Parametric eq's question!

[Andiio]

x = 3sin t, y = 4 cos t, t E [-pi/2, pi/2]

-> x^2/9 + y^2/16 = 1

But when you sketch it, the answer shows the upper half of the ellipse (i.e. an upside down 'U')

But if the domain of the values of t is [-pi/2, pi/2], shouldn't it be the right side of the ellipse instead of the top half?
I thought of changing the domain to [0,pi] but have no idea.. :S

Also, for x = sec t, y = tan t, t E (-pi/2, pi/2);
--> x^2 - y^2 = 1
But the solution denotes that: "For x = sec t, t E (-pi/2, pi/2), the range is [1,infinity).
---> How do they arrive at that range? I kind of understand it because the domain of (-pi/2, pi/2) is the right branch of the hyperbola, but just need some confirmation.

Thanks!

[brightsky]

x = 3sint, y = 4cost

t = arcsin(x/3), y = 4cos(arcsin(x/3)) = 4sqrt(9 - x^2)/3, starting at (-3, 0) and ending at (3,0), which is in essence a upside down U.

[Andiio]

x = 3sint, y = 4cost

t = arcsin(x/3), y = 4cos(arcsin(x/3)) = 4sqrt(9 - x^2)/3, starting at (-3, 0) and ending at (3,0), which is in essence a upside down U.

About the first one, I got it
Sketched cos and sin graphs over 2 cycles, then you restrict both graphs at [-pi/2,pi/2], which leaves you with Ran(y) and Ran(x); thus Ran(y) = Ran(cart), Ran (x) = Dom(cart), giving me an upside down 'U'!

Help with the second one would be appreciated though!
Thanks!

[brightsky]

For your second one, I'm pretty sure the range is R.

But for x = sec(t) it's [1, infty] because t E [-pi/2, pi/2].

[Andiio]

For your second one, I'm pretty sure the range is R.

But for y = sec(t) it's [1, infty] because t E [-pi/2, pi/2].

Yeah it is [1, infinity], but did you come to that conclusion after drawing the sec graph?

[brightsky]

You can do that. Or you can just work it out using some algebra.

y = sec(t) --> y = 1/cos(t)
Since -pi/2 =< t =< pi/2
Then 0 =< cos(t) =< 1
So 0 =< 1/cos(t) =< infty [Note that y gets bigger as cos(t) gets smaller positively so as cos(t) approaches 0, 1/cos(t) approaches a very very big number while on the opposite end, when cos(t) approaches 1, 1/cos(t) approaches 1]

[Andiio]

Ah dw, I drew the sec graph and figured it out; I look so stupid right now

Interval between -pi/2 and pi/2, y intercept of the sec graph is (0,1), so the ran is [1,infty)!

Thanks

[Andiio]

Don't you mean:

   0 < cos(t) < 1
Thus infinity < 1/cos(t) < 1?

[brightsky]

Don't you mean:

   0 < cos(t) < 1
Thus infinity < 1/cos(t) < 1?

Noo..that doesn't even make sense logically. :p

Re: Note that y gets bigger as cos(t) gets smaller positively so as cos(t) approaches 0, 1/cos(t) approaches a very very big number while on the opposite end, when cos(t) approaches 1, 1/cos(t) approaches 1

[Andiio]

Oops...misinterpreted what you typed xD

mmm but isn't the inverse/reciprocol of 0 is undefined, and the inv/recip of 1 is 1?
:\

[brightsky]

Oops...misinterpreted what you typed xD

mmm but isn't the inverse/reciprocol of 0 is undefined, and the inv/recip of 1 is 1?
:\

Yep, but algebra (especially in inequations) is weird. Don't think it's 'legal' to just do 1 over the whole inequation like that. xD

[Andiio]

yeah LOL, and the cos of -pi/2 and pi/2 are both 0, aren't they..? :|

[brightsky]

yeah LOL, and the cos of -pi/2 and pi/2 are both 0, aren't they..? :|

Yep, but visualise it on a cosine graph. The "t" is the x-axis, and we have restrictions at -pi/2 and pi/2. The graph goes from 0 at t = -pi/2 up to 1 and then down again to 0.

[Andiio]

Mmm yeah, haha!
You sure love breaking the laws of algebra, don't you brightsky?

[luken93]

If you need anything else I asked a similar Q here:
http://vce.atarnotes.com/forum/index.php/topic,34901.0.html