Question 1059222: find the point on the curve y= (x-1)^1/2 that is closest to point (2,0)
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! find the point on the curve y= (x-1)^1/2 that is closest to point (2,0)
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Simplify:
find the point on the curve y= x^1/2 that is closest to point (1,0)
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Points on the curve are (x,sqrt(x))
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d^2 = diffy^2 + diffx^2 = (sqrt(x) - 0)^2 + (x-1)^2 = 2x^2 - 2x + 1
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d = sqrt(2x^2 - 2x + 1)
dd/dx = (1/2)*(2x^2 - 2x + 1)^(-1/2)*(4x-2) = 0
2x-1 = 0
x = 1/2
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Point (1/2,sqrt(1/2))
Move it all 1 to the right
--> point is (3/2,sqrt(1/2)
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Distance = sqrt(3)/2, btw
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