# Homework Help: Identifying and Drawing Surfaces

1. Jul 29, 2010

### jegues

1. The problem statement, all variables and given/known data
See figure

2. Relevant equations

3. The attempt at a solution

Just to give the readers some background on my current situation,

Recently I've been doing some independent study on some of the material that will be covered in upcoming math analysis course I'm taking, and I'm stuck on this question.(See figure)

The only form that I've seen surfaces described in is the following,

$$z = f(x,y)$$

I tried to throw this question into that format as well,

$$z = \sqrt{-x^{2}+2x-3-y} +1$$

but this doesn't get me any closer to "identifying" or drawing the surface.

Any ideas?

NOTE: This is my first attempt at a question like this

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2. Jul 29, 2010

### lanedance

how squaring & about rearranging, can you simplify any more
$$(z-1)^2 +(x^2-2x+1)= -(y+2)$$

3. Jul 29, 2010

### lanedance

ps... how about thinking y = f(x,z)

4. Jul 29, 2010

### Staff: Mentor

This form is very useful. For fixed values of y, with y < -2 (why?), the cross-sections are all circular, with centers along the line (1, y, 1) and with varying radii.

5. Jul 29, 2010

### jegues

So should I be "Identifying" and drawing the surface from this form,

$$(z-1)^2 +(x^2-2x+1)= -(y+2)$$

Or should I simplify it so that,

$$y = f(x,z)$$

?

I've never sketched a surface before so I don't really know what to look for.

6. Jul 29, 2010

### Staff: Mentor

The form in the first equation, above, is about as good as any.

To sketch a surface you don't want to just plot points. If possible you want to sketch cross-sections, which will give you a better idea of the shape of the overall surface.

For this equation, the best thing to do is sketch five or six cross sections for selected values of y (i.e., in planes parallel to the x-z plane). As already mentioned, there are some y values that aren't allowed.

For example, if y = -3, what geometric figure is described?

7. Jul 29, 2010

### jegues

If y = -3,

$$(z-1)^{2} +(x-1)^{2}= 1$$

This describes a circle with radius of 1 whos origin resides at the point (1,1) in the x-z plane.

Is this correct?

What other sketchs can I pull out of this equation?

Sorry if its not obvious for me, again, this is my first attempt at a problem like this.

8. Jul 29, 2010

### Staff: Mentor

No. The center of the circle is at (1, -3, 1), a point in the plane y = -3. This isn't the x-z plane. What other information about the circle can you get from this equation?
Pick several other values of y and do what you did above. For each value of y you get a different cross section of your surface.

9. Jul 29, 2010

### jegues

Okay, so the center of the circle is at (1, -3, 1). I'm not sure what else I could tell you...

With this equation,

$$(z-1)^{2} +(x-1)^{2}= 1$$

The radius of the circle would be 1, no?

Is there any way that I can quickly figure out which y values will not work? Or do I simply keep picking simply values of y and identifying new cross sections?

10. Jul 29, 2010

### Staff: Mentor

The radius is the other piece of information I was looking for.

The equation you're working with is
$$(z-1)^2 +(x^2-2x+1)= -(y+2)$$

The left side is always nonnegative, so the right side must likewise be nonnegative. The y values to check are those for which -(y + 2) >= 0.

11. Jul 29, 2010

$$(z-1)^{2} +(x-1)^{2}= -(y+2)$$ is a circular paraboloid that opens from (1,-2,1) in the negative y direction.

Since you're studying this on your own, you should at the very least familiarize yourself with all the typical quadratic surfaces and their basic properties. Here:http://en.wikipedia.org/wiki/Quadric_surface

12. Jul 29, 2010

### jegues

This makes perfect sense. Thank you for pointing it out, I would have missed it.

Thanks for the link, it helped alot with the visualization of these surfaces. Are students often introduced to this before attempting questions like this?

Thanks again.

13. Jul 29, 2010

### jegues

Another question,

The values for y must range from -2 to infinity then, correct?

How do you sketch something that goes off to infinity in one direction?

14. Jul 29, 2010

No problemo. uhmm...well there's really nothing new here to be introduced. It all builds on elementary material. There's no new formulas or concepts really. We're mainly just categorizing the surfaces. You would still have arrived at the same result had you only sketched out several cross sections.

For each of the quadric surfaces, I'd highly recommend inspecting the 3 cross sections created by keeping x, y, and z constant, respectively. Do this about 5 to 10 times for each of the quadric surfaces, and you should start getting the hang of it.

15. Jul 29, 2010

### Char. Limit

I doubt you're expected to graph the entire function. Just pick a reasonable endpoint for y, like 6 or something.

16. Jul 29, 2010

No, y ranges from -2 to NEGATIVE infinity.

Like I said, this builds up on regular plane curves. We're just "going 1 dimension higher." Yay!! But, seriously, it's the same concepts. How do you draw a parabola that goes to infinity only in the negative y direction? Now, for our surface, it's the same thing except we have circular cross sections when we keep y constant, with increasing radii as we decrease y. Try to visualize this yourself. Practice will help.

Step 1: Let x = 1. What type of curve do we trace?
Step 2: Let z = 1. Same as above.
Step 3: Fix y at several different values. etc.

17. Jul 29, 2010

### jegues

I'm not sure how to do this, can you give me a little push to get me started?

EDIT: You beat me to it!

If x=1 then the equation would describe an quadratic, correct? If z=1 then it would still be a quadratic?

18. Jul 29, 2010

look at my last post ^^

Basically, we want to "simplify" the surface into a planar curve since we're (you're :D) probably more familiar with those. Plus, looking at the different cross sections usually gives an insight into the surface's properties.

i.e., if a certain cross section of the surface is a parabola, then the surface itself must have a "min/max/saddle point."

19. Jul 29, 2010

### jegues

Bump, it edited my previous post to reflect the new info you gave me.

Am I close?

20. Jul 29, 2010

mhmm...and in both cases, the parabola opens up in the negative y direction. Now that you know what your surface looks like, can you visualize these two curves that lie on the surface? If not, consult more notes on wikipedia, wolframalpha, paul's notes, google, etc. :). Don't worry, you'll get the hang of it as you see more and more of them.