Halils Help Thread for Specialist

[Halil]

Hi there guys,

Let this be my thread to put up questions for specialist which i need help with. Having alot of trouble with extended response questions but good with the others.

Ill post up every question that I'll be needing help with

[onur369]

Good on you Deborah, thumbs up. All the best with Spesh this year.

[Halil]

Thanks Owner Boner. It's going to help out with my methods I guess, so it will be worth it. Im studying really good for it so InsAllah I'll get a decent mark. 30+ would be jizzing

[onur369]

LOL jizzing :/ Finished Ch1 Methods In 4 days. So I'm putting in tonnes of effort for methods. Im gnna finish Short Answer of Ch Review and I finished the prac test you gave to us on Friday :p

[Halil]

Got a problem here.

Chapter is Coordinate geometry.

When you have a question such as this one, and you have restrictions for the 'T' values inside the sine and cosine such as [-(pie/2),(pie/4)] what does it actually mean or do?

'Find the cartesian Equation when:

y=3+2sin(t) and x=2+3cos(t)'    t = [0, 2pie]

[TrueTears]

restriction on y(t) places restriction on the range of cartesian

for x(t) places restriction on the domain of cartesian

for the actual q

(y-3)/2 =sint and (x-2)/3=cost

square both sides of the above and add, rest is obvious

[kamil9876]

I assume you are asking for the significance of the domain of . To illustrate this note that the set of (x,y) values in here is an ellipse. If however our domain was instead then it would only be a semi-ellipse. You can basically think about it as some guy subbing in t=0, t=0.1... (subbing in all possible values of t in the domain) and plotting each point and eventually getting a shape. Practically there is no general recipe to see what the restriction of a domain does(inspection usually needed) but for circles it is easy as is usually the set of all possible angles made with an axis and hence for ellipses it should be easy too as an ellipse is just dilated circle.

e.g:

, with would be the part of the semi circle about the origin in the 2nd and 3rd quadrant (with (0,-1) removed, why?)

, with would then be the exact same shape, just stretched by a factor of 3 in the x-axis and a factor of 2 in y-axis, hence a semi-ellipse in the 2nd and 3rd quadrant.

[Halil]

Made an error there sorry. Its sec not  sin (t). So its a hyperbola graph.
Similiar thing with the hyperbola graphs?

[Halil]

Another quick question, how do i find the x intercepts of a hyper bola function/graph.
I have a test tomorrow on it, and some how the book is not precisely explaining it.

I tried cross multiplying but it gives a linear function which means you will obtain a single intercept. But you need two, because the graph will intersect at two points.
The book uses the quadratic formula, however does not show how it obtained the result. In a normal function you require the a,b,c values, however on this type of function you pretty much only have a,b values..

Anyone?

[onur369]

y=0 , For example lets say.
The equation is y= 2/(x+2) -4
Make y=o, Then move over the -4.
4= 2/(x+2)
Cross multiply the 4 and (x+2)
That will be 4x+8=2
4x=-6
x=-6/4
x= -3/2

[jasoN-]

where a, b, c, d are all constants

x-int, y=0:

[Halil]

Onur, read my question brother. You need two x intercepts, your solution which i already explained gives you 1 intercept.

Thanks for the help Jason, i think its right

[onur369]

Just realized... >.<

[Halil]

Actually, Jason, that formula aint working buddy.

[pi]

Don't use formulas for asymptotes and intercepts... Understand how they come about in the first place (limits, zeroes, stc.). That way you'll be able to tackle harder/'out-of-the-box' questions come exam time.