Question

Determine the point of the parabola y=x² closest to (6,3).

Answer 1

the distance from f(x) to the point is given by
\(D = \sqrt{(6-x)^2-(3-y)^2}\)

we need to minimize this, but really we can just minimize the stuff under the square root, because that would do the same thing

do we minimize \((6-x)^2-(3-y)^2\) you with me?

Answer 2

yes

Answer 3

if we replace y with x^2 we get

\((6-x)^2-(3-x^2)^2=27 - 12 x + 7 x^2 - x^4\)

Answer 4

take the derivative we get
\(-12+14x-4x^3\)
and we set that equal to 0

Answer 5

gr that should be a plut

Answer 6

\(D = \sqrt{(6-x)^2+(3-y)^2}\)
minnimize
\(D =(6-x)^2+(3-y)^2=45 - 12 x - 5 x^2 + x^4\)

Answer 7

the derivaitve of that is \(2 (-6-5 x+2 x^3)\) set equal to 0 we get x = 2

Answer 8

I did this:\[D(x,y)= \sqrt{(x-6)²+(y-3)²}\] so, if \[D(x,y) \ge D(xo,yo)\],
\[D²(x,y)\ge D²(xo,yo)\] . Once y=x²,
\[D²(x,y)=(x-6)²+(x²-3)²\] The derivative of D² gives 4x³-10x-12 and I equaled it to 0. now I can't go any further...I found x=2, but the answer is (1,1). I don't know if it's correct =S

Answer 9

the distance from (2,4) to (3,6) is much closer than the distance from (1,1) to (3,6)

Answer 10

lol, never mind. really stupid. lol Thanks @zzr0ck3r