Question

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer 1

Question:

Find the point (,) on the curve
`y = \sqrt x` that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

`(x,y) = ((5)/(2),(√(10))/(2)})`

Explanation:

`y = \sqrt x` can be represented as:
`(x,y)`

Substitute
`\sqrt x` for
`y`

`(x,y) = (x,\sqrt x)`

So, next:

Calculate the distance between
`(x,\sqrt x)` and
`(3,0)`

Distance is calculated as:

`d = √((x_1-x_2)^2 + (y_1 - y_2)^2)`

So:

`d = √((x-3)^2 + (\sqrt x - 0)^2)`

`d = √((x-3)^2 + (\sqrt x)^2)`

Evaluate all exponents

`d = √(x^2 - 6x +9 + x)`

Rewrite as:

`d = √(x^2 + x- 6x +9 )`

`d = √(x^2 - 5x +9 )`

Differentiate using chain rule:

Let

`u = x^2 - 5x +9`

`(du)/(dx) = 2x - 5`

So:

`d = \sqrt u`

`d = u^(1)/(2)`

`(dd)/(du) = (1)/(2)u^{-(1)/(2)}`

Chain Rule:

`d' = (du)/(dx) * (dd)/(du)`

`d' = (2x-5) * (1)/(2)u^{-(1)/(2)}`

`d' = (2x - 5) * \frac{1}{2u^{(1)/(2)}}`

`d' = (2x - 5)/(2\sqrt u)`

Substitute:
`u = x^2 - 5x +9`

`d' = (2x - 5)/(2√(x^2 - 5x + 9))`

Next, is to minimize (by equating d' to 0)

`(2x - 5)/(2√(x^2 - 5x + 9)) = 0`

Cross Multiply

`2x - 5 = 0`

Solve for x

`2x =5`

`x = (5)/(2)`

Substitute
`x = (5)/(2)` in
`y = \sqrt x`

`y = \sqrt{(5)/(2)}`

Split

`y = (\sqrt 5)/(\sqrt 2)`

Rationalize

`y = (\sqrt 5)/(\sqrt 2) * (\sqrt 2)/(\sqrt 2)`

`y = \frac{\sqrt {10}}{\sqrt 4}`

`y = \frac{\sqrt {10}}{2}`

Hence:

`(x,y) = ((5)/(2),(√(10))/(2)})`