Which point on the graph of #y= sqrt(x)# is closest to the point #(7,0)#?
1 Answer
Apr 5, 2018
At
Explanation:
We can express the curve
#f(t) = (t^2, t)#
where
The distance between
#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
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]}