Which point on the graph of #y= sqrt(x)# is closest to the point #(7,0)#?

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Answer 1 · 2018-04-05T13:11:23

At #(13/2, sqrt(13/2))#

We can express the curve #y = sqrt(x)# parametrically as:

#f(t) = (t^2, t)#

where #t in [0, oo)#

The distance between #f(t)# and #(7, 0)# is:

#d(t) = sqrt((t^2-7)^2+(t-0)^2)#

#color(white)(d(t)) = sqrt(t^4-13t^2+49)#

So:

#d/(dt) d(t)^2 = 4t^3-26t = 2t(2t^2-13)#

This has a positive zero when #t = sqrt(13/2)#

Then:

#f(sqrt(13/2)) = (13/2, sqrt(13/2))#

graph{(y-sqrt(x))((x-7)^2+y^2-0.01)((x-13/2)^2+(y-sqrt(13/2))^2-0.01) = 0 [-7.08, 12.92, -3.44, 6.56]}