Minimum Distance from Point to Curve?

Question

Find the minimum distance from the point (3,2) to the curve y = x² + 2.

I know how to find the distance from a point to a line. The distance from the point to the curve should be the same as the distance from the point to the tangent of the curve at the nearest point on the curve. But since I don’t know either the nearest point on the curve or the slope of the curve at that point, I am not sure how to proceed.

Answer 1

Ok what is the distance, from (3,2) to(x, y)?

its distance = sqrt ( (3-x)^2 + (2-y)^2 ) right?

So since y = x² + 2, you can substitute this into the above so that you have a function of only x, no y in it

d = sqrt ( (3-x)^2 + (2 - x² - 2)^2 )

Now, how do you minimize d? I assume you know how to do differential calculus because you should for this question

the first derivative is 1/2*(-6+2*x+4*x^3)/sqrt(9 - 6*x + x^2+ x^4) you set it to 0 and solve for x, You get 3 roots, one of which is real, the other are complex. Discard the complex roots, and take x = 1, which will satisfy the derivative = 0

Substitute x = 1 into d = sqrt ( (3-x)^2 + (2 - x² - 2)^2 ),

and you get d = sqrt(5) or 2.236067977

in fact x = 1, y = 3, is the point on the curve closest to the point (3, 2)

Answer 2

d( (3,2) , (x,y) ) = sqrt( (3-x)^2 + (2-y)^2)
we find the minimum of (3-x)^2 + (2-y)^2 when y = x^2 +2:
f(x) = (3-x)^2 + (2-y)^2 = 9 - 6x + x^2 + 4 -4y +y^2
= 9 - 6x + x^2 + 4 -4(x^2+2) + (x^2+2)^2
= 5 -3x^2 -6x + (x^2+2)^2.
f'(x) = -6x -6 + 2( x^2+2) (2x)
= -6x -6 + 4x^3 +8x
= 2x +4x^3-6 =0
2x^3 +x -3 = (x-1)(2x^2 +2X +3) =0 only when x=1, since it is a polynomial of degree 4, and it has to have a minimum, then x=1is the minimum,
so the closes point in the curve y=x^2 +2 to the point (3,2) is the point: (1, 3) .

Answer 3

x = 1, y = 3

Answer 4

Distance to any point: r=sqrt((x-3)^2 +(y-2)^2); minimal r is the very distance you need. Assume z=r^2 =(x-3)^2 +(y-2)^2, then deriving by x: z’ =2(x-3) +2(y-2)y’ =0;
Or z’=2x-6 +2(xx+2 –2) (2x) =0, z’= 2xxx +x-3=0; by the way z’’ =6xx+1 >0 means that z’ has only one real root x1 and z(x1) is a minimum of z; thence x1 = 1, r(x1) = 2.236;