Question

Help me with this please.

Find a point on the curve y=x^2 that is closest to the point (0,2).

Answer 1

ok just the distance formula, or rather its square

Answer 2

you have a point (0,2) and a point on the curve (x,x^2)
and the distance between two is
\[D(x)=(x-0)^2+(x^2-2)^2\]

Answer 3

ok that is not really the distance, because we didn't use the square root, so it is the square of the distance. you do get to use calculus right?

Answer 4

help me also

Answer 5

\[D(x)=x^4-3x^2+2\] if my algebra is correct, (it is early) and you want the minimum. take the derivative, set = 0 and solve
\[D'(x)=4x^3-6x\]
put
\[4x^3-6x=0\]
\[2x(2x-3)=0\] so choices are
\[x=0\] or
\[x=\frac{3}{2}\]

Answer 6

clearly 0 works because it is the same point. so i guess you are supposed to say
\[(\frac{3}{2},\frac{9}{4})\]
again it is early so make sure to check my work

Answer 7

i look at this way\[\]the shortest distance is a line from the point that make a perpendicular to the tangent of the curve

Answer 8

good morning. how did i do?

Answer 9

mornin, i see you survived the aftershocks aswell :)

Answer 10

yes and the tsunami

Answer 11

y=x^2; y' = 2x defines our slope; now we need to develop a notion for the slope from the point (0,2) to a given point on the line y = x^2, or rather: (x,x^2)\[\]

Answer 12

are the
lines working

Answer 13

see my qus also

Answer 14

pls help me

Answer 15

(x,x^2)
-(0, 2)
--------
(x, x^2-2); slope =\(\cfrac{x^2-2}{x}\), right?

Answer 16

grrr.....

Answer 17

uhm...I'll process this first, but yeah we're using calculus.:))..Thanks for the help SATELLITE 73..I got disconnected.

And thank you amistre64. I understand now. :)

Answer 18

per slope is flip and negative right?

Answer 19

hmm i am following but getting a different equation than the one i solved.

Answer 20

2x = \(-\cfrac{x}{x^2-2}\), and solve for x is my thought

Answer 21

oh i see my dumb mistake when i was using your method. yea this will do it and get the exact same equation too right?

Answer 22

2 = -1/(x^2-2)

Answer 23

x^2-2=-1/2

Answer 24

\[2x^3-4x=-x\]
\[2x^3-3x=0\] etc

Answer 25

x^2 = 1.5

Answer 26

x= sqrt(1.5); maybe?

Answer 27

ok now we diverge

Answer 28

its prolly my brain math

Answer 29

\[2x=-\frac{x}{x^2-2}\]

\[2x^3-4x=-x\]
\[2x^2-3x=0\]
\[x(2x-3)=0\]
\[x=\frac{3}{2}\]

Answer 30

i divided off the xs first

Answer 31

hmmm well you lose a root that way but lets see what happens, because there is more than one way to skin a moose

Answer 32

x = sqrt(3/2)

Answer 33

now i am thoroughly confused!

Answer 34

\[2x = \frac{-x}{x^2-2}\]
\[2 = \frac{-1}{x^2-2}\]
\[2(x^2-2) = -1\]
\[2x^2-4+1 = 0\]
\[2x^2-3 = 0\]
\[x^2 = \frac{3}{2}\]

Answer 35

i cleared the fraction and got
\[2x^3-4x=-x\] or
\[2x^3-3x=0\] which made me happy because it is the same equation i got form the distance formula

Answer 36

oh wait i see

Answer 37

you cannot divide by x!

Answer 38

becasue x could be zero roght?

Answer 39

my professor's answer was y=3/2 , x=+- (sq rt of 3/2)does this make sense?

Answer 40

http://www.wolframalpha.com/input/?i=2x+%3D+-x%2F%28x^2-2%29

Answer 41

yes it makes sense and i am a moron

Answer 42

i was solving for y and got
\[\frac{3}{2}\]

Answer 43

when you find the points of interest; we plug them in to see which ones work

Answer 44

-sqrt(3/2) aint it since its further away from (0,2) to begin with

Answer 45

ok i see my mistake
\[D'(x)=4x^3-6x\]

Answer 46

\[4x^3-6x=0\]
\[2x(x^2-3)=0\]
\[x=0, x=\pm\sqrt{\frac{3}{2}}\]

Answer 47

i factored incorrectly ! lol

Answer 48

:) factoring is for losers anyways, i say we always just go with the gut

Answer 49

@amistre i still think distance formula is easier, but that is only because it is what i am used to. use the distance formula, take the derivative, set = 0 and solve. but i do like that you can do this without calc at all

Answer 50

of course it helps if you solve correctly!

Answer 51

well, i used calc to find the derivative to compare slope with; but that was all

Answer 52

oh right. so no savings as far as calc goes.

Answer 53

x^2 --> 2x , ow! i think i broke a sweat on that one ;)

Answer 54

@wengzie in my own defense i did make several "it is early" disclaimers. method is right algebra is wrong
@amistre, i differentiated a polynomial as well. it is the hidden machinery that is heavy, not the technique

Answer 55

your method looks svelter tho, mine tends to drag knuckles. I do my math like a do my (......), theres alot of yelling and screaming, and in the end noone is happy

Answer 56

spoken like a true mathematician!

Answer 57

im quite confused now..hahaha...i'll follow ur flow of conversation
..:)))

Answer 58

@AMISTRE64 and @SATELLITE73: You both are awesome!!! Thank you!!! I got it alright. :)