jessica
  • jessica
Find the minimum distance between the curves y=-x^2 and y=(6-x)^2
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
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.)
jessica
  • jessica
So I found that there is a minimum at x=3, but that doesn't mean that the distance is 3 right? it means that the minimum between the curves occurs at x=3. So I plug it in to each curve and get the points (3,9) and (3,-9) and from there find the distance between those two points... is that right?
jessica
  • jessica
That doesn't feel right.

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