domingo, 8 de enero de 2017

1421. Distance between two parabolae

    Pepe Chapuzas had drawn two parabolae that didn't cut each other on the blackboard. One was convex and the other concave... It was a challenge...
    Dear Teacher:
    What's the distance between the parabolae if their equations are


f(x)  =  x2/3 + 47/3
g(x)  =  − x2/6 + 10x/3

    Susi Fogata did the calculations...

    Dear Teacher:
    These parabolae don't cut each other because the equation


 x2/3 + 47/3  =  − x2/6 + 10x/3
or better
3x2  20x + 94  =  0


hasn't any solutions because its discriminant

202  4·3·94  =  − 728  <  0

    If  S (s, s2/3 + 47/3)  is a point of the parabola  f  and  T (t, t2/6 + 10t/3)  is a point of the parabola  g , and are the nearest ones, then the slopes of  f  and  g  at  S  and  T  are equal...  

 f ' (s)  =  g ' (t)
2s/3  =  − t/3 + 10/3
t  =  10 − 2s

    On the other hand, the vector  TS  is perpendicular to  f  at  S ...

(s − t, f(s) − g(t)) · (1, f ' (s))  =  0
(s − t, s2/3 + 47/3 + t2/6 − 10t/3) · (1, 2s/3)  =  0
(s − (10−2s), s2/3 + 47/3 + (10−2s)2/6 − 10(10−2s)/3) · (1, 2s/3)  =  0
(3s − 10, s2/3 + 47/3 + 50/3 − 20s/3 + 2s2/3 − 100/3 + 20s/3) · (1, 2s/3)  =  0
(3s − 10, s2  1) ·  (1, 2s/3)  =  0
3s − 10 + 2s3/3  2s/3  =  0
 2s3 + 7s  30  =  0
    s = 2  is the only solution because  42  4·2·15  =  − 104  <  0 .
    So, we have the points  S (2, 17)  and  T (6, 14)  and the vector  TS = (4, 3) , hence the distance is  |TS| = √(16+9) = √25 = 5 .

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