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k8lyn911

  • one year ago

Calculus III: Find an equation of a sphere if one of its diameters has endpoints (3, 5, 6) and (5, 7, 8).

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  1. misty1212
    • one year ago
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    HI!! this is not as hard as it looks

  2. misty1212
    • one year ago
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    find the center first, the average in each coordinate

  3. misty1212
    • one year ago
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    you got that?

  4. anonymous
    • one year ago
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    anticipation is killing me...

  5. k8lyn911
    • one year ago
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    Sorry! Working through some other homework problems. Which equations do we use for that?

  6. misty1212
    • one year ago
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    just like the midpoint in two dimensions take the average in each coordinate

  7. k8lyn911
    • one year ago
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    So (x1 - x2, y1 - y2, z1 - z2)?

  8. misty1212
    • one year ago
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    so, the average

  9. misty1212
    • one year ago
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    add up and divide by two

  10. misty1212
    • one year ago
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    "no" i meant just like finding the midpoint of \((3,5)\) and \((7,11)\) in two dimensions add and divide by two in each coordinate in my example the midpoint would be \((5,8)\)

  11. misty1212
    • one year ago
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    clear or no?

  12. k8lyn911
    • one year ago
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    ( (3+5)/2, (5+7)/2, (6+8)/2) = (4, 6, 7)?

  13. misty1212
    • one year ago
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    right

  14. misty1212
    • one year ago
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    before we continue, do you remember how to find the equation of a circle given two endpoints of the diameter? this is identical, except with one extra coordinate

  15. misty1212
    • one year ago
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    the equation is going to be \[(x-4)^2+(y-6)^2+(z-7)^2=r^2\]

  16. k8lyn911
    • one year ago
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    Not exactly...

  17. misty1212
    • one year ago
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    because the center is \((4,6,7)\)

  18. misty1212
    • one year ago
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    equation for a circle with radius \(r\) and center \((h,k)\) is \[(x-h)^2+(y-k)^2=r^2\]

  19. misty1212
    • one year ago
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    this is exactly the same, but with one more coordinate

  20. k8lyn911
    • one year ago
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    Oh, so it's like the other problems we've had, but backwards! So we add these values squared and take the square root to find r?

  21. misty1212
    • one year ago
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    ok lets go slow

  22. misty1212
    • one year ago
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    maybe i am not sure what exactly you mean we got the center, we need the radius

  23. misty1212
    • one year ago
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    actually we don't need the radius, we just need the square of the radius, so forget the square root business

  24. k8lyn911
    • one year ago
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    Okay.

  25. misty1212
    • one year ago
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    same distance formula as with two points, only now each point has three coordinates instead of two

  26. misty1212
    • one year ago
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    square of the distance between \((3, 5, 6) \) and \( (4, 6, 7)\) is \[((4-3)^2+(6-5)^2+(7-6)^2\] a pretty easy calculation in this case

  27. misty1212
    • one year ago
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    for the actual distance it would be \[\sqrt{(4-3)^2+(6-5)^2+(7-6)^2}\]

  28. misty1212
    • one year ago
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    you get 3 pretty much instantly right?

  29. k8lyn911
    • one year ago
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    Yes.

  30. misty1212
    • one year ago
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    final answer:\[(x-4)^2+(y-6)^2+(z-7)^2=3\]

  31. misty1212
    • one year ago
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    a nice picture http://www.wolframalpha.com/input/?i=%28x-4%29%5E2%2B%28y-6%29%5E2%2B%28z-7%29%5E2%3D3

  32. k8lyn911
    • one year ago
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    Oh, I see. Thank you so much for explaining it to me! :)

  33. misty1212
    • one year ago
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    \[\huge \color\magenta\heartsuit\]

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