2 Answers | Add Yours
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).
We’ve answered 301,108 questions. We can answer yours, too.Ask a question