Parametric domain/range

[dekoyl]

I don't completely understand Essential's explanation
My question is: when we find the domain or range of a cartesian equation using the parametric form, why do we use the range of the trigonometric function? For example:


The domain of the graph (ellipses) when sketched would be [-3,3] because the range of x = 3sin(t) is [-3,3] for . And same goes for the range of the graph (range of y = 4cos(t) is [0,4] for 

Thanks 

[unknown id]

^ This is because the domain of x(t) [ie. the defined t values] determines the values for which x(t) is defined (its range); that is, it sets up the domain (ie. the defined values of x) of the graph of y(x), where y is in terms of x.

domain t sets up the range of x(t)
range of x(t) sets up the domain of y(x)

[shinny]

The domain of the actual graph is basically what values x can equal. The only restriction we have on this is in regards to its relation with t. Using its relation with t, we know that the range of x(t) is [-3,3], so these are the actual values that x can equal. The same applies for y in regards to the range of the actual graph. I'll elaborate further if you still don't understand.

[dekoyl]

The domain of the actual graph is basically what values x can equal.

Yep.

The only restriction we have on this is in regards to its relation with t.

Yep.

Using its relation with t, we know that the range of x(t) is [-3,3], so these are the actual values that x can equal.

Not so yep

Could you possibly elaborate, shinny? =[
Why is the domain of x(t) or y(t) never used? Why only the range?


Thanks unknown id and shinny for trying to help a slow boy out

[Collin Li]

^ This is because the domain of x(t) [ie. the defined t values] determines the values for which x(t) is defined (its range); that is, it sets up the domain (ie. the defined values of x) of the graph of y(x), where y is in terms of x.

domain t sets up the range of x(t)
range of x(t) sets up the domain of y(x)

This explanation is gold. Try to understand it.

Basically, you use the restrictions on to find out how and are restricted - since you have and .

Then, the restrictions on - the range of - become the domain of .

The restrictions on - the range of - become the range of .

[shinny]

Might be easier if I just list them out systematically
1. The domain of a typical graph is the values that x can take
2. The range of a typical graph is the values y can take
3. The parametrics state that x=g(t), y=h(t)
4. The values that x can take (i.e. the domain!? link this back to statement 1) is equal to the values that g(t) can take, because x=g(t)
5. The values that g(t) can take is basically the range of g(t)
6. Therefore the domain of x is the range of g(t)
Apply a similar train of thought for y to find the range

I might have to write up an example or something to make it even more obvious if you don't get this =\

[dekoyl]

F''' yeah. Thanks everyone I get it

[dekoyl]

Are parametric stuff usually in the non-calculator able exam?

Sorry double post =P

[shinny]

It can be in either exam as it'd be part of a larger vector calculus question.