Since this problem has already been solved by minimizing the distance function between the given point and the curve, I am going to use a different approach.

The point *P * on the curve `y = sqrt(x)` is closest to the point (3,0) if the line connecting the two points is perpendicular...

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Since this problem has already been solved by minimizing the distance function between the given point and the curve, I am going to use a different approach.

The point *P *on the curve `y = sqrt(x)` is closest to the point (3,0) if the line connecting the two points is perpendicular to the curve (that is, perpendicular to the tangent line to the graph of `y = sqrt(x)` at *P*).

The two lines are perpendicular if their slopes `m_1` and `m_2` are negative reciprocals: `m_1*m_2 = -1`.

The slope of the tangent line to the graph of `y = sqrt(x)` at the point `P` is the value of the derivative *y*'(*x*) at this point:

`y'(x) = (sqrt(x))' = 1/(2sqrt(x))`

At point `P` , `y'(x_P) = -1/(2sqrt(x_p))`. So, `m_1 =- 1/(2sqrt(x_P))`.

The line connecting the points `P ` and (3, 0) has the slope that can be found using the following slope formula:

`m_2 = (0 - y_P)/(3 - x_P) = sqrt(x_P)/(3 - x_P)`.

Plugging these into `m_1*m_2 = -1`, we get the following:

`-1/(2sqrt(x_P)) * sqrt(x_P)/(3 - x_P) = -1`

which simplifies to the following:

`1/(2(3 - x_P)) = 1`

`2(3 - x_P) = 1`

`3 - x_P = 1/2`

`x_P = 3 - 1/2 = 5/2`

So, the *x-*coordinate of the point closest to (3,0) is 5/2, and the *y-*coordinate is as follows:

`y = sqrt(5/2) = sqrt(10)/2`.

**The point on the given curve closest to (3,0) is** `(5/2, sqrt(10)/2)`.

### Videos

This is a minimization problem hence, you need to find a function that models the distance between a point `A(x,y)` that lies on the given curve and the fixed point `(3,0).`

You need to use the distance formula such that:

`d = sqrt((x - 3)^2 + (y - 0)^2)`

You need to write the distance formula in terms of one variable, hence, you need to use the equation of the curve to write y in terms of x such that:

`y = sqrt x => y^2 = x`

Substituting x for `y^2` in equation of distance yields:

`d = sqrt((x - 3)^2 + x) => d = sqrt(x^2 - 6x + 9 + x)`

`d(x) = sqrt(x^2 - 5x + 9)`

You need to find derivative of this function such that:

`d'(x) = ((x^2 - 5x + 9)')/(2sqrt(x^2 - 5x + 9))`

`d'(x) = (2x - 5)/(2sqrt(x^2 - 5x + 9))`

You need to solve the equation `d'(x) = 0` to find the extreme values of the distance function such that:

`(2x - 5)/(2sqrt(x^2 - 5x + 9)) = 0 => 2x - 5 = 0 => 2x = 5= > x = 5/2`

You need to find y coordinate at `x = 5/2` such that:

`y = sqrt x => y = sqrt(5/2) => y = (sqrt10)/2`

**Hence, evaluating the point that lies on the curve `y = sqrt x` and it is closest to the fixed point (3,0) yields`A(5/2 ; (sqrt10)/2).`**

**Further Reading**