Question

Find the minimum distance between the curves y=-x^2 and y=(6-x)^2

Answer 1

WOOT MY FAVE TOPIC! OPTIMIZATION!

Answer 2

let me refresh my memory real quick.

Answer 3

so glad you can help i've been stuck for like 2 hours.

Answer 4

what was the distance formula again? sorry haven't done this in almost 2 years

Answer 5

nevermind ill get one of my review books

Answer 6

k workin on it

Answer 7

sorry the distance formula is \[D= \sqrt{(x1-x2) ^{2}+(y1-y2)^{2}}\]

Answer 8

okay well how far did u get in this problem? I've never done problems with curves. Only distance to the point. Do u have anoy worked out examples?

Answer 9

I mean the problems i used to solve were find the minimum from curve to a point

Answer 10

Right that's where I'm having the problem
so I've created a formula by subtracting the two curves from eachother... hold on i'll put all my work on here, i thought i had gotten to an answer, but my answer didn't make sense so now i'm back at square 1

Answer 11

k

Answer 12

Let D(x) = | [(6 – x)^2] - [-x^2] | be the
function representing the Distance between those two given functions.

Both of the given functions are continuous.
So D(x) is continuous.
If D(x) is 0 at some point, then that’s our minimal distance.

Otherwise, D(x) is never zero.
And so it would be always strictly positive or strictly negative for any x.

Can it ever be true that those two curves cross, making D(x) = 0?
We’d like to solve –x^2 = (6-x)^2.
This gives –x^2 = 36 – 12x + x^2.
Or x^2 – 6x + 18 = 0.
This yields only complex roots, and so the curves never cross for any x.

Therefore, either D(x) is always negative, or it is always positive.
(It can’t go between being positive and negative or else it would have to be zero at some point.)

Let’s pick a test point, some easy point to evaluate D(x).
Like D(0). D(0) = 36.

So D(x) will always be positive.
Moreover, we want to minimize D(x), and we can drop the absolute value bars now.

[If D(x) was always negative, we would then want to maximize D(x) to find the minimal distance (I don’t know if that makes sense to you).]

So minimize it using calculus. Find the critical points of D’(x).
Then test to see whether those critical points are minima
(x = c is a minimum if f’(c) = 0 (or x = c is a cusp) and
f’(x) < 0 for x slightly less than c and f’(x) > 0 for x slightly bigger than c).

Then choose the smallest minimum. (I think there’s only one minimum in this case.)

Answer 13

I found my minimum to be x=3
but then i didn't know if that three needed to be plugged into the original equations, and then the distance found between those two points. That didn't make sense becuase the points were (3,9) and (3,-9) and the distance from that is way far

Answer 14

im getting x = 3 too

Answer 15

Does that thought process make sense though?
What does the 3 mean... Do I just find D(3)? 18 doesn't seem like the right answer

Answer 16

Look at the graphs of those 2 equations and you'll see what why it doesn't make sense

Answer 17

k brb lemme find my calculator

Answer 18

the minimum is 1.5

Answer 19

wait no

Answer 20

how did you get that answer?

Answer 21

nevermind i just looked at the graph and my first guess was 1.5, but then if u move up along one and down the other u'll see that there are maybe closer point

Answer 22

what class is this?

Answer 23

multivariable calc

Answer 24

my teacher is ridiculous. we've never been over anything like this... and i have no idea what i am doing.

Answer 25

im checkin the stuff, i think i gotta get my other textbook, i took calc 3 last semester.. worst class ever.

Answer 26

it's been better then linear algebra if you ask me... just wish i had a better teacher

Answer 27

i actually have to go to a meeting, and i'll be back in like an hour... but i really really appreciate any help that you could throw out there for me

Answer 28

sorry to peace like this

Answer 29

okay where can i send the stuff to?

Answer 30

you could put it on here, or email me at vballninja@gmail.com

Answer 31

k

Answer 32

thanks so much!!!!!

Answer 33

np, ill do my best

Answer 34

i found some stuff, looking thru it

Answer 35

we were differentiatin it wrong

Answer 36

how are we supposed to differentiate it?

Answer 37

http://in.answers.yahoo.com/question/index?qid=20100218090950AAGvf4z
see there it's like we need first to find derivative with respect to x1 then x2

Answer 38

hey can u go on twiddla? it'll be easier to write there i think

Answer 39

http://www.twiddla.com/494934

Answer 40

hey jessica click on the link above, im waitin for u on twiddla

Answer 41

ill wait here too just in case

Answer 42

sorry i'm in the meeting now so i'm like trying to be discrete

Answer 43

oh okay, fine then just wait till its over, i got plenty of time right now