Shorest distance between a curve to a point?

Question

Find the points on the graph of f(x)=(x^2)+1 that are closest to the point (0,3).

**I do not know how to calculate shortest distance between a curve to a point.

Answer 1

If a pont P is on the graph, then P =(x, x^2 +1) and the distance from P to (0,3) is D = sqrt((x - 0)^2 + (x^2 +1 -3)^2) = sqrt((x^2 + x^4 - 4x^2 + 4) = sqrt(x^4 - 3x^2 + 4).

To find the minimum value of D, a square root, is equivalent to finding the minimum value of P(x) = x^4 - 3x^2 + 4). Since this is a polynomial of even degree with positive leading coefficient, it actually has a minimum.

P'(x) = 4x^3 - 6x = x(4x^2 - 6), so the roots of P' are 0, sqrt(6)/2 and - sqrt(6)/2.

P''()x) = 12x^2 - 6 = 6(2x^2 -1), so that P'' is negative at x= 0 and positive at the other roots of P'. It follows sqrt(6)/2 and - sqrt(6)/2 are local minimuns and since they yields the same value for P, both are global minimums.

It follows there are 2 points on the graph that are shortes to (0.3): (sqrt(6)/2 , f(sqrt(6)/2)) and (-sqrt(6)/2 , f(sqrt(6)/2)))

Answer 2

d = √((x-0)^2 + (y-3)^2)
d = √(x^2 + y^2 - 6y +9)
d = √(x^2 + (x^2 + 1)^2 - 6(x^2 +1) +9)
d = √(x^2 + x^4 + 2x^2+1 - 6x^2 -6 +9)
d = √( x^4-3x^2 + 4)
dd/dx = 1/[2*√( x^4-3x^2 + 4)]*(4x^3 - 6x)
To minimize let dd/dt = 0
1/[2*√( x^4-3x^2 + 4)]*(4x^3 - 6x) = 0
x = 0 or x^2 = 3/2
You have three possible points:
(0,1) ( √6/2, f(√6/2)) and ( - √6/2, f(-√6/2))
To finish the problem we need to substitute to find y for the second two points then use the distance formula to find d. Select the smallest of these three optimal values, where the derivative is zero.
If you are lucky, the minimum distance will be from (1.0). more likely, looking at the graph, the best point is:
( √6/2, f(√6/2))
Good Luck!

Answer 3

I would say the shortest distance from the graph y=x^2 + 1 to the point (0,3) is sqrt 2, because that would be the distance from the point (1,2) and (-1,2) on the graph y=x^2 +1, the closest next to that would be a distance of 2 from the point (0,1)

Answer 4

The distance from a point on the curve x,y is sq rt((x-0)^2 + (y-3)^2)) ......................Pythags theorem.

As y =x^2 + 1 this = sq rt(x^2 +(x^2-2)^2)

Which = sq rt (x^2 +x^4 - 4x^2 +4) = sq rt (x^4 - 3x^2 +4)

Differentiate this function and set it = 0 to find turning point which will almost certainly be minimum. This will give you value for x. You can then calculate y.