Post by: golden on December 03, 2010, 07:37:14 pm
I've looked through videos and haven't seen much on hyperbola parameters.
With this question:
Determine the:
1. Equation.
2. Domain and Range.
x = sect, y = tant, t E (
This is my working out:
tan^2t + 1 = sec^2t
y^2 + 1 = x^2
x^2 - y^2 = 1
How would I determine the domain and range?
Thanks in advance. + Karma for those who help!
EDIT:
Asymptotes:
y = x, y = -x.
Post by: dptjandra on December 03, 2010, 07:57:59 pm
Range is the set of values that y can take...so similarly, y = tan(t). if we sketch for t: (pi/2, 3pi/4) --> range = y E (-infinity, -1)
Is that the right answer? (My trig values are a bit rusty)
In short, you find the range for each component equation to find the values for x (the domain) and the values for y (the range).
Post by: golden on December 03, 2010, 08:13:52 pm
Domain is the set of values that x can take...that is, between t: (pi/2, 3pi/4), what values are x = sec(t)? --> you have to quickly sketch it out to see that x E (-infinity, -rt(2))
Range is the set of values that y can take...so similarly, y = tan(t). if we sketch for t: (pi/2, 3pi/4) --> range = y E (-infinity, -1)
Is that the right answer? (My trig values are a bit rusty)
In short, you find the range for each component equation to find the values for x (the domain) and the values for y (the range).
The question has been updated. It was 3pi on 2 not on 4. Please excuse that.
The answers are:
Domain: (-infinity, -1)
Range: R.
The graph drawn is the left half of the hyperbola.
Is drawing out the only main way of figuring it out? There isn't a faster way?
Thanks for the answer!
Post by: dptjandra on December 03, 2010, 08:28:41 pm
Problem with not sketching is the cyclic nature of the trigonemtric function. To illustrate, imagine that I subbed in the values of t to find x and y:
t = pi/2 --> x = undefined (-ve or +ve infinity), y = undefined (-ve or +ve infinity).
t = 3pi/2 --> x = undefined (-ve or +ve infinity), y = undefined (-ve or +ve infinity).
This isn't very instructive...If you sketch both graphs, however, you will see that x = sec(t) gives a single "upside down U" branch and y = tan(t) gives one full "tan" branch.
If you're very experienced with trigonemtric values (especially the tan graph), you'll probably recognise that for tan, pi/2 and 3pi/2 are two adjacent asymptotes, and therefore all real numbers are covered by y = tan(t). I find it harder to do this with a sec(t) graph in my head, though one could arguably recognise that similarly, pi/2 and 3pi/2 are adjacent asymptotes. You then know that cos(t) is negative between pi/2 and 3pi/2, so you know that sec(t) will be negative too. The minimum magnitude of a sec(t) graph is 1 (ie. |1| or |-1|). This is pretty unnecessarily complicated, and I usually draw a very, very rough sketch to sort it out. I don't know of any "easier" way of doing it...so if others know if there's a faster way of working out the domain than sketching, I'd like to learn it too.
In either case, I would argue that the rough sketching method (especially with a calculator handy to check) provides an extra layer of security, since there is no guarantee that the two asymptotes will always be adjacent anyway and the equations given in the parametric can be much more complicated...
Post by: golden on December 03, 2010, 08:35:37 pm
With the new values, simply do the same process going up to 3pi/2. You get the answers you're looking for.
Problem with not sketching is the cyclic nature of the trigonemtric function. To illustrate, imagine that I subbed in the values of t to find x and y:
t = pi/2 --> x = undefined (-ve or +ve infinity), y = undefined (-ve or +ve infinity).
t = 3pi/2 --> x = undefined (-ve or +ve infinity), y = undefined (-ve or +ve infinity).
This isn't very instructive...If you sketch both graphs, however, you will see that x = sec(t) gives a single "upside down U" branch and y = tan(t) gives one full "tan" branch.
If you're very experienced with trigonemtric values (especially the tan graph), you'll probably recognise that for tan, pi/2 and 3pi/2 are two adjacent asymptotes, and therefore all real numbers are covered by y = tan(t). I find it harder to do this with a sec(t) graph in my head, though one could arguably recognise that similarly, pi/2 and 3pi/2 are adjacent asymptotes. You then know that cos(t) is negative between pi/2 and 3pi/2, so you know that sec(t) will be negative too. The minimum magnitude of a sec(t) graph is 1 (ie. |1| or |-1|). This is pretty unnecessarily complicated, and I usually draw a very, very rough sketch to sort it out. I don't know of any "easier" way of doing it...so if others know if there's a faster way of working out the domain than sketching, I'd like to learn it too.
In either case, I would argue that the rough sketching method (especially with a calculator handy to check) provides an extra layer of security, since there is no guarantee that the two asymptotes will always be adjacent anyway and the equations given in the parametric can be much more complicated...
I get it now! Thanks!
Post by: golden on December 03, 2010, 08:36:34 pm
Post by: dptjandra on December 03, 2010, 08:38:07 pm