Graph y = x
3/3 - x
2/2
Imagine a small circle sitting inside the local minimum. Gravity pulls in the negative y direction so it is quite stable there.
Now increase the size of the circle. At some point it will become too large to fit in this hollow and will be forced to roll away down into the third quadrant.
At what radius does this happen?
The circle is going to be of the form
(y-r)^2+x^2=r^2
substitute in y=x^3/3-x^2/2 and solve for r
r=(4x^4-12x^3+9x^2+36)/(24x-36)
Now the desired value of r will be when there is a single solution for x which will happen at the local minimum of the above equation which can be solved for by looking for where the derivative is zero
Doing that gives the value of r as
r=(1/4)*(1+(228-37*sqrt(37))^(1/3)+(228+37*sqrt(37))^(1/3))
Which agrees with Charlies approximate solution
|
|
Posted by Daniel
on 2018-08-15 18:55:58 |
Exact Answer
