Parametrics help (1 Viewer)

[verdades]

So I've done this once, at the start of this term. Went over it again with my teacher.
I've gone back and redone most of the simple locus from 2 unit year 11 work.
But I still don't quite get it.

Could someone please explain the what/why/how for me?

 

[lyounamu]

So I've done this once, at the start of this term. Went over it again with my teacher.
I've gone back and redone most of the simple locus from 2 unit year 11 work.
But I still don't quite get it.

Could someone please explain the what/why/how for me?

Um, give us the question? It's just hard to explain.

Only thing I can really say is that parametrics is another way of representing the function by using another unknown such as t or p or q or whatever.

 

[verdades]

Just, parametrics in general.
The entire concept.
I just found the link to the Biki for it, though.

So I s'pose x=2at and y=at^2.
x^2=4ay.

I suppose it shows my lack of understanding that I'm not even sure what to ask xD.

 

[gurmies]

Basically, the entire parametrics topic is just a mechanism to solve problems where you have a parameter (t or whatever) connecting up x and y.

 

[verdades]

Basically, the entire parametrics topic is just a mechanism to solve problems where you have a parameter (t or whatever) connecting up x and y.

Express both in terms of the parameter, get yourself a simulatenous equation and solve for x or y?

And use calculus [or dy/dt x dt/dx] for gradient of tangent, recipricol of tangent gradient for normal?

 

[gurmies]

^ Yep thats perfect. Suppose I gave you the co-ordinates:

(2at, at^2)

Well, you know that x = 2at, and y = at^2

t = x/2a

Subbing into y, y = a(x^2/4a^2)

= x^2/4a

x^2 = 4ay...

Similar mechanism for finding out other loci. Always remember though, if you have something like:

(2at, 3a), and you start to think...hmm..what's the locus of this, I can't connect t..well that's becsaue the y co-ordinate is independent of the parameter. Hence locus is y = 3a.

 

[Timothy.Siu]

just do questions, ur not gonna learn it just from people explaining it to you
and read the textbook, if u cant do a question ask,

 

[verdades]

xD thanks everyone.

 

[Zeber]

locus's is mainly about eliminating the parametrics ie. t

as y is the variable that varies with x, and a is a constant.

 

[pyrodude1031]

Simply speaking, the whole idea behind parametrics is the ability to express the (x,y) coordinates in terms of an alternative coordinate , which for the matter of HSC, a single parameter, say t.

What this means is that...ermmm.. take this situation. Say, a rich man like Bill Gates own houses across the globe. He chooses to live in each house according to that day's temperature t. And, we use (x,y) to define the longitude and latitude of the location of his residence. Hence, we can say his resort is located at (x,y) which is a function of t. i.e. (x,y) = (f(t),g(t)) or x = f(t) & y = g(t).

Questions in the hsc, will often give you a number of conditions, tangents crossing and normals crossing etc.. THe aim is to express the intersection point (or whatever the condition is) in terms of the parameters (p,q if two tangents exist). and then using algebraic simplification, simulataneous equations, roots of quadratics, discriminants to find the equation(s) which define the point of intersection traces as it moves.

As previous replies , x = 2ap & y = ap^2 defines a point which traces out the parabola x^2 = 4ay.

Another example is x = acost & y = bsint defines an ellipse of equation x^2/a^2 + y^2/b^2 =1.

2005 HSC Graduate (Killara)
EngAdv 82 MX1 98 MX2 95 Phys 93 Chem 90 FrCont 67
B.E. (Biomedical) B.Sc. (Advanced Mathematics) USYD