Parametric Dom/Ran Q's?

[luken93]

Although there are a few threads already on this, could someone explain how the Essentials Q's get the values for their domain and range?

Especially Q2, are you meant to know what sec graphs look like to find out what the available values that x can take from the limitations of t?

And how do the values of the sec/tan graphs link in to the cartesian dom/ran?

Thanks

[/0]

Yeah, it's essential you know what sec graphs look like, as well as all other fundamental trig graphs.

The domain of the corresponding cartesian equation is the 'range of x', and the range of the corresponding cartesian equation is the 'range of y'. So in question 2 u need to know the range of and in the given interval

[jimmy999]

A good way to do this is to sketch x against t and y against t for the given t-domain. Then the maximum and minimum x and y values will give the domain and range respectively.

For sketching sec, it's useful to remember what it looks like, however you can always construct it yourself. First sketch cos, then sketch the reciprocal of that (If you don't know how to do that, basically large values become small, small values become large).

To convert sec/tan into the cartesian x/y equations, you'll have to use an appropriate trig identity (Look on the formula sheet and you'll find it easily).

[luken93]

A good way to do this is to sketch x against t and y against t for the given t-domain. Then the maximum and minimum x and y values will give the domain and range respectively.

OK so from what I gather from your posts, if I have

x = 2 sec (t) and y = 2 tan (t)

I'd graph each over the t domain [- pi/2 , pi/2], and then the range x values on the sec = domain, and the range of y values on tan = range?

[dptjandra]

yup

[luken93]

yup

argh its such a wast of time haha, i spose i just need to learn the sec graph and then i should kinda be able to remember the range of each...

[dptjandra]

Yeh - especially if you have a calculator, you'll find it pretty manageable.  Also, you need to be able to graph it anyway for exam 1 so it's good practice. It's faster than the alternative mental strategy I discussed here: http://vcenotes.com/forum/index.php/topic,34790.0.html at least

[98.40_for_sure]

Hey luke, are you working through essentials from start to end? I remember we did parametrics near the end along with vector calculus because it's easier to pick it up after all the other stuff which leads to it, namely reciprocal graphs and trig identities.

[QuantumJG]

With 8

First thing to note:



So we have:





So:



So the shape is an ellipse centered at (2,4) with semi major axis of 3 and semi minor axis of 2

a) 0 < t < 0.25

    2 < x < 5

    4 < y < 6

So it's the top right quarter of the ellipse.

b) 0 < t < 0.5

    2 < x < 5

    2 < y < 6

So it's the right half of the ellipse.

c) 0 < t < 1.5 well since the period of each graph is 1 that is just going around again.

   -1 < x < 5
 
    2 < y < 6

So it's the whole elipse.

Parametrising a path is very useful in vector calculus.
   

[QuantumJG]

with the next one.

Know that:

[luken93]

Yerp trig identities are fine to find the cartesian eqs etc, it was only finding the domain and range that was a problem as Essentials seem to pluck it out of thin air...

Hey luke, are you working through essentials from start to end? I remember we did parametrics near the end along with vector calculus because it's easier to pick it up after all the other stuff which leads to it, namely reciprocal graphs and trig identities.

Nah mate, its in Chap 1 (Toolbox), but we aren't tackling it until later in the year but we were asked to do Chap 1 for Hol HW

[luken93]

Thanks everyone, I'm pretty sure I know what to do now, just gonna take longer than I thought haha

[recovered]

With 8

First thing to note:



So we have:





So:



So the shape is an ellipse centered at (2,4) with semi major axis of 3 and semi minor axis of 2

a) 0 < t < 0.25

    2 < x < 5

    4 < y < 6

So it's the top right quarter of the ellipse.

b) 0 < t < 0.5

    2 < x < 5

    2 < y < 6

So it's the right half of the ellipse.

c) 0 < t < 1.5 well since the period of each graph is 1 that is just going around again.

   -1 < x < 5
 
    2 < y < 6

So it's the whole elipse.

Parametrising a path is very useful in vector calculus.
   

hey mate  i just wanted to knw is there a way to work the answers backwards i mean to get the vector function from the answer that they give you
can it work for any question
cheers