Question

Find the point (x,y) on the graph of y=√x nearest the point (4,0)
Can you also please explain "why" you do each step and not just "how" to solve the problem? Thank you!

Answer 1

The distance formula is:

\[d = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}\]

But minimizing d^2 will minimize what's inside the square root, so we can rewrite that as

\[d^2 = (x1-x2)^2 + (y1 - y2)^2\]

plug in your values

\[d^2 = (4 - x)^2 + (0 - \sqrt{x})^2\]

which gives us

\[d^2 = (4-x)^2 + x\]

Expand

\[d^2 = 16 - 8x + x^2 + x\]
which becomes
\[d^2 = x^2 - 7x + 16\]

Take the derivative of both sides

\[2*d*d/dx = 2x - 7\]

But since you want the shortest possible distance, d/dx = 0 which indicates a minimum

So you have

\[0 = 2x-7\]

So

x = 7/2

Plug that back into your equation for y to get the point.