parameterization of a curve

Printable View

February 8th 2009, 06:16 AM
sonia1

parameterization 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))
February 8th 2009, 06:22 AM
Danny

Quote:

Originally Posted by sonia1

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))

The first is a straight line parallel to the xy plane moved 4 units in the z direction and the second, an ellipse in the yz plane.
February 8th 2009, 06:27 AM
sonia1

how did you find that out, could u please show me
February 8th 2009, 06:52 AM
Danny

Quote:

Originally Posted by sonia1

how did you find that out, could u please show me

Sure. In the first case

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

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

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

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

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

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

$y^2 + \frac{z^2}{3^2} = 1$ , the formula for an ellipse in the yz plane. Hope that helps.
February 8th 2009, 07:06 AM
sonia1

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)
February 8th 2009, 07:46 AM
Danny

Quote:

Originally Posted by sonia1

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)

You'll notice with the first equation and the identity

$\sin^2t + \cos^2 t = 1$ then $36\sin^2t + 36\cos^2 t = 36$ so comparing gives

$x = 6 \sin t,\;\;\;y = 6 \cos t$ . Now sub. these into z and simplify.
February 8th 2009, 07:51 AM
sonia1

thanks..i simplified it and i get z= (1/pi) arc 6tan^2 t, is that correct
February 8th 2009, 08:05 AM
sonia1

regarding the before question, could u please show me a drawing of what the ellipse actually looks like.
February 8th 2009, 08:25 AM
Danny

Quote:

Originally Posted by sonia1

thanks..i simplified it and i get z= (1/pi) arc 6tan^2 t, is that correct

Not $6\tan^2 t$ but $\tan t$
February 8th 2009, 08:28 AM
sonia1

r u sure it should be tan^2 t because (6sint/6cost) =tan t

and then we get 1/pi arc tan^2t right
February 8th 2009, 08:33 AM
Danny

1 Attachment(s)

Here's a picture of the ellipse.
February 8th 2009, 08:40 AM
sonia1

thankz for your help,

did u get tan^2 t then not tant
February 8th 2009, 08:57 AM
Danny

Quote:

Originally Posted by sonia1

thankz for your help,

did u get tan^2 t then not tant

From your earlier post,

$x^2 + y^2= 36 \;\; \text{and}\;\; z= \frac{1}{\pi}\tan^{-1}(\frac{x}{y})$

if $x = 6\sin t, y =6\cos t$ then $\frac{x}{y} = \frac{6\sin t}{6\cos t} = \tan t$
February 8th 2009, 09:12 AM
sonia1

thankz for your help!
February 8th 2009, 06:25 PM
mr fantastic

Quote:

Originally Posted by sonia1

thankz for your help!

You can thank danny (and add to his reputation) by clicking the Thanks button (bottom right).