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

  • Thread starter undrcvrbro
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In summary: No problem. I hope you start to enjoy your math more as you get better at it.In summary, the equation of the sphere with a diameter having endpoints (-1, -16, -1) and (17, 2, 17) is (x+3)^2 + (y+8)^2 + (z-8)^2 = 243.
  • #1
undrcvrbro
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Homework Statement


Find the equation of a sphere if one of its diameters has endpoints: (-1, -16, -1) and (17, 2, 17).


Homework 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

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.
 
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  • #2
undrcvrbro said:

Homework Statement


Find the equation of a sphere if one of its diameters has endpoints: (-1, -16, -1) and (17, 2, 17).


Homework 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
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.
undrcvrbro said:

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]
This is correct. You could also write it as [itex]9\sqrt{3}[/itex]
undrcvrbro said:
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]
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.
undrcvrbro said:
So then plugging into the equation of a sphere...

(x-3)^2 + (y+7)^2 + (z-8)^2 = (15.588)^2
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.
undrcvrbro said:
They want the entire equation equal to zero
An equation can't be equal to zero. An expression can be equal to zero.
undrcvrbro said:
(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.
 
  • #3
Mark44 said:
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.
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

Mark44 said:
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.
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.

Mark44 said:
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.
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.
 
  • #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?
 
  • #5
Check your arithmetic on the x-coordinate where you are finding the center of the diameter.
(-1 + 17)/2 != 3
 
  • #6
undrcvrbro said:
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?

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.
 
  • #7
Mark44 said:
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.

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.
 
  • #8
undrcvrbro said:
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.
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.
undrcvrbro said:
Anyway, thanks a lot. I think I needed a daily dosage of nit picking to realize the foolish mistake I made.
 

1. What is the equation of a sphere?

The equation of a sphere is (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2, where (h,k,l) represents the center of the sphere and r represents the radius.

2. How do you know if you went wrong when finding the equation of a sphere?

If you are given the center and radius of the sphere, you can plug those values into the equation and see if it satisfies all points on the sphere. If it doesn't, then you may have made a mistake in your calculations.

3. What are common mistakes when finding the equation of a sphere?

Some common mistakes include using the wrong formula, forgetting to square the radius, or making errors in the calculations of the center and radius.

4. Can I use the equation of a sphere to find the volume or surface area?

Yes, you can use the equation of a sphere to find the volume using the formula V = (4/3)πr^3 and the surface area using the formula A = 4πr^2.

5. Are there any other ways to represent the equation of a sphere?

Yes, instead of using the center and radius, you can also represent the equation of a sphere using its diameter (d) with the formula (x - a)^2 + (y - b)^2 + (z - c)^2 = (d/2)^2, where (a,b,c) is the center of the sphere.

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