can someone help please:

Find what kind of curves are given by the following representations and draw the curves:

1. r(t) = (2t-5, -3t+1, 4)

2. r(t) = (0, -cos(t), 3sin(t))

Printable View

- Feb 8th 2009, 06:16 AMsonia1parameterization of a curve
can someone help please:

Find what kind of curves are given by the following representations and draw the curves:

1. r(t) = (2t-5, -3t+1, 4)

2. r(t) = (0, -cos(t), 3sin(t)) - Feb 8th 2009, 06:22 AMJester
- Feb 8th 2009, 06:27 AMsonia1
how did you find that out, could u please show me

- Feb 8th 2009, 06:52 AMJester
Sure. In the first case

$\displaystyle x = 2t -5,\;\;y= -3t+1,\;\;z=4$

*z*isn't changing,*x*and*y*are. Eliminating*t*gives

$\displaystyle \frac{x+5}{2} = \frac{y-1}{-3}$ a straight line.

In the second, $\displaystyle x = 0$, it's not changing and

$\displaystyle y = - \cos t,\;\; z = 3 \sin t$

Eliminating*t*using the fact that $\displaystyle \cos^2 t + \sin^2 t= 1$ gives

$\displaystyle y^2 + \frac{z^2}{3^2} = 1$, the formula for an ellipse in the*yz*plane. Hope that helps. - Feb 8th 2009, 07:06 AMsonia1
yes thanks alot,

i'm also stuck on a different question on the same topic

wonder if you can help:

find a parametric representation of the curve :

x^2 + y^2= 36 and z= (1/pi) arctan(x/y)

i.e. find a representation in the form x=x(t), y=y(t), z=z(t) - Feb 8th 2009, 07:46 AMJester
- Feb 8th 2009, 07:51 AMsonia1
thanks..i simplified it and i get z= (1/pi) arc 6tan^2 t, is that correct

- Feb 8th 2009, 08:05 AMsonia1
regarding the before question, could u please show me a drawing of what the ellipse actually looks like.

- Feb 8th 2009, 08:25 AMJester
- Feb 8th 2009, 08:28 AMsonia1
r u sure it should be tan^2 t because (6sint/6cost) =tan t

and then we get 1/pi arc tan^2t right - Feb 8th 2009, 08:33 AMJester
Here's a picture of the ellipse.

- Feb 8th 2009, 08:40 AMsonia1
thankz for your help,

did u get tan^2 t then not tant - Feb 8th 2009, 08:57 AMJester
- Feb 8th 2009, 09:12 AMsonia1
thankz for your help!

- Feb 8th 2009, 06:25 PMmr fantastic