If you do, then you question makes no sense.