Question

Find the point P on the graph of the function below:
y=sqrt{x} closest to the point (3,0)

Answer 1

`\begin{cases} (x-3)^(2) + y^(2) = R^(2) \\ y = √(x) \implies y^(2) = x \end{cases} \\ \\ (x-3)^(2) + x = R^(2) \\ \\ x^(2) - 6x + 9 + x - R^(2) = 0 \\ \\ x^(2) - 5x + 9 - R^(2) = 0 \\ \\ \Delta = 0 \\ \\ \Delta = (-5)^(2) - 4 \cdot 1 \cdot (9 - R^(2)) = 25 - 36 + 4R^(2) = 4R^(2) - 11 \\ \\ 4R^(2) - 11 = 0 \\ \\ R^(2) - (11)/(4) = 0 \\ \\ R^(2) = (11)/(4)`

`(x-3)^(2) + x = (11)/(4) \\ \\ x^(2) - 6x + 9 + x - (11)/(4) = 0 \\ \\ x^(2) - 5x + (25)/(4) = 0 \\ \left (x - (5)/(2) \right)^(2) = 0 \\ \\ x = (5)/(2) \\ \\ y = \sqrt{ (5)/(2) } = (√(10))/(2) \\ \\ \boxed{P = \left ( (5)/(2), (√(10))/(2) \right ) }`

Answer 2

Distance between (x,y) and (3,0) is:

D = sqrt{ (x-3)^2 + y^2 }

y = sqrt(x)

y^2 = x

Then substitute D, that would be

D = sqrt{ (x-3)^2 + x } = sqrt(x^2 - 6x + 9 + x}

D = sqrt(x^2 - 5x + 9)

Minimize D by finding the derivative with respect to x, equating it to 0 and solving for x.

D = sqrt(x^2 - 5x + 9)

find derivative and set it to 0 in order to minimize D

D' = 1/2(x^2 - 5x + 9)^(-1/2) * (2x - 5) = 0

D' will be zero when numerator = 0

2x - 5 = 0

x = 5/2

Put x = 5/2 back in y=sqrt{x}

y = sqrt(5/2)

the points of the curve that is nearest to (3,0) will be (5/2, sqrt(5/2))

So the answer will be (2.5, 1.58)