Find Largest Dropping Circle in Parabola & x^4 Graphs

[Saint Medici]

Here's an interesting problem (interesting to me, at least) that my professor gave me last year (outside of class...it had nearly nothing to do with the subject we were studying). It's of two parts:

The first part is fairly simple. Suppose you have the graph f(x)=x^2. What is the radius of the largest circle that you can "drop" into the parabola so that one point at the bottom of the circle touches the bottom tip of the parabola. So, said again for redundancy's sake, what is the radius of the largest circle that can fit inside a standard parabola and still touch the bottom.

The second part took me more time, but that's because I kept going down the same dead-end road over and over again. Basically, it's the same as the above, except using f(x)=x^4 instead of the standard parabola. So, once again for redundancy's sake, what is the radius of the largest circle that can fit inside the graph f(x)=x^4 and still touch the bottom-most point. If I remember correctly, this one will touch at more than one point.

I'm curious to see how many ways there are to do this. I know of two ways already: the method that I used and the method my professor used. Anyway, I'd be very interested to see what you guys come up with.

 

[AKG]

My approach would be to maximize r such that:

\forall x \in [-r,r],\ r - \sqrt{r^2 - x^2} \geq x^4

r - x^4 \geq \sqrt{r^2 - x^2}

r^2 - 2rx^4 + x^8 \geq r^2 - x^2

-2rx^4 \geq -x^2 - x^8

2rx^2 \leq 1 + x^6

2r \leq x^4 + x^{-2}

Now, to maximize r, 2r should be equal to x^4 + x^{-2} when x^4 + x^{-2} reaches it's minimum. Now, realize that the steps above included dividing by x², which means that it doesn't include the case when x is zero. of course, we know that the circle touches the graph at x=0, so we don't really care about that point. Now, to find the minimum value for the right side on the interval (0,r], it can be done simply by finding where it's derivative is zero:

4x^3 -2x^{-3} = 0

2x^3 = x^{-3}

2x^6 = 1

x = (1/2)^{1/6}

At this point, the right side expression is:

[(1/2)^{1/6}]^4 + [(1/2)^{1/6}]^{-2}

= (1/2)^{2/3} + 2^{1/3}

This is to be equal to 2r, so:

r = (1/2)^{5/3} + (1/2)^{2/3}

That's your final answer.

 

[wisredz]

Hi there,
I'm new here and found this forum while searching for a solution for the first one. not the harder one. I'm a high school student who studies calculus, and I came across this problem while I was studying the limit in edward and penny's book. Now I finished the derivative from thomas' calculus(but didn't finish the aplications of derivative, so have no idea about maximum-minimum problems). I still can't solve the one that involves y=x^2. Could someone please help me about this?

 

[wisredz]

noone? please help, i do not understand the first part of your solution where

\forall x \in [-r,r],\ r - \sqrt{r^2 - x^2} \geq x^4

Could you please help?

 

[PhilG]

noone? please help, i do not understand the first part of your solution where

\forall x \in [-r,r],\ r - \sqrt{r^2 - x^2} \geq x^4

Could you please help?

This is for the second problem involving y = x^4. That inequality says that the lower half of the circle is above the graph of y=x^4, except for points where they are tangent. You can solve the first problem involving y=x^2 without calculus. It is helpful to draw a picture of the circle and parabola. For the circle to have its lowest point at the origin, it will have the equation

x^2 + (y-R)^2 = R^2

Whether or not the circle fits inside the parabola y=x^2 depends on how many points they have in common. They always have (0,0) in common, since we constructed the circle that way. By symmetry about the y-axis, there will always be an odd number of such common points. But there can't be more than 4, since substituting x^2 in for y in the equation of the circle gives a 4th degree equation. So the circle and parabola can have either 1 or 3 points in common. If there are 3 points, the two points other than (0,0) have the same y-coordinate, which is positive. Further, they must be intersections, not tangency points. If they were points of tangency, you could make the circle slightly smaller or bigger to get 5 total common points. Therefore, the only way that the circle fits inside the parabola is if they have only the origin in common. Putting x^2 = y in the equation for the circle, you can find the condition on R so that this is satisfied.

 

[wisredz]

Okay, now I understand the whole solution. Thanks a lot, I hadn't realized that was the lower half of the circle...