Sphere Intersecting a Paraboloid

[vlad4232]

Homework Statement

Hi, I am trying to solve the following problem, and seem to just be going in circles.
A sphere of radius=4 is "dropped" into a paraboloid with equation z=(x^2)+(y^2).
Find the distance "a" from the origin to the center of the sphere at the point where it will
"get stuck" or stop falling further into the paraboloid.

Homework Equations

sphere radius=4
paraboloid z=(x^2)+(y^2)
What I believe is the eq. for the sphere (x^2)+(y^2)+((z-a)^2)=16

I should add that my teacher has solved it and said that the value for a=16.25 and that trying to solve
the problem assuming that the sphere will stop when the cross section of the paraboloid is equal in diameter to
the diameter of the sphere will yield an incorrect answer (a=16)

The Attempt at a Solution

Well I have tried to solve for z and set them equal to each other, but this yields 1 equation with 3 variables. I have been trying to come up with other equations but thus far have only thought of z=a+4 ==> (x^2)+(y^2)=a+4
Which I don't even think is correct.
I also tried to convert to cylindrical coordinates after solving for z and setting the two equations to each other, getting
0=(r^4)-(r^2)-[2(r^2)a]+(a^2)-16
Which might yield something if i knew what to do with it.
Any suggestions would be very helpful. Thank You

 

[Sourabh N]

First thing you need to realize is the configuration when this is possible and why your assumption "the sphere will stop when the cross section of the paraboloid is equal in diameter to the diameter of the sphere" is not necessary. Once you can imagine this, I believe you can proceed.

I used spherical coordinates for the points of intersection (circle of intersection) of the paraboloid and sphere.

Tell me if any of this helps.

 

[vlad4232]

Right I don't assume that the sphere will stop when the diameters are equal. I understand that it should happen before that. But, are you saying that the method of setting the equation of the sphere and paraboloid equal to each other and then trying to solve for the variable "a" incorrect?

I have converted to spherical coordinates and this is where I am at the moment.
Again 2 Unknowns and only 1 equation.
+or-sqrt[(r*sin(theta))^2+16]-(rsin(theta))^2= a

Am I right to expect an answer in the form of a= (r*sin(theta))^2 ?

Thank you for your help though. This problem is nagging, isn't even for a grade, but I really want to solve it.

 

[vlad4232]

I believe I may have it.
-I substituted z=x^2+y^2 into the eq. for the sphere
yielding z+(z-a)^2-16=0
-then used the quadratic formula to solve for z, with z^2-(2a+1)z+(a^2-16)=0
assuming that a is a constant.
-Now I have replaced the enormous expression for z into z+(z-a)^2-16=0
and I have an equation solely in terms of "a"
-now to solve for "a" i'll have to wait till I can get to Maple because the expression is really huge and contains several sqrts.

On the right track?

 

[Sourabh N]

Looks like you're on track. http://twitpic.com/4euybo/full" a quick pic of the cross section of the set up. The best way AFAIT is to write the parametric equation of sphere and make it satisfy the equation of paraboloid. You will get an equation is terms of a and $$\theta$$. Extremize a and you will get two solutions for a. Now you see that a starts from a large value and is decreasing and so, among the two solutions, the larger one is correct. [Spoiler - The two solutions are what you and what your professor got]