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Noob88
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Joined: 23 Nov 2005
Posts: 239
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 Posted: 04 Sep 2007 03:25:06 pm Post subject: |
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Ok, so for homework tonight in Calculus we are supposed to find Cartesian equations for a curve that contains the parametrized curve. These problems started off rather easily but as they started using sin, cos, tan, sec, etc. it got rather confusing.
Ex:
x=t, y=sqrt(t), t>=0
y=sqrt(x) [found by solving for Y]
There were quite a few easy ones like that, and now I have
x=tan(t), y=-2sec(t), -pi/2 < t < pi/2
To be honest, I really have no idea where to start now. A hasty reply would be appreciated :biggrin: |
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Weregoose
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Joined: 25 Nov 2004
Posts: 3976
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| Posted: 04 Sep 2007 03:44:18 pm Post subject: |
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tan‾¹(x)=t
Go for it.
Last edited by Guest on 25 Sep 2010 03:13:40 pm; edited 1 time in total |
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thornahawk
μολών λαβέ
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Joined: 27 Mar 2005
Posts: 569
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| Posted: 04 Sep 2007 03:46:47 pm Post subject: |
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I was slightly confused by your post, so I'll assume you meant "turn parametric Cartesian equations into vanilla Cartesian equations". :)
The first example you gave is trivial, of course. :D
The second one... that needs a basic trig formula:
sec²(t)-tan²(t)=1
Solve for sec(t) and tan(t) in those two equations and substitute.
There are a lot of other tricks to convert between the two representations (parametric and plain Cartesian), so if you get stuck on something, just post here. :)
thornahawk
Last edited by Guest on 04 Sep 2007 03:48:25 pm; edited 1 time in total |
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DarkerLine
ceci n'est pas une |
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Joined: 04 Nov 2003
Posts: 8328
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| Posted: 04 Sep 2007 03:48:39 pm Post subject: |
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Another way to do it besides the one thornahawk mentioned (if you don't end up seeing the trig identity right away):
You could start off with solving to get t=tan-1(x) and y=-2sec(tan-1(x)). However, that might not be simplified enough for your teacher.
To simplify this, imagine a right triangle with the perpendicular sides 1 and x. The tangent of the angle next to the 1 is opposite/adjacent = x/1=x (This was our goal in choosing the sides 1 and x for the triangle). So this angle is tan-1(x). Now we take the secant of that angle: hypotenuse/adjacent = √(x2+1)/1 = √(x2+1) to get our final answer:
y=-2√(x2+1)
Last edited by Guest on 04 Sep 2007 04:45:30 pm; edited 1 time in total |
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Noob88
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Joined: 23 Nov 2005
Posts: 239
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| Posted: 04 Sep 2007 03:54:08 pm Post subject: |
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Alright thanks... That helped out a lot with that problem but the whole chapter itself is really confusing to me I just can't seem to get it  |
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Weregoose
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Joined: 25 Nov 2004
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| Posted: 04 Sep 2007 03:57:35 pm Post subject: |
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Keep posting, if it will help make something click. 
Last edited by Guest on 04 Sep 2007 03:57:53 pm; edited 1 time in total |
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Noob88
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Joined: 23 Nov 2005
Posts: 239
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| Posted: 04 Sep 2007 04:19:18 pm Post subject: |
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Ok to dumb it down just a bit, I know sec²(t)-tan²(t)=1, but could you show me the steps you would go about in getting the final equation? I'm more of a trig person than a geometry person so that kind of stuff makes more sense to me. Sorry to be such a bother but it would be really beneficial to me to see all the steps to get from the beginning to the answer
[EDIT] Ahh I'm catching on now... so is there some kind of rule/law that says sec(tan-1(x)) = √(x2+1)... This is the only part I really need explained to me now please
Wait a second... The graphs of y=-2sec(tan-1(x)) and y=√(x2+1) are showing up as different graphs on my calculator... Now I'm completely confused...
I cleaned up your post a wee bit - thornahawk
Last edited by Guest on 04 Sep 2007 10:42:34 pm; edited 1 time in total |
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DarkerLine
ceci n'est pas une |
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Joined: 04 Nov 2003
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| Posted: 04 Sep 2007 04:46:24 pm Post subject: |
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Well, since sec(tan-1(x))=√(x2+1), what should -2sec(tan-1(x)) be?
Noob88 wrote: Ahh I'm catching on now... so is there some kind of rule/law that says sec(tan-1(x)) = √(x2+1)... This is the only part I really need explained to me now please
sec²(t)-tan²(t)=1
Solve this equation for sec(t), and see what you get.
Last edited by Guest on 04 Sep 2007 04:48:50 pm; edited 1 time in total |
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Noob88
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Joined: 23 Nov 2005
Posts: 239
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| Posted: 05 Sep 2007 05:06:14 pm Post subject: |
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| Ok I got it thanks guys, It's really just a matter of me learning my trig identities a bit better I suppose. |
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