# Thread: A Hard One - Ball in a Container

1. ## A Hard One - Ball in a Container

Here's a problem that my teacher specifically assigned to me to make it as hard as possible.

A ball with a radius of 1 is dropped into a parabolic container whose surface is described by rotating y=x^2 around the y-axis. When the ball comes to rest, how far is its bottom from the bottom of the container?

(The image above isn't the exact image that's on the problem, but it's an accurate re-creation.)

The image above is missing point a (which is at the center of the circle and on the y-axis. The slanted line going from point a (the center) to the point labeled (x,y) = 1 because it's a radius.

Okay, so I'm not 100% sure how to start this problem. Is it even possible to get an exact value, or will I have to find the distance between the two bottoms in terms of x?

2. Originally Posted by Lord Darkin
Here's a problem that my teacher specifically assigned to me to make it as hard as possible.

A ball with a radius of 1 is dropped into a parabolic container whose surface is described by rotating y=x^2 around the y-axis. When the ball comes to rest, how far is its bottom from the bottom of the container?

(The image above isn't the exact image that's on the problem, but it's an accurate re-creation.)

The image above is missing point a (which is at the center of the circle and on the y-axis. The slanted line going from point a (the center) to the point labeled (x,y) = 1 because it's a radius.

Okay, so I'm not 100% sure how to start this problem. Is it even possible to get an exact value, or will I have to find the distance between the two bottoms in terms of x?
I would start here:

1. Find the points of intersection.

And yes, you are given plenty of info to solve this problem.

3. How do I find the point of intersection? (I assume you mean the radius and point (x,y) right?) I can't seem to figure out how to get an exact value and I don't know the equation of the circle since I don't know where the center of the circle is.

4. Originally Posted by Lord Darkin
How do I find the point of intersection? (I assume you mean the radius and point (x,y) right?) I can't seem to figure out how to get an exact value and I don't know the equation of the circle since I don't know where the center of the circle is.

Also, remember that it is implied that the center is

$\displaystyle (0,y)$.

Theres a bunch a ways to approach this problem. You could find the length of the line through the points of intersecton, and then find the the height of the triangle that is formed with this line and the center of the circle because the other two sides will have a length of one unit.

Think outside the box.

5. Sorry, I am not fully understanding your last statement. (I will get it eventually though .)

Do you mean find this triangle?

6. Originally Posted by Lord Darkin
Sorry, I am not fully understanding your last statement. (I will get it eventually though .)

Do you mean find this triangle?
Yes! Find the coordinates for the vertex and you're there!

7. Originally Posted by Lord Darkin
Here's a problem that my teacher specifically assigned to me to make it as hard as possible.

A ball with a radius of 1 is dropped into a parabolic container whose surface is described by rotating y=x^2 around the y-axis. When the ball comes to rest, how far is its bottom from the bottom of the container?

(The image above isn't the exact image that's on the problem, but it's an accurate re-creation.)

The image above is missing point a (which is at the center of the circle and on the y-axis. The slanted line going from point a (the center) to the point labeled (x,y) = 1 because it's a radius.

Okay, so I'm not 100% sure how to start this problem. Is it even possible to get an exact value, or will I have to find the distance between the two bottoms in terms of x?
The problem is essentially to find the distance between the circle and the origin. This distance will be k - 1 where k is the y-coordinate of the centre of the circle. This is how I'd do it:

The equation of the circle is $\displaystyle x^2 + (y - k)^2 = 1$. Substitute $\displaystyle y = x^2$ and re-arrange the resulting equation into the form of a quadratic in y being equal to zero. Now get the discriminant of the quadratic and make it equal to zero (why?). Hence solve for k.

8. $\displaystyle x^2 + (y - k)^2 = 1$

$\displaystyle y + (y - k)^2 = 1$

$\displaystyle y + y^2 -2yk + k^2-1 = 0$

$\displaystyle y^2 + y -2yk + k^2 -1 = 0$

I set the quadratic equal to zero.

$\displaystyle y^2 + y -1 = (-k)^2 + 2yk$

Discriminant (I think I did somethign wrong here):

$\displaystyle b^2 - 4ac = (-k)^2 + 2yk$

$\displaystyle 1^2 - 4(1)(-1) = (-k)^2 + 2yk$

$\displaystyle 5 = (-k)^2 + 2yk$

Just wondering, but why do I need the discriminant? It provides information about the nature of the roots but I already know that the root is equal to zero.

9. Originally Posted by Lord Darkin
$\displaystyle x^2 + (y - k)^2 = 1$

$\displaystyle y + (y - k)^2 = 1$

$\displaystyle y + y^2 -2yk + k^2-1 = 0$

$\displaystyle y^2 + y -2yk + k^2 -1 = 0$

I set the quadratic equal to zero.

$\displaystyle y^2 + y -1 = (-k)^2 + 2yk$ Mr F says: Write this in the form a y^2 + by + c = 0. The discriminant is b^2 - 4ac.

Discriminant (I think I did somethign wrong here):

$\displaystyle b^2 - 4ac = (-k)^2 + 2yk$

$\displaystyle 1^2 - 4(1)(-1) = (-k)^2 + 2yk$

$\displaystyle 5 = (-k)^2 + 2yk$

Just wondering, but why do I need the discriminant? It provides information about the nature of the roots but I already know that the root is equal to zero.
..

10. I thought this was set in ay^2 + b y + c = 0 mode? Or do I need to do something about the k's?

y^2 + y -2yk + k^2 -1 = 0

11. Originally Posted by Lord Darkin
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Just wondering, but why do I need the discriminant? It provides information about the nature of the roots but I already know that the root is equal to zero.
The roots of the quadratic are the vertical coordinates of the contact points of the circle and of the parabola. Since both contact points have the same $\displaystyle y$ coordinate, the discriminant will be zero - i.e. both roots of the quadratic will be equal.

In general, the discriminant of a polynomial is zero if and only if the polynomial has a repeated root.

12. Originally Posted by Lord Darkin
I thought this was set in ay^2 + b y + c = 0 mode? Or do I need to do something about the k's?

y^2 + y -2yk + k^2 -1 = 0
=> y^2 + y(1 - 2k) + k^2 - 1 = 0. So a = 1, b = 1 - 2k and c = k^2 - 1. You do not seem to have realised this.

13. Originally Posted by Lord Darkin
I thought this was set in ay^2 + b y + c = 0 mode? Or do I need to do something about the k's?

y^2 + y -2yk + k^2 -1 = 0
Here $\displaystyle y$ is the root of the polynomial and $\displaystyle k$ is just a constant. Writing your equation as $\displaystyle y^2 + y(1-2k) + k^2-1 = 0$, you see that $\displaystyle a=1,\ b=1-2k,\ c=k^2-1$.

You might be a bit confused because usually when solving a quadratic we are solving for its roots. But here the roots are equal to the $\displaystyle y$ coordinate of the contact point between the circle and the parabola - that is not what you are looking for. You are looking for the value of $\displaystyle k$ for which both roots of the polynomial $\displaystyle y^2 + y(1-2k) + k^2-1 = 0$ are equal, and that is why you must set the discriminant of the polynomial equal to 0 and then solve for the value of $\displaystyle k$ which makes this possible.

14. Sorry about that, I should have gotten that sooner.

15. No worries, happens to the best. Good luck!