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Circle Inscribed in a Parabola?

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the largest circle centered on the positive y-axis which touches the origin and which is above y=x^2

    2. Relevant equations

    equation of a circle: r^2=(x-a)^2+(y-b)^2
    equation of a circle centered on the y-axis: x^2+(y-b)^2=r^2
    equation of a parabola: y=x^2

    3. The attempt at a solution

    For the life of me, I couldn't figure out how to do this problem. I "eyeballed" the graphs and realized that a circle with a radius of 0.5 perfectly fits the parameters required by the problem statement, but I have no idea how to do this mathematically.

    A radius of 0.5 yields the equation:
    x^2+(y-0.5)^2=0.5^2
    x^2+y^2-y+0.25=0.25
    x^2+y^2-y=0

    I've tried (rather aimlessly) taking the derivative of this function, yielding:
    2x+2yy'-1y'=0

    However, I realize that I should use the generic equation for a circle, which gives me:
    x^2+(y-b)^2=r^2
    x^2+y^2-yb+b^2=r^2

    As I'm sure you all can tell, I'm not making much progress as to an actual solution to this problem.
    I was hoping that you guys could give me some direction and a hint or two.

    Any help would be appreciated! Thanks!
     
  2. jcsd
  3. Mar 12, 2014 #2

    jbunniii

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    So far so good. You are given one additional constraint: the circle must touch the origin. What does that tell you about how ##r## and ##b## are related to each other?

    OK, can you write down an inequality that describes the region above the parabola?
     
  4. Mar 12, 2014 #3

    Dick

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    Another way to think about it is that you are looking for the 'osculating circle' to y=x^2 at (0,0). It will have the same curvature as y=x^2 at the origin. y'' must be the same for the two curves there.
     
  5. Mar 14, 2014 #4

    utkarshakash

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    I have some other kind of approach which uses the concept of projectiles in physics (I know it sounds weird but it works :approve:). Here goes the solution:

    Let us assume the parabola to be y= -x^2. I simply did this because trajectory of a projectile is always a concave down parabola and also because changing the equation won't have any effect on the original question. Now lets say the projectile is fired with a velocity of v at an angle Θ from horizontal.
    [itex] x= v cos \theta t \\
    y = v sin \theta t - gt^2 / 2 [/itex]

    If you eliminate x and y from the equations, you get the eqn of trajectory which is

    [itex] y = x tan \theta - \dfrac{g sec^2 \theta}{2 v^2} x^2 [/itex]

    But the original trajectory is y = -x^2. Comparing the two equations

    tanΘ = 0 and [itex]\dfrac{g sec^2 \theta}{2 v^2} = 1[/itex]

    The radius of curvature is given by r= v^2 /g.

    Solving the above eqns you can see that r indeed comes out to be 1/2!

    But I'd also like to point out that this has no physical significance as Θ can never be zero in a projectile and if it is zero, then it would be wrong to call it a projectile motion. But mathematically I don't see any problem with this approach. Now, it's upto you whether you want to go with this method or not.

    Cheers. :smile:
     
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