VCE Specialist 3/4 Question Thread!

[Zues]

and in this case wit the reciprocal, why have they just changed the y value and not the x?

[Zues]

Not totally sure on what you said Zezima with As such, the output value - the y value - not the input (x-value) should be reciprocated.

[M_BONG]

Not totally sure on what you said Zezima with As such, the output value - the y value - not the input (x-value) should be reciprocated.

If you think about it, you don't actually flip the x-value because it's an input value.

Say I have a relation f(x) = x and g(x) is its reciprocal g(x)= 1/x

What you sub in as x (the input) doesn't actually matter - anything except x = 0 in this case because denominators cannot equal zero. It's the y-coordinate (the output) that changes because you have already flipped the relation.

So going with the two relations, I sub in x=2.

when I sub x =2 into f(x), I get 2
with I sub x = 2 into g(x), I get 1/2.

Do you see how only the y-coordinate changes?

Don't worry, this first chapter is pretty tricky - screwed around with me as well.

[Zues]

does this applie to y intercepts etc?

yeap that makes sense now.

quick question, what do they mean "in the same quadrant" in terms of stuff like f(x) = 1/ g(x) because both are in the same quadrants.

Also, why do these have 1 top 1 bottom 1 top (which i understand) and this other one has 2 in the bottom ?

[IndefatigableLover]

Find the Cartesian Equation of the curves whose vector equations are given:



So I managed to get it to it's parametric equation however I just have a bit of trouble getting to the final answer...

My parametric equations were:
and
But whatever I do, I end up getting some weird answer or end up with two values of for 't' but don't know which one to cancel (if that's right)?

[M_BONG]

Find the Cartesian Equation of the curves whose vector equations are given:



So I managed to get it to it's parametric equation however I just have a bit of trouble getting to the final answer...

My parametric equations were:
and
But whatever I do, I end up getting some weird answer or end up with two values of for 't' but don't know which one to cancel (if that's right)?

How are you so keen? I learned parametric equations in term 3, lol.

Is the equation y = x^2 + 2?

Btw, this is pretty tricky...
HINT: square the x component. Then realise that by squaring the x-component you get something remotely similar to the y-component. Then substitute/manipulate it so it becomes identical to the y-value.

[IndefatigableLover]

How are you so keen? I learned parametric equations in term 3, lol.

Is the equation y = x^2 + 2?

Btw, this is pretty tricky...
HINT: square the x component. Then realise that by squaring the x-component you get something remotely similar to the y-component. Then substitute/manipulate it so it becomes identical to the y-value.

Yeah it is actually.. could you show me the substituting/manipulating part please? That's the part I don't get and I've been stuck on it for a while LOL

And weirdly the school made us use Mathematica to work out parametric equations this year already so I'm just brushing up I suppose since we went over it a few weeks back

[cosine]

Hey everyone, has anyone got some advice for a bloke who is really keen to absolutely destroy specialist next year?
It would be great if anyone could share their personal opinions/experience and actually how hard it is?
Much appreciated!

[keltingmeith]

Yeah it is actually.. could you show me the substituting/manipulating part please? That's the part I don't get and I've been stuck on it for a while LOL

Our parametric equations are and , so let's just do random things and hope for the best.

y is all squared, so let's try squaring the x part, getting . At this point, two thirds of the RHS looks like y - so, let's make ALL of it look like way. Transposing, we get . Finally, we can simplify the first fraction, and we get . Important to note the parametric equation isn't defined when t=0, but if you take limits to figure out what values of x/y aren't defined, you'll get infinities, so we don't need to worry about that point this time.

If you have a TI-nspire, it also has a parametric graphing option, so if you get stuck you can always use that and have a guess at the Cartesian equation.

And weirdly the school made us use Mathematica to work out parametric equations this year already so I'm just brushing up I suppose since we went over it a few weeks back

... Mathematica...? Instead of your CAS? O.o

[IndefatigableLover]

Our parametric equations are and , so let's just do random things and hope for the best.

y is all squared, so let's try squaring the x part, getting . At this point, two thirds of the RHS looks like y - so, let's make ALL of it look like way. Transposing, we get . Finally, we can simplify the first fraction, and we get . Important to note the parametric equation isn't defined when t=0, but if you take limits to figure out what values of x/y aren't defined, you'll get infinities, so we don't need to worry about that point this time.

If you have a TI-nspire, it also has a parametric graphing option, so if you get stuck you can always use that and have a guess at the Cartesian equation.

... Mathematica...? Instead of your CAS? O.o

Haha thanks for that EulerFan101 (didn't look at this till now but solved just before :')

And yeah Mathematica! We have a teacher who's properly trained in giving lessons with Mathematica so we learn how to use it and when our exams come up, you have the option of using Mathematica during Exam 2 (if you want or just use the CAS or both) I know I can't use it properly since I spend more time on the Help files than actual programming LOL (but there's some people at school who are really good at it) ._.

[keltingmeith]

And yeah Mathematica! We have a teacher who's properly trained in giving lessons with Mathematica so we learn how to use it and when our exams come up, you have the option of using Mathematica during Exam 2 (if you want or just use the CAS or both) I know I can't use it properly since I spend more time on the Help files than actual programming LOL (but there's some people at school who are really good at it) ._.

Impressive! :O
Mathematica's definitely stronger and more useful, but the CAS is (imo) faster, and you don't need all the fancy Mathematica functions for VCE.

Plus, the smarties can program their CAS for extra help - can't do that with Mathematica.

[IndefatigableLover]

Impressive! :O
Mathematica's definitely stronger and more useful, but the CAS is (imo) faster, and you don't need all the fancy Mathematica functions for VCE.

Plus, the smarties can program their CAS for extra help - can't do that with Mathematica.

Haha definitely agree with that LOL (thankfully I don't need to use it for my VCE exams next year :') and yeah all them macros for your CAS (there's some floating around at our school but I definitely know that MHS makes good use of those programs for extra help )

[Zues]

with this question how is the range [-3,3]. the domain says [0, pi] and thus shouldnt the range be [0,3]

[keltingmeith]

with this question how is the range [-3,3]. the domain says [0, pi] and thus shouldnt the range be [0,3]

So, t goes between 0 and pi - but the angle isn't t, it's 2t, and 2t goes between 0 and 2pi, so you've got a full revolution - ergo, the range is [-3, 3].

[Zues]

ok thanks, but i thought with the 2t it would mean pi/2 hence quater of a circle? or is it because t is [0,pi] hence 2t is [0, 2pi]? which thought process is correct?

also,
x = 2-4cos(t)

how to solve for cos(t) (make this the subject), i know this is a very simple question but id go along the lines of

x-2/(-4) = cost
get rid of negative, so becomes x+2/4, why is their meant to be a negative before the "x"?

whats actually negative in this case, the whole expression or just the 4, i know its the former thus -x-2, devide by negative is x+2 ?
answer attached,