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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Halil on February 06, 2011, 08:12:36 pm

Title: Halils Help Thread for Specialist
Post by: Halil on February 06, 2011, 08:12:36 pm: 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
Title: Re: Halils Help Thread for Specialist
Post by: onur369 on February 06, 2011, 08:14:24 pm: Good on you Deborah, thumbs up. All the best with Spesh this year. :D
Title: Re: Halils Help Thread for Specialist
Post by: Halil on February 06, 2011, 08:28:32 pm: 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
Title: Re: Halils Help Thread for Specialist
Post by: onur369 on February 06, 2011, 08:31:12 pm: 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
Title: Re: Halils Help Thread for Specialist
Post by: Halil on February 07, 2011, 09:29:46 pm: 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]
Title: Re: Halils Help Thread for Specialist
Post by: TrueTears on February 07, 2011, 09:37:16 pm: 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
Title: Re: Halils Help Thread for Specialist
Post by: kamil9876 on February 07, 2011, 09:46:45 pm: I assume you are asking for the significance of the domain of $t$ . To illustrate this note that the set of (x,y) values in here is an ellipse. If however our domain was instead $[0,\pi]$ 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 $t$ 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:

$x=cost$ , $y=sint$ with $t \in [\pi/2, 3\pi/2)$ would be the part of the semi circle about the origin in the 2nd and 3rd quadrant (with (0,-1) removed, why?)

$x=3cost$ , $y=2sint$ with $t \in [\pi/2, 3\pi/2)$ 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.
Title: Re: Halils Help Thread for Specialist
Post by: Halil on February 08, 2011, 09:23:42 pm: Made an error there sorry. Its sec not sin (t). So its a hyperbola graph.
Similiar thing with the hyperbola graphs?
Title: Re: Halils Help Thread for Specialist
Post by: Halil on February 10, 2011, 07:00:05 pm: 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?
Title: Re: Halils Help Thread for Specialist
Post by: onur369 on February 10, 2011, 07:08:03 pm: 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
Title: Re: Halils Help Thread for Specialist
Post by: jasoN- on February 10, 2011, 07:09:39 pm: $\frac{(x-a)^2}{b} - \frac{(y-c)^2}{d} = 1$ where a, b, c, d are all constants
$\frac{(x-a)^2}{b} - \frac{(y-c)^2}{d} = 1$
x-int, y=0:
$\frac{(x-a)^2}{b} - \frac{(0-c)^2}{d} = 1$
$\frac{(x-a)^2}{b} - \frac{c^2}{d} = 1$
$\frac{(x-a)^2}{b} = 1 + \frac{c^2}{d}$
$\therefore x = \pm \sqrt{b(1+\frac{c^2}{d}}) + a$
Title: Re: Halils Help Thread for Specialist
Post by: Halil on February 10, 2011, 07:15:08 pm: 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
Title: Re: Halils Help Thread for Specialist
Post by: onur369 on February 10, 2011, 07:15:39 pm: Just realized... >.<
Title: Re: Halils Help Thread for Specialist
Post by: Halil on February 10, 2011, 07:24:43 pm: Actually, Jason, that formula aint working buddy.
Title: Re: Halils Help Thread for Specialist
Post by: pi on February 10, 2011, 08:36:09 pm: 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.
Title: Re: Halils Help Thread for Specialist
Post by: Halil on February 15, 2011, 06:31:13 pm: Question is:

Find the exact value of the following

$csc(((5*π)/(12)))$

I turned it into 1/sin(5pie/12)

then used $(π/6 + π/4)$

In the end i got

$((1)/(((√(6)+√(2))/(4))))$

which also equals to:

$((4)/((√(6)+√(2))))$

However, answer is just $(√(6)-√(2)$

How? What did I do wrong?
Title: Re: Halils Help Thread for Specialist
Post by: Halil on February 15, 2011, 06:32:38 pm: Sorry i dont know why it came out like that, the &#8730 ones means square root and the 960 ones are pie
Title: Re: Halils Help Thread for Specialist
Post by: Halil on February 15, 2011, 06:43:07 pm: Exactly same problem and and answer for sec (pie/12)
Title: Re: Halils Help Thread for Specialist
Post by: pi on February 15, 2011, 06:43:40 pm: Fixed it for you.

For next time:
π or pie (no) or pi = \pi
√ = \sqrt{...}
For fractions = \frac{...}{...}

Quote from: Halil on February 15, 2011, 06:31:13 pm
Question is:

Find the exact value of the following

$csc(\frac{5\pi}{12})$

I turned it into $\frac{1}{sin(\frac{5\pi}{12})}$

then used $(\frac{\pi}{6} + \frac{\pi}{4})$

In the end i got

$\frac{1}{\sqrt{6}} + \frac{\sqrt{2}}{4}}$

which also equals to:

$\frac{4}{\sqrt{6}+\sqrt{2}}$

However, answer is just $\sqrt{6}-\sqrt{2}$

How? What did I do wrong?

Quote from: Halil on February 15, 2011, 06:43:07 pm
Exactly same problem and and answer for sec (pie/12)

Its 'pi'...
Title: Re: Halils Help Thread for Specialist
Post by: pHysiX on February 16, 2011, 02:34:15 pm: Hint: rationalise $\frac{4}{\sqrt{6}+\sqrt{2}}$ and you shall have your answer. Remember, to be mathematically elegant, we never want surds in the denominator.