Find the minimum distance from the point (3,0) to y=(x)^1/2
Make a sketch
Let the point of contact be P(x,y)
here is an easy way to do this, based on the property that the tangent at P must be perpendicular to the line from P to (3.0)
slope of function is (1/2)x^(-1/2) = 1/(2√x)
slope of the line of shortest distance using the grade nine way = y/(x-3) = √x/(x-3)
so using the negative reciprocal of perpendicular line property
√x/(x-3) = - 2√x
1/(x-3) = -2
-2x + 6 = 1
-2x = -5
x = 5/2 and y = √(5/2) or √10/2
shortest distance = √((5/2-3)^2 + (√10/2 - 0)^2)
= √(1/4 + 5/2) = √11/2
check my arithmetic