Matching Equations to Spheres: Solving the Mystery

[Turbodog66]

I have been given a problem with 4 equations, that need to be matched up to the corresponding image. I have worked the equations already and determined their center, but for the life of me I cannot seem to figure out which graph goes with which equation. The images are not that easy to read which is most of my issue. Here is an example:

x2−4x+y2−4y+z2−2z = -35/4 (original equation)

(x-2)^2 + (y-2)^2 + (z-1)^2 -9 = -35/4

(x-2)^2 + (y-2)^2 + (z-1)^2 = -35/4 + 9 >> √1/4 >> 1/2
C = (2,2,1) r = 1/2

This would give me a center of (2,2,1). However, none of the spheres I am provided appear to match this. Am I blind or does my math appear to be off?

 

[mfb]

(D) could fit. There are sign errors in your absolute values (doesn't influence the center, but influences the radius) - with your current number you would not get a sphere because the equation has no solution.

 

[Turbodog66]

(D) could fit. There are sign errors in your absolute values (doesn't influence the center, but influences the radius) - with your current number you would not get a sphere because the equation has no solution.

I see that, I have it written down correctly on my notebook, resulting in a r of 1/2. I appreciate your response, at this point I am more concerned with getting the correct values than matching to the correct graph. There are 4 total equations, and if even one of the choices are wrong it marks the entire problem wrong. Prevents guessing I suppose, but not very helpful otherwise.

 

[Ray Vickson]

I have been given a problem with 4 equations, that need to be matched up to the corresponding image. I have worked the equations already and determined their center, but for the life of me I cannot seem to figure out which graph goes with which equation. The images are not that easy to read which is most of my issue. Here is an example:

x2−4x+y2−4y+z2−2z = -35/4 (original equation)

(x-2)^2 +4 (y-2)^2 +4 (z-1)^2 +1 = -35/4

(x-2)^2 + (y-2)^2 + (z-1)^2 + 9 = -35/4

This would give me a center of (2,2,1). However, none of the spheres I am provided appear to match this. Am I blind or does my math appear to be off?View attachment 104840

$x^2 - 4x = x^2 - 4x + 4 - 4 = (x-2)^2 -4$, etc. So, you should have $-9$ as the constant on the left of your final equation, not the $+9$ that you wrote. I cannot see what work you did to get your solution, so I don't know what steps you took.

 

[Turbodog66]

$x^2 - 4x = x^2 - 4x + 4 - 4 = (x-2)^2 -4$, etc. So, you should have $-9$ as the constant on the left of your final equation, not the $+9$ that you wrote. I cannot see what work you did to get your solution, so I don't know what steps you took.

My apologies for the typo, I corrected it in the original post. Ultimately, I have what I believe is the correct values, but I am not seeing a sphere whose center is (2,2,1) with a radius of 1/2. The attachment is the image with the spheres, I'm not sure if it shows up embedded in the first post or not.

 

[Ray Vickson]

My apologies for the typo, I corrected it in the original post. Ultimately, I have what I believe is the correct values, but I am not seeing a sphere whose center is (2,2,1) with a radius of 1/2. The attachment is the image with the spheres, I'm not sure if it shows up embedded in the first post or not.

I cannot see any sphere with center (2,2,1), or even close to it. Are you sure you copied down the original problem correctly?

 

[Turbodog66]

I cannot see any sphere with center (2,2,1), or even close to it. Are you sure you copied down the original problem correctly?

This is the original problem, directly from my assignment. #1 is the one in question. I will say that I failed to use "^" on the 3 variables to plainly indicate that they are squared, but I treated them as such and completed the squares.

Just in case the attachment doesn't work, x^2 - 4x +y^2 - 4y +z^2 - 2z = -35/4

 

[Ray Vickson]

This is the original problem, directly from my assignment. #1 is the one in question. I will say that I failed to use "^" on the 3 variables to plainly indicate that they are squared, but I treated them as such and completed the squares.
View attachment 104848
Just in case the attachment doesn't work, x^2 - 4x +y^2 - 4y +z^2 - 2z = -35/4

One of these equations fits one of the diagrams perfectly (as far as my eyes can make out). I can't say more without giving a way the solution.

 

[Turbodog66]

One of these equations fits one of the diagrams perfectly (as far as my eyes can make out). I can't say more without giving a way the solution.

I agree, I have worked out the other 3 equations and was able to match 2 of them with a diagram fairly easily. But I still have to figure out which one matches the 1st equation(all 4 equations are supposed to be represented in the diagram). Are you saying you do see one with (2,2,1)? If so that's good enough for me, and I will keep staring until I see it, haha. In any case I appreciate the help.

 

[mfb]

I see a sphere where (2,2,1) as center would fit. You cannot extract the coordinates from the images reliably (2 vs. 3 dimensions), but you can go the opposite way: see where (2,2,1) is in each image, and see if that matches the center of the sphere in the plane.

 

[Ray Vickson]

I agree, I have worked out the other 3 equations and was able to match 2 of them with a diagram fairly easily. But I still have to figure out which one matches the 1st equation(all 4 equations are supposed to be represented in the diagram). Are you saying you do see one with (2,2,1)? If so that's good enough for me, and I will keep staring until I see it, haha. In any case I appreciate the help.

As I said, I don't see any sphere with center near (2,2,1), but, apparently, others can.

 

[SammyS]

As I said, I don't see any sphere with center near (2,2,1), but, apparently, others can.

Neither can I see any sphere with center near (2,2,1) .

 

[mfb]

Do we see the same image?

 

[Ray Vickson]

Do we see the same image?

View attachment 104851

To me it looks like the center is at about (0,1.5,0). However, as you said before, some of the figures are not very clear.

 

[Turbodog66]

Do we see the same image?

View attachment 104851

One thing is certain, this problem is more of a test of your visual skills rather than testing your ability to solve an equation.. I have reached out to my TA to see if there is any help they can provide as well. I appreciate the help everyone

 

[mfb]

To me it looks like the center is at about (0,1.5,0). However, as you said before, some of the figures are not very clear.

You cannot uniquely determine the center of a 3D point from a 2D drawing without any additional help. The sphere in D is at the place where a sphere around (2,2,1) has to appear, and no other sphere fits.

 

[Ray Vickson]

You cannot uniquely determine the center of a 3D point from a 2D drawing without any additional help. The sphere in D is at the place where a sphere around (2,2,1) has to appear, and no other sphere fits.

Assuming, of course, that the person setting the problem has not given an erroneous statement.

 

[mfb]

Why?
Sorry, I don't understand the problem. We are supposed to find the right sphere for this equation, there is exactly one sphere that fits. What needs further discussion?

 

[Ray Vickson]

Why?
Sorry, I don't understand the problem. We are supposed to find the right sphere for this equation, there is exactly one sphere that fits. What needs further discussion?

Whether it is correct that every one of the equations will, in fact, fit one of the displayed spheres. In other words, is the wording of the problem accurate?

Sometimes problem statements given in this Forum have turned out to be wrong, more often than not because of typos, or whatever. Certainly, typos are not so rare as to be dismissable right away.

 

[mfb]

Whether it is correct that every one of the equations will, in fact, fit one of the displayed spheres. In other words, is the wording of the problem accurate?

Just one of the four equations has been posted here, but this one fits to exactly one sphere.

Certainly, typos are not so rare as to be dismissable right away.

If the problem statement doesn't seem to be right: sure. But here there is no problem with the problem statement. If you have to pick the answer to "5+5" and the options are "9", "10" and "15", you wouldn't expect a problem with the problem statement, right?

 

[Turbodog66]

Here is the problem in its entirety. I didn't include everything initially because I did not have a problem with the other 3. My intention was to present it as if it was simply a problem with a single equation and provided with 6 possible solutions, knowing that one of them must be correct.

 

[SammyS]

Do we see the same image?

View attachment 104851

Now I can see it.

Thanks.