find the point on the curve y=sqr.(x) that is the closest to the point (3,0).
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)`.
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).`
Any point on the curve is (x,y) = (x, sqrtx).
The distance d between the points (3,0) and (x, sqrtx) is given by:
d^2 = (3-x)^2+(sqrtx-0)^2.
d^2 = 9 -2sqrtx+x+x
d^2 = 9 - 2sqrtx +2x..(1)
When d is minimum, d^2 = is also minimum.
d^2 to be minimum, (9 - 2sqrtx +2x)' = 0 and (9 - 2sqrtx +2x)'' > 0.
=> -2/(2sqrtx) +2 = 0
=> -x^(-1/2) +1 = 0
=> x = 1.
(9 - 2sqrtx +2x)'' = (-)(-1/2)x^(-3/2) = (1/2)x^(-3/2) = 1 at x= 1.
Put x = 1 in (1), d^2 = (9-2sqrt1+2*1)
d^2 = 9-2+2 = 9
d = 9^(1/2) = 3.
So 3 is the shortest distance between the curve y = sqrtx and the point (3,0).