Find the point on the curve y=√x which is closest to the point (19/2 , 0)

Answer 1

Write d = [(x − 9.5)² + (√x − 0)²]^0.5.

Then d = (x² − 19x + 9.5² + x)^0.5 or d = (x² − 18x + 9.5²)^0.5.

Next, d' = 0.5(2x −18) ÷ (x² − 18x + 9.5²)^0.5 or (x − 9) ÷ (x² − 18x + 9.5²)^0.5.

With d' = 0 at x = 9, (x,√x) or (9,3) is a critical point.

Now obtain d" or [(x² − 18x + 9.5²)^0.5(1) − (x −9)²(x² − 18x + 9.5²)^-0.5] ÷ (x² − 18x + 9.5²); this simplifies to 9.25/(x² − 18x + 9.5²)^1.5 which is greater than 0 at x=9.

The distance d from y = √x to (9.5,0) is then minimum or least at (x,√x) equal to (9,3).

Answer 2

For this you need to minimize the distance, (or the square distance since if the squared distance has a minimum so does the distance), from the point in question to a generic point on the curve. In other words we need to minimize the function (which is the distance squared):

f(x) = (p_x-x)² +(p_y-y)² = (19/2-x)²+(√x)² = x + 19²/4 - 19x + x²

with respect to x. The minimum conditon yields:

∂_xf(x) = 0 = 2 x - 18

In other words x = 9, so the point on the curve wich is closest to the point p = (19/2,0) is:

q = (9,3)

Answer 3

y=sqrx

y'= 1/2sqrx

m= negative inverse = -2sqrx

y=mx + b

y=-2xsqrx + b

0 =-2(9.5)sqr9.5 + b

b = 2(9.5)sqr9.5

y=-2xsqrx + 2(9.5)sqr9.5

sqrx =-2xsqx+ 19sq9.5

2xsqx + sqx = 19sq9.5

sqrx(2x+1) = 19sq9.5

sqrx= 19sq9.5/(2x+1)

x = 361sq9.5/(4x^2 + 4x + 1)

4x^3 +4x^2 +x -3430 = 0

x=about 9.2

y =sqr9.2

point on the curve closest to (19/2,0) is (9.2, sqr9.2) = (9.2,3.03)

Answer 4

y=sqrx

y'= 1/2sqrx

m= negative inverse = -2sqrx

y=mx + b

y=-2xsqrx + b

0 =-2(9.5)sqr9.5 + b

b = 2(9.5)sqr9.5

y=-2xsqrx + 2(9.5)sqr9.5

sqrx =-2xsqx+ 19sq9.5

2xsqx + sqx = 19sq9.5

sqrx(2x+1) = 19sq9.5

sqrx= 19sq9.5/(2x+1)

x = 361sq9.5/(4x^2 + 4x + 1)

4x^3 +4x^2 +x -3430 = 0

x=about 9.2

y =sqr9.2

point on the curve closest to (19/2,0) is (9.2, sqr9.2) = (9.2,3.03)