## jessica 5 years ago Find the minimum distance between the curves y=-x^2 and y=(6-x)^2

1. bahrom7893

WOOT MY FAVE TOPIC! OPTIMIZATION!

2. bahrom7893

let me refresh my memory real quick.

3. Jessica

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

4. bahrom7893

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

5. bahrom7893

nevermind ill get one of my review books

6. bahrom7893

k workin on it

7. Jessica

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

8. bahrom7893

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?

9. bahrom7893

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

10. Jessica

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

11. bahrom7893

k

12. Jessica

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.)

13. Jessica

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

14. bahrom7893

im getting x = 3 too

15. Jessica

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

16. Jessica

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

17. bahrom7893

k brb lemme find my calculator

18. bahrom7893

the minimum is 1.5

19. bahrom7893

wait no

20. Jessica

how did you get that answer?

21. bahrom7893

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

22. bahrom7893

what class is this?

23. Jessica

multivariable calc

24. Jessica

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

25. bahrom7893

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

26. Jessica

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

27. Jessica

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

28. Jessica

sorry to peace like this

29. bahrom7893

okay where can i send the stuff to?

30. Jessica

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

31. bahrom7893

k

32. Jessica

thanks so much!!!!!

33. bahrom7893

np, ill do my best

34. bahrom7893

i found some stuff, looking thru it

35. bahrom7893

we were differentiatin it wrong

36. Jessica

how are we supposed to differentiate it?

37. bahrom7893

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

38. bahrom7893

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

39. bahrom7893
40. bahrom7893

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

41. bahrom7893

ill wait here too just in case

42. Jessica

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

43. bahrom7893

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