- jessica

i know i've asked this a couple of times already, but i've been working on it for 5 hours and i really need help: Find the minimum distance between the curves y=-x^2 and y=(6-x)^2

- jamiebookeater

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- anonymous

Did you understand what I wrote a couple hours ago?

- jessica

yes, but i got my min as x=3 which doesn't make sense

- jessica

if you plug that into the original equations you get the points (3,9) and (3,-9)

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## More answers

- jessica

the distance between those is 18, which can't be the minimum distance because if you look at the graphs, the distance between the critical points is 6 which is less than 18

- anonymous

The distance between their critical points is 6?
What points are their critical points?

- jessica

just by looking at the graph one is at the orgin and the other is around (6,0)

- anonymous

Oh I see. Yeah that is confusing.
Distance defined on functions is defined pointwise.
So we say that the distance between f(x) and g(x) at point x = c is |f(c) – g(c)|.
So even if there is some other point d such that |f(c) – g(d)| is less than |f(c) – g(c)|,
that’s not the minimum distance because distance.
It’s like if you imagine a vertical line scanning across the graphs.
Then asking when that vertical passes through the two closest points.

- jessica

not sure i understand... so x=3 is correct, and the distance is 18?

- anonymous

That is quite tricky wording.
You're right that the distance between (0, 0) and (6, 0) is 6, which is less than 18,
but that's not the distance between the two functions *at the same value of x*

- anonymous

Yeah, x = 3 with distance 18 if you measure distance pointwise.

- jessica

oh ok, so what we did was find the minimum distance with the same x, but if i can have different values of x.. i'm in a whole different ball park

- jessica

i need to set up y1=-x1^2 and y2=(6-x2)^2 then use the distance formula then take a partial derivative in terms of x1 and x2? is that right

- anonymous

Yeah, I think you would use multivariable calculus and Lagrange multipliers.

- anonymous

Actually, now that you mention it, x = 3 doesn't seem surprising anymore.
The vertices are at (0, 0) and (6, 0) as you said.
And their shapes are their same. And so there's a symmetry around the point (3, 0). So might expect their minimum distance to be at x = 3.

- jessica

um, I understand what you mean by the symmetry thing, so what if i found the minimum distance between the point (0,3) and each of the curves, since there is symmetry would that give the the minimum distance between the two curves?

- anonymous

There's symmetry around the point (0, 3)?
Do you mean (3, 0)?

- jessica

yes i do, sorry my bad

- anonymous

In general, that’s not necessarily true.
Some functions might be periodic or behave wildly.
But for parabolas pointing in opposite directions,
if you can find the point of symmetry in the graph,
then that point will be contained by the line segment which is
the shortest distance between the two points (defined pointwise).
But in general, you’d want to differentiate the
pointwise distance (or difference) function D(x) and find the minimum that way.

- jessica

but will that give you a minimum distance with the same x value, or different x values on each curve?

- bahrom7893

Hey jessica u still here?

- jessica

yes

- bahrom7893

k i found something for u

- bahrom7893

i asked a question on yahoo answers:
http://answers.yahoo.com/question/index?qid=20110224154533AAw62vr
and nevermind the questions description, i wrote it like that on purpose, so that they'd respond faster

- jessica

oh thank you so much !!!!!!! :) stil gotta read it all, but this is so helpful

- bahrom7893

welcome, i find yahoo answers more helpful than this place in higher math, most ppl here are in calc 1 or 2 highest.. i found calc 3 super hard, even though i got 5s in both AP calc ab and bc exams and had a 100 avg in bc calc in hs...

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