Find the points on the curve y=x2+2 closest to the point (0,3). Round to the nearest two decimal places. I know this requires the distance formula.
The distance z is z = √(x^2 + (y-3)^2) = √(x^2 + (x^2+2)^2)
= √(x^4 + 5x^2 + 4)
dz/dx = x(2x^2+5)/√(x^4 + 5x^2 + 4)
Clearly, dz/dx = 0 only at x=0.
So, (0,2) is the closest point to (0,3)
Makes sense, since the vertex of the parabola lies on the y-axis.
that answer is not the one I need, I need two that are on either side of the 0
SKETCH A GRAPH !!!!
y = x^2 + 2
a point on that curve has form
( x , x^2+2)
so your distance formula would be
d^2 = [ ( x-0)^2 + (x^2 -1)^2 ]
lets call d^2 = z and minimize z
z = x^2 + x^4 -2 x^2 +1 = x^4 -x^2 +1
min or max when dz/dx = 0
dz/dx = 0 = 4 x^3 - 2x = 2x [ 2x^2 -1 ]
so when x = 0 or when x = +/- 1/sqrt 2
well which?
when x = 0, the distance is 1 from 2 to 3 obviously
what about if x = 1/sqrt 2
y = 1/2 + 2 = 2.5 so from (1/sqrt2 , 2.5) to (0,3)
d^2 = [ (0 -1/sqrt 2)^2 + .5^2 ]
d^2 = 1/2 + 1/4 = 3/4
d = (1/2) sqrt (3) = 0.866
so that is the WINNER
x = +/- 1/sqrt 2 (plus or minus due to symmetry of that parabola about y axis)
y = 2 1/2
Well, use the normal.
The slope of the normal at (h,k) on the curve is -1/(2h)
So, where on the curve does the line
y-k = -1/(2h) (x-h)
go through (0,3) ?
Go with Damon. I made a typo.
Whew :)
Using the normal, you want the line from (0,3) to (x,y) to have slope -1/(2x).
That is,
(3-y)/x = 1/(2x)
(1-x^2)/x = 1/(2x)
1-x^2 = 1/2
x^2 = 1/2
x = ±1√2
as Damon showed using the distance formula.