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Equation of a sphere(where did I go wrong?)

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the equation of a sphere if one of its diameters has endpoints: (-1, -16, -1) and (17, 2, 17).


    2. Relevant equations
    midpoint formula= x1+x2/2 + y1+y2/2 + z1+z2/2
    equation of a sphere= (x-a)^2 +(y-b)^2 +(z-c)^2

    3. The attempt at a solution
    Okay this problem is driving me nuts. It seems so simple and I keep goofing it up somehow.

    First, the magnitude of the diameter:

    =[tex]\sqrt{(17-(-1))^2 + (2-(-16))^2 + (17-(-1))^2}[/tex]
    =[tex]\sqrt{972}[/tex]

    Radius= (1/2)diameter
    =[tex]\sqrt{972}/2[/tex]

    So I think this is where I'm confusing things. For the coordinates I used to mid point formula:

    [tex]\frac{(-1)-17}{2}=3[/tex]
    [tex]\frac{(-16)-2}{2}=-7[/tex]
    [tex]\frac{(-1)-17}{2}=8[/tex]


    So then plugging into the equation of a sphere...

    (x-3)^2 + (y+7)^2 + (z-8)^2 = (15.588)^2

    They want the entire equation equal to zero
    (x-3)^2 + (y+7)^2 + (z-8)^2 - 243 = 0

    But apparently this answer is wrong. Where did I go wrong? I've gone through this problem multiple times and I can't figure it out.
     
  2. jcsd
  3. Sep 7, 2009 #2

    Mark44

    Staff: Mentor

    This can't possibly be the equation of anything, since it is not an equation. An equation is the statement of the equality of two things; hence, there has to be an = in it.
    This is correct. You could also write it as [itex]9\sqrt{3}[/itex]
    This is where you went wrong. All three of your values are incorrect. All three values are actually negative. You seem to be having trouble doing subtraction.
    Two things here: As noted above, your point for the center of the sphere is incorrect. Also, the radius is not equal to 15.588. Use the exact value. If I square the value you show for the radius, I get 242.985744.
    An equation can't be equal to zero. An expression can be equal to zero.
     
  4. Sep 7, 2009 #3
    I'm sorry. Apparently my attention to detail has gotten sloppy as I've moved on in math.
    Would this be better?:

    r^2= (x-a)^2 +(y-b)^2 +(z-c)^2

    Sorry, that's also my mistake. The midpoint formula is actually (x1 + x2)/2. So my numbers should be correct, I just made a mistake while typing it.

    Once again, that's my negligence to detail. I apologize. I meant:

    15.55634919.

    So can you spot an issue with my work then? I do suspect it deals with how I obtained the center of the sphere.
     
  5. Sep 7, 2009 #4
    Woah...I need to stop staying up so late studying. Apparently ((-1)+17)/2 does not equal 3....


    I guess my lack of attention to detail seems to be a recurring theme, eh?
     
  6. Sep 7, 2009 #5

    Mark44

    Staff: Mentor

    Check your arithmetic on the x-coordinate where you are finding the center of the diameter.
    (-1 + 17)/2 != 3
     
  7. Sep 7, 2009 #6

    Mark44

    Staff: Mentor

    Most of the things I pointed out weren't serious enough to keep you from find the solution to this problem. I was just trying to help you say the words that represent what you are doing.
     
  8. Sep 7, 2009 #7
    I realize that. It's sad that I can be in Calculus 3 and not use simple terminology correctly. I think a lot of that deals with my youth. I really didn't apply myself until my sophomore year of high school. I was lucky enough to work hard and get into a decent honors college. I guess those early years of slacking off when I was young kept me from fully understanding the basics.

    Anyway, thanks a lot. I think I needed a daily dosage of nit picking to realize the foolish mistake I made.
     
  9. Sep 7, 2009 #8

    Mark44

    Staff: Mentor

    That's water under the bridge, so there's no undoing it. I'm sure there are lots of people in this situation, but when you spot an area that you're weak in, put in a little extra time and effort and strengthen your skills.
     
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