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Homework Help: Minimum veritcal distance between two graphs.

  1. Sep 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Calculate the minimum vertical distance between the cubic y=x^3-x^2-4x+4 and the parabola y=-2x^2+16x-30, when x is positive.

    2. Relevant equations
    y=x^3-x^2-4x+4
    y=-2x^2+16x-30

    3. The attempt at a solution
    I have no idea at all how to work this out, so perhaps someone could walk me through it?
     
  2. jcsd
  3. Sep 4, 2010 #2
    Made an attempt:
    Since the x intercepts of each graph closest to each other are 2 and 3, the line must cut through 2.5. The coordinates for graphs are now (2.5,y1) and (2.5,y2). Substituting 2.5 into the equations gives the the y coordinates as 3.375 and -2.5. No need to use 2.5 to calculate distance since the gradient doesn't exist, so 3.374+2.5=5.875.

    Not sure if this is right, but if it is could someone tell me how to work it out in a more calculus kind of way since this doesn't seem at all like calculus to me.
     
  4. Sep 4, 2010 #3

    Mentallic

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    It's a nice attempt, but not exactly right. You can't make the assumption that since they're between 2 and 3 then it must be 2.5.
    Take this graph, [tex]y=\frac{8x-16}{5-2x}[/tex]. It's x-intercept is 2 but it has an asymptote at 2.5 so this is an extreme example of how it doesn't work.

    If we have y=f(x) and y=g(x) then we let them equal each other so we have f(x)=g(x) then y=f(x)-g(x), we are finding x where the graphs of f(x) and g(x) are equal, or in other words, where f(x)-g(x)=0. But since in our case they don't cross, we are trying to find x where they are closest to each other. We are trying to find where y is a minimum.
     
  5. Sep 4, 2010 #4
    How do you find the x intercept?
     
  6. Sep 4, 2010 #5

    Mentallic

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    There is no x-intercept.
    Again, in the function y=f(x)-g(x) where y is a minimum.
     
  7. Sep 4, 2010 #6
    Whats g(x)? Is it f(x) for the 2nd equation?
     
  8. Sep 4, 2010 #7

    Mentallic

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    Yep! Sorry I was assuming you knew about functions considering you're doing calculus already.
     
  9. Sep 4, 2010 #8
    I do, but we don't use g(x), only f(x). Could you show me the first steps of working to get this? I'm pretty confused still.
     
  10. Sep 4, 2010 #9

    Mentallic

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    Well, what's y=f(x)-g(x) ?
     
  11. Sep 4, 2010 #10
    x^3+x^2-20x+34?
     
  12. Sep 4, 2010 #11

    Mentallic

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    Right, so we have the graph y=x3+x2-20x+34. This graph tells us the distance between our function f(x) and g(x) since we are finding the difference between them. So at x=2, it will give us y=6 so that means the distance at x=2 between f(x) and g(x) is 6. How do you find where the minimum distance is?
     
  13. Sep 4, 2010 #12
    I have no idea :(
     
  14. Sep 4, 2010 #13

    Mentallic

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    Do you know calculus?
     
  15. Sep 4, 2010 #14
    Yea but I'm having a mental block at the moment.
     
  16. Sep 4, 2010 #15

    Mentallic

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    It's really simple! On the graph it tells us the distance between the two original curves f(x) and g(x). At x=2 we have y=6 which tells us the distance between the two curves is 6, at x=3 the distance between them is 10. Obviously the minimum distance between them is then the minimum point on the graph!
    If I asked you to find the minimum of the parabola y=x2+2x using calculus, what would you do?
     
  17. Sep 4, 2010 #16
    EDIT: Never mind with that, bit late
     
  18. Sep 4, 2010 #17

    Mentallic

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    Don't think about finding what the actual distance is yet, we just want to find the x-value where the minimum occurs, which is where calculus comes into play. After we do this we'll substitute it back into equation to find the distance.
     
  19. Sep 4, 2010 #18
    Is the minimum the turning point, so I need to find when the gradient equals zero?
     
  20. Sep 4, 2010 #19

    Mentallic

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    You should be more confident in your abilities, rather:
    "The minimum is the turning point, so I need to find when the gradient equals zero!"

    :tongue:
     
  21. Sep 4, 2010 #20
    f'(x)=3x^2+2x-20
    I can't seem to manage factorizing this now, although I may be going in the wrong direction.
     
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