jessica
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Did you understand what I wrote a couple hours ago?
jessica
  • jessica
yes, but i got my min as x=3 which doesn't make sense
jessica
  • jessica
if you plug that into the original equations you get the points (3,9) and (3,-9)

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jessica
  • 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
  • anonymous
The distance between their critical points is 6? What points are their critical points?
jessica
  • jessica
just by looking at the graph one is at the orgin and the other is around (6,0)
anonymous
  • 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
  • jessica
not sure i understand... so x=3 is correct, and the distance is 18?
anonymous
  • 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
  • anonymous
Yeah, x = 3 with distance 18 if you measure distance pointwise.
jessica
  • 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
  • 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
  • anonymous
Yeah, I think you would use multivariable calculus and Lagrange multipliers.
anonymous
  • 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
  • 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
  • anonymous
There's symmetry around the point (0, 3)? Do you mean (3, 0)?
jessica
  • jessica
yes i do, sorry my bad
anonymous
  • 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
  • jessica
but will that give you a minimum distance with the same x value, or different x values on each curve?
bahrom7893
  • bahrom7893
Hey jessica u still here?
jessica
  • jessica
yes
bahrom7893
  • bahrom7893
k i found something for u
bahrom7893
  • 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
  • jessica
oh thank you so much !!!!!!! :) stil gotta read it all, but this is so helpful
bahrom7893
  • 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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