Need some math help

[Aramatheis]

Hey everyone!

I have a calculus exam coming up (tomorrow, actually) and I'm having trouble figuring out a practice question. I was hoping somebody here could lend a hand  
Please and Thank You!


So the question has to do with sketching a curve/surface.

Sketch; x=cos(4t), y=sin(4t), z=1

----------------------------------------------------------------------------------

Now there was a previous question that was similar to the above. The info given was x=2tcos(t), y=3tsin(t), z=t^2
It ended up being a paraboloid, centered on the z-axis, with a swirling curve inside its boundaries.

It was solved by using the identity sin^2x + cos^2x = 1

which ended up giving (x/2)^2 + (y/3)^2 = t^2 = z

and in turn,  (x^2)/4 + (y^2)/9 = z  (the corresponding paraboloid function)

From there, it was determined that z increases, while x and y increase/decrease periodically, creating the swirl.




So when I try a similar method of solving the (initial) question, I can't seem to formulate a surface equation.
I guess I'd like somebody to help me solve this or push me in the right direction.

Thanks DTF!

[ricky]

calling jamesman.

[Aramatheis]

 

[Jamesman42]

Using the same method as your other example,

x^2 + y^2
= (cos(4t))^2 + (sin(4t))^2 (Given identities)
= 1 (Pythagorean identity)

So x^2 + y^2 = 1

Since z = 1 and no other number, this is a circle of radius 1 in space moved up one unit on the z-axis. It's basically a 2-D curve but moved up one unit in 3-D. With different values of t, however, you move faster around the circle, because of the factor of 4 from the parametric equations. x^2 + y^2 = 1 just tells you the "shadow" of the space curve on the x-y plane.

Let me know if anymore questions.

I'll try to find a way to get a picture of what I mean for you.

[alirocker08]

...And think, I'm just having troubles with circle theorems. Good luck on your exam!

[Aramatheis]

Alright, that's a wonderful response Jamesman, but...
I screwed up.

z =\= 1
 it's actually z=t

...And think, I'm just having troubles with circle theorems. Good luck on your exam!

That's ok, I'm not even sure what you mean by circle theorems..
I don't really know the names for most of the theorems/applications I work on   

[Jamesman42]

If z = t, then the change that happens is you still have the same shadow, but you now have a helix increasing in the positive z direction.

[Jamesman42]

That's ok, I'm not even sure what you mean by circle theorems..
I don't really know the names for most of the theorems/applications I work on   

She probably means theorems of circles, like "The measure of an inscribed angle in a circle is equal to half the measure of its intercepted arc."

[Aramatheis]

so it's a circle, centered on the z-axis, up 1 on the z axis (from the origin), with a helix spiraling upwards?

Thank you!


(I'll be gone for lunch soon, so I won't be able to respond for a little while)

That's ok, I'm not even sure what you mean by circle theorems..
I don't really know the names for most of the theorems/applications I work on   

She probably means theorems of circles, like "The measure of an inscribed angle in a circle is equal to half the measure of its intercepted arc."

Yeah, I looked it up ^^

[Jamesman42]

The shadow on the x-y plane is the circle of radius 1 centered at the origin...but the actual curve itself is a helix. Because z = t and not z = 1 now, forget the upshift of 1 on the z-axis, and forget that it isn't a circle, but that it is a helix now. In short, just a helix of radius 1.

As t increases, you still move fast on the helix as you go up because of the 4 on the trig functions from the parametric equations.

[Aramatheis]

Would it resemble this, except upside down?

[kári]

I don't think that the distance between two "swirls" gets less though.

[Jamesman42]

Would it resemble this, except upside down?

*helix*

Sorta....it's just a helix going up counter-clockwise. Like kari said, no variable spaces between the swirls.

[Aramatheis]

It's ok, I was at a help session today, and we ended ended up covering that question.


Thanks for your help though, it is very much appreciated