This curve is the graphical representation of the square root function:

Let A be the point of coordinates (2,0). Find the coordinates of the point B of the curve that is the closest to A.

The distance between two points A(xA,yA) and B(xB,yB​) is given by the formula:

\(\begin{array}{rcl} z &=& \sqrt{ (x_B-x_A)^2 + (y_B-y_A)^2} \qquad x_B =x, \quad y_B=\sqrt{x}, \quad x_A =2, \quad y_A = 0 \\ z &=& \sqrt{ (x-2)^2 + (\sqrt{x}-0)^2} \\ z &=& \sqrt{ (x-2)^2 + \sqrt{x}^2} \\ z &=& \sqrt{ (x-2)^2 + x } \\ z &=& \sqrt{ x^2-4x+4+x } \\ \mathbf{z} & \mathbf{=} & \mathbf{ \sqrt{ x^2-3x+4 } }\\ \end{array}\\ \begin{array}{rcl} \text{The mimimum dictance is, when } z' = 0 \\ \text{We can also use } z^2 = u \text{ and set } u' = 0\\ \end{array}\\ \begin{array}{rcl} u = z^2 &=& x^2-3x+4\\ u' &=& 2x-3 = 0\\ 2x-3&=& 0\\ 2x&=& 3\\ \mathbf{x}&\mathbf{=}& \mathbf{\frac32}\\ \mathbf{y}&\mathbf{=}& \mathbf{\sqrt{\frac32} } \end{array}\\\)

Point  \(B (\frac32, \sqrt{\frac32} )\)  is the closest to A

   \(\text{The nearest distance is: }\\ \begin{array}{rcl} z &=& \sqrt{ x^2-4x+4+x } \\ &=& \sqrt{ (\frac32)^2-3\cdot \frac32+4 } \\ &=& \sqrt{1.75}\\ &=& 1.32287565553 \end{array}\\\)

i cant answer that,and dont know if you answered....

I did, this morning. 

HINT: The distance between two points A(x_A,y_A) and B(x_B,y_B​) is given by the formula:

\(AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\)

I got to the same result. That's good!

20/20

Plus a brownie:

Here's another way to do this without using too much Calculus

Note that.....at ponit B, a line through (2,0) can be constructed that will be perpendicular to the tangent line on y = sqrt(x) at point B.......and the shortest distance bettween a point and a line is a perpendicular line drawn from the point to the given line

Where sqrt(x)  is differentiable, the slope will be 1 /[2sqrt(x)]

So, the slope of a perpedicular line will be -2sqrt(x)

And the equation of such a such a line through (2,0) will be

y = -2sqrt(x)(x - 2)   ......setting this equal to y = sqrt(x), we have

-2sqrt(x)(x - 2)  = sqrt(x)

-2sqrt(x)(x - 2) - sqrt(x)  = 0

sqrt(x) [ -2(x - 2) - 1]  = 0

sqrt(x) [ -2x + 3]  = 0

So......sqrt(x) = 0   and x =  0 [reject]      or    -2x + 3 = 0   and x = 3/2

This is the x coordinate of B..... and .the y coodinate of B  = sqrt(3/2)  

Houston, we've got a problem. I solved the problem using another method and I got another result (I don't remember this result, I think it was 1.25 or something like this); I did this with two of my friends and they got to this result too, using the same method ><

I found the answer my teacher gave me:

A(2,0); let B be the point of coordinates \((x,\sqrt{x}) \\\text{With } x \geq 0\)

The formula to calculate the distance between to points M(x_M,y_M) and N(x_N,y_N) is \(MN=\sqrt{(x_N-x_M)^2+(y_N-y_M)^2}\)

So \(AB=\sqrt{(x-2)^2+(\sqrt{x}-0)^2} \\AB^2=(x-2)^2+\sqrt{x}^2 \\AB^2=x^2-4x+4+x=x^2-3x+4\)

We end up with a quadratic polynomial (the expression has the form ax²+bx+c, here a=1, b=-3 and c=4). Since a=1>0, the polynomial has a minimum. The formula to calculate the extremum E (minimum or maximum) of a quadratic polynomial ax²+bx+c is \(E=\frac{-b}{2a}\)

So the x coordinate of the minimum M of the polynomial (the x coordinate of B) is \(\frac{3}{2}=1.5\)

And the minimum (y coordinate of B) is \(\sqrt{3/2}\)

So the coordinates of B are \((\frac{3}{2}; \sqrt{3/2})\)

Ah, and I'm silly too, I never found a distance AB of 1.25; here's the result I got:

\(AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2} \\AB=\sqrt{(\frac{3}{2}-2)^2+(\sqrt{\frac{3}{2}}-0)^2} \\AB=\sqrt{\frac{1}{4}+\frac{3}{2}}=\sqrt{\frac{7}{4}}\)

So heureka your result for distance seems to be wrong, but I didn't take away some points because I didn't ask for the distance in the question; you gave me the correct coordinates of B, so that's OK.