k8lyn911
  • k8lyn911
Calculus III: Find an equation of a sphere if one of its diameters has endpoints (3, 5, 6) and (5, 7, 8).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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misty1212
  • misty1212
HI!! this is not as hard as it looks
misty1212
  • misty1212
find the center first, the average in each coordinate
misty1212
  • misty1212
you got that?

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anonymous
  • anonymous
anticipation is killing me...
k8lyn911
  • k8lyn911
Sorry! Working through some other homework problems. Which equations do we use for that?
misty1212
  • misty1212
just like the midpoint in two dimensions take the average in each coordinate
k8lyn911
  • k8lyn911
So (x1 - x2, y1 - y2, z1 - z2)?
misty1212
  • misty1212
so, the average
misty1212
  • misty1212
add up and divide by two
misty1212
  • misty1212
"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 \((5,8)\)
misty1212
  • misty1212
clear or no?
k8lyn911
  • k8lyn911
( (3+5)/2, (5+7)/2, (6+8)/2) = (4, 6, 7)?
misty1212
  • misty1212
right
misty1212
  • misty1212
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
misty1212
  • misty1212
the equation is going to be \[(x-4)^2+(y-6)^2+(z-7)^2=r^2\]
k8lyn911
  • k8lyn911
Not exactly...
misty1212
  • misty1212
because the center is \((4,6,7)\)
misty1212
  • misty1212
equation for a circle with radius \(r\) and center \((h,k)\) is \[(x-h)^2+(y-k)^2=r^2\]
misty1212
  • misty1212
this is exactly the same, but with one more coordinate
k8lyn911
  • k8lyn911
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?
misty1212
  • misty1212
ok lets go slow
misty1212
  • misty1212
maybe i am not sure what exactly you mean we got the center, we need the radius
misty1212
  • misty1212
actually we don't need the radius, we just need the square of the radius, so forget the square root business
k8lyn911
  • k8lyn911
Okay.
misty1212
  • misty1212
same distance formula as with two points, only now each point has three coordinates instead of two
misty1212
  • misty1212
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
misty1212
  • misty1212
for the actual distance it would be \[\sqrt{(4-3)^2+(6-5)^2+(7-6)^2}\]
misty1212
  • misty1212
you get 3 pretty much instantly right?
k8lyn911
  • k8lyn911
Yes.
misty1212
  • misty1212
final answer:\[(x-4)^2+(y-6)^2+(z-7)^2=3\]
misty1212
  • misty1212
a nice picture http://www.wolframalpha.com/input/?i=%28x-4%29%5E2%2B%28y-6%29%5E2%2B%28z-7%29%5E2%3D3
k8lyn911
  • k8lyn911
Oh, I see. Thank you so much for explaining it to me! :)
misty1212
  • misty1212
\[\huge \color\magenta\heartsuit\]

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