Question

What is the shortest possible distance from a point on the parabola y=x^2 to the point (0,1)?

Answer 1

a perpendicular line

Answer 2

and wont it be 2 points?

Answer 3

distance, ok, gotta learn to read

Answer 4

still can use that though right?

Answer 5

(x,x^2)
-(0, 1 )
----------
x,x^2-1

distance = sqrt((x^2)+(x^2-1)^2)

Answer 6

id prooly try to get the point thru calculus and then find the distance

Answer 7

make your life easy and ignore the radical.

Answer 8

dD/dx = 4x^3 - 2x
= 2x (2x^2 - 1) = 0

x = 0, x^2 = 1/2 then

right?

Answer 9

yeah i am getting a different answer

Answer 10

i get
\[x=\pm\frac{\sqrt{2}}{2}\]

Answer 11

same here; or rather the shortest distance is about; .866...

Answer 12

and x = 0 is a local max

Answer 13

sorry 'bout that.

Answer 14

here is an alternate method, not that the other one doesn't work.
you get the slope as
\[\frac{x^2-1}{x}\] and it must be perpendicular to the tangent line at
\[y=x^2\] so you must have
\[\frac{x^2-1}{x}=-\frac{1}{2x}\] solve for x and get
\[x=\pm\frac{\sqrt{2}}{2}\]

Answer 15

\[d=\sqrt{(x-0)^2+(x^2-1)^2}\]
\[d^2=x^2+(x^2-1)^2 \]
\[(d^2)'=2x+2(x^2-1)(2x) =(2x)(1+2(x^2-1))=2x(2x^2-2+1)=2x(2x^2-1)\]
\[=> x=0, x=\frac{1}{\sqrt{2}}, x=\frac{-1}{\sqrt{2}}\]

Answer 16

and we need to check to see they are max or mins

Answer 17

at x=1 the distance is obviously 1 :)

Answer 18

err... at x=0

Answer 19

using my method (he boasts) i don't need to check

Answer 20

Well, D'' = 12x^2 - 2 hence D''(0) = -2 < 0 and hence x = 0 is a local max.

Answer 21

\[f(x)=x^2;\ f'(x)=2x;\ perp.slope=-\frac{1}{2x}\]
\[y = -\frac{1(x)}{2x}+\frac{1(0)}{2x}+1\]
\[y = -\frac{1}{2}+1\]
\[y = \frac{1}{2};\ x^2=\frac{1}{2};\ x=\frac{1}{\sqrt{2}}\]