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=12√x|x=α=12√α and →ut=ddxg(x)=2x
Note that the tangent slopes are parallel iff the tangent slopes are equal
So 12√α=2x⇔4x√α=1⇔16x2α=1⇔α=116x2 (*)
But the slope of PQ is m=yQ−yPxQ−xP=x2+1−√αx−α
Now to have (QP)⊥(C) we need to find the normal slopes
So the normal slope of (C) is 2xm′=−1⇔m′=−12x
Observe that the normal slope of (C) is parallel to (PQ)
Hence m=m′⇔x2+1−√αx−α=−12x⇔x2+1−14xx−116x2+12x=0
⇔4x3+4x−14x16x3−116x2+12x=0⇔16x2(4x3+4x−1)4x(16x3−1)+12x=0
⇔8x2(4x3+4x−1)+16x3−12x(16x3−1)=0
⇔32x5+32x3−8x2+16x3−1=0⇔32x5+48x3−8x2−1=0
Solving to x=0.331695⇔α=116x−2=116(0.331695)−2=1.46797
Hence the minimum distance from P to Q is PQ=√(xQ−xP)2+(yQ−yP)2
=√(1.46797−0.331695)2+(√1.46797−(0.331695)2−1)2=1.14081
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