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minimum distance between two curves parabola and any function


Exercise:

Find the minimum distance between the curves (C):f(x)=x&(C):g(x)=x2+1

Note that the distance is not horizontal or vertical line ( see the figure )

Solution: Let P be a point on the parabola (C)

So the coordinates of P will be P(α,α)

And Q be a point on the curve (C)

So the coordinate of Q are Q(x,x2+1)

So the slopes of the tangent lines to (C)&(C) are:

vt=ddxf(x)|x=α=12x|x=α=12α and ut=ddxg(x)=2x

Note that the tangent slopes are parallel iff the tangent slopes are equal

So 12α=2x4xα=116x2α=1α=116x2     (*)

But the slope of PQ is m=yQyPxQxP=x2+1αxα

Now to have (QP)(C) we need to find the normal slopes

So the normal slope of (C) is 2xm=1m=12x

Observe that the normal slope of (C) is parallel to (PQ)

Hence m=mx2+1αxα=12xx2+114xx116x2+12x=0

4x3+4x14x16x3116x2+12x=016x2(4x3+4x1)4x(16x31)+12x=0

8x2(4x3+4x1)+16x312x(16x31)=0

32x5+32x38x2+16x31=032x5+48x38x21=0

Solving to x=0.331695α=116x2=116(0.331695)2=1.46797

Hence the minimum distance from P to Q is PQ=(xQxP)2+(yQyP)2

=(1.467970.331695)2+(1.46797(0.331695)21)2=1.14081

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