Cylinders and quadric surfaces

[ineedhelpnow]

ok so this is the part I am really struggling with. we need to be able to recognize an ellipsoid, cone, elliptic paraboloid, hyprboloid of one sheet, hyperbolic paraboloid, hyperboloid of two sheets given an equation. he's going to give us an equation of one and ask us to identify and sketch the graph. please help me! and i would reaaaalllly appreciate it if anyone can let me know if they know how to graph these type of graphs on the ti nspire.

here are just some random examples of what we'll be asked on the test.
View attachment 3219View attachment 3216View attachment 3217View attachment 3218

 

[I like Serena]

ok so this is the part I am really struggling with. we need to be able to recognize an ellipsoid, cone, elliptic paraboloid, hyprboloid of one sheet, hyperbolic paraboloid, hyperboloid of two sheets given an equation. he's going to give us an equation of one and ask us to identify and sketch the graph. please help me! and i would reaaaalllly appreciate it if anyone can let me know if they know how to graph these type of graphs on the ti nspire.

here are just some random examples of what we'll be asked on the test.

1. Press [DOC]→Insert→Graphs.

2. Press [MENU]→View→3D Graphing.

3. Type a function in the entry line, z1(x,y)=sin(x)·cos(y) and press [ENTER] to graph it.

4. Use your Touchpad keys to rotate the graph.

For fun, press [A] to auto-rotate the graph; pressing [R] allows you to manually rotate the graph again using your Touchpad arrow keys. Press [$\div$] to shrink the box or [x] to magnify the box.​

5. Press [MENU]→Trace→zTrace.

Hold the [SHIFT] key down and use your Touchpad arrow keys to trace the graph.​

6. Explore some new color options of your 3D graph by right-clicking on the graph, [CTRL][MENU]→Color→Custom Plot Color.

(Movie)

 

[ineedhelpnow]

thanks ILS but its not that simple. you see this equations that I am dealing with are squared so i have to solve it in terms of z and with the square root and everything it make the graph totally incomplete and messed up. is there any way to get the full 3d graph?
(ill show you what i mean in a sec)

 

[ineedhelpnow]

lucky you guys. you get to see more pictures of the pretty screen ;)

ok so this is the graph of #26 (its in the 2nd pic i think). i have to solve and put them in separately. is there any way to put it all in at one shot? to get a COMPLETE graph.
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[I like Serena]

lucky you guys. you get to see more pictures of the pretty screen ;)

ok so this is the graph of #26 (its in the 2nd pic i think). i have to solve and put them in separately. is there any way to put it all in at one shot? to get a COMPLETE graph.

You'll get a better picture with:
$$z1(x,y)=\sqrt{x^2+2y^2}$$
$$z2(x,y)=-\sqrt{x^2+2y^2}$$
(Smile)
But I'm afraid that when all coordinates are squared, you can't do it in one shot. (Worried)

Anyway, the ones that have an unsquared coordinate can be done in one shot.

 

[ineedhelpnow]

it looks perfect!...but its not the same graph... (:D i would put it on auto rotation for you ILS but its only a screen capture so i can't (Giggle))

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- - - Updated (Malthe) - - -

you did intend for it to be the graph of another equation, right (Wondering)?

 

[I like Serena]

it looks perfect!...but its not the same graph... (:D i would put it on auto rotation for you ILS but its only a screen capture so i can't (Giggle)

Yeah. It's rotated with respect to the axes.
So you'll have to rotate your screen to get a proper view. (Doh)
Hmm... that doesn't really matter if you only want to recognize the shape, does it? (Wink)

you did intend for it to be the graph of another equation, right ?

It's the same graph - just rotated. (Emo)

 

[ineedhelpnow]

just to be safe though. i don't want to mess up on the test. and how will i know how to change the equation so it comes out looking like a pretty one?

- - - Updated (Malthe) - - -

i have a question! i have a question! i have a question!

is there any way to convert the equation into a parametric equation? that's way easier to graph. (Angel) (i think we have a problem...addicted to smilies (Nod))

 

[I like Serena]

just to be safe though. i don't want to mess up on the test. and how will i know how to change the equation so it comes out looking like a pretty one?

If one of the coordinates is not squared, then reorder the coordinates (switch them around) to make the non-squared one the z-coordinate.

For instance: rearrange $x=y^2+4z^2$ to $z=x^2+4y^2$.

The graph will remain the same, just rotated.If all of them are squared, you will have to introduce a square root.
Pick the z-coordinate as the one that at least gives a square root of something that is always positive. That is, make sure you take the square root of the sum of 2 squares.

i have a question! i have a question! i have a question!

is there any way to convert the equation into a parametric equation? that's way easier to graph.

You are already converting the equations to parametric equations.
The problem is that your TI Nspyre is limited to parametric equations of the form $z=f(x,y)$. (Swearing)
Wolfram|Alpha doesn't have such limitations. (Muscle)

 

[ineedhelpnow]

buuut i caaaant uuuse wolfram alpha on the test. (Crying) no these graphs i posted are in function form. i can do it in parametric.

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[I like Serena]

buuut i caaaant uuuse wolfram alpha on the test. (Crying)

I guess you'll just have to learn a bit more math then. (Devil)

no these graphs i posted are in function form. i can do it in parametric.

Oh.
Well...
Then I think you can graph $y^2=x^2+2z^2$ in one shot with:
$$xp(t,u)=t\cos(u)$$
$$yp(t,u)=t$$
$$zp(t,u)=\frac {t} {\sqrt 2} \sin(u)$$

 

[ineedhelpnow]

i literally just burst out laughing when i graphed that.
ignoring the big fat thing in the middle, is the graph represented by the two cones?

View attachment 3228

 

[I like Serena]

i literally just burst out laughing when i graphed that.
ignoring the big fat thing in the middle, is the graph represented by the two cones

Yes. (Wink)
I recognize it by the way the cones are built from rings and radial lines.

The big fat thing in the middle is the graph $z(x,y)=\sqrt{x^2+2y^2}$ that is apparently still lying around.

 

[ineedhelpnow]

LOL oh. well when i get rid of that part... i end up with 3/4 of a graph...

 

[I like Serena]

LOL oh. well when i get rid of that part... i end up with 3/4 of a graph...

What!
A 3/4 graph?
What does it look like? (Wondering)