Maths Methods 3/4 Help Thread 2011

[pi]

what happens when a square root is in the denominator for graphs?

eg. "The Maximal domain of the function with rule f(x)=9/rootx^2-1 is"

Domain is when bit inside the root exists and the fraction exists. ie. when (x^2 - 1) > 0. Solve this via a quick sketch graph.

[xZero]

It's basically the same as 1/x^2, it has a slower rate of decrease but you have to account for the sq root when finding the domain. For the q above, the max domain is -infinity<x<-1, 1<x<infinity (assuming both x^2 and 1 are in the sq root)

[iNerd]

Is there any easy, systematic, formulaic way of sketching insane trig graphs with x-translations.

Eg: y = -2 Sin (x/6 - pi/3) -2 in the domain [-pi,2pi]

[xZero]

you can either sketch the graph with dilation factor and correct period first then add the translations in or the way i like to do is draw a new set of axes on the axes given which has the correct x-translations (so if the graphs been shifted by 1 unit to the right, draw the new y-axis at x=1) and just draw a normal trig graph about the new axes.

[Matt-Ong]

Hey, I have a question. I can't really understand it so if you can help me, it'll be really appreciated.

Question:
P(x)= x^2 - 4x - 13/2 (parabola)
T(x)= (-24/(x-2)^2) + 5  (truncus)
A particle moves along a path that can be described using the Cartesian Equations described by the functions P(x) and T(x). The particle's horizontal distance is x and the particle's verticle distance is y.
So that; y= P(t) and x= T(t)

i) Write down your equations for y and x
ii) Fina an equation relating the horizontal and vertical distances so that you are expressing y in terms of x.
iii) Hence find the rate of change of y with respect to x.

[brightsky]

i) quite trivial, just substitute x with t in the original equations
ii) we know that y = t^2 - 4t - 13/2. we have two solutions for t,
t = 2 - sqrt(y + (21/2)) or;
t = 2 + sqrt(y + (21/2))

substitute them the first into x = T(t):
x = (-24/((2 - sqrt(y + (21/2)))-2)^2) + 5 = (10y + 57)/(2y+21) which is the same as if we substitute the second one due to the square.
rearrange to get y in terms of x:
y = 3(19 - 7x)/2(x-5)
iii) hence dy/dx = 24/(x-5)^2 after some calculation

[ech_93]

I'm not sure how to do this question, as you're not allowed to use your calculator for it....

26=22-4cos[pi(t+4)]/12. Find t.


Just to clarify; the stuff in the [ ] is all divided by 12.
Thanks!

[thushan]

you know that cos[pi(t+4)/12] = -1

so pi(t+4)/12 = (2n+1)pi, and you can then solve for the general solution

[ech_93]

Sorry, I don't understand!
If I'm trying to find the first value of t, how do i do it?
(t is meant to be hours after 6am, and the answer is 2 pm)



edit. do i just put 1 in for n, then solve?

Okay, i think i know how to do it now. But, where did you get (2n+1)pi from?

[thushan]

Oh sorry, I was giving the general solution.

[Tobias Funke]

With matrices - what happens to the transformations if it's been inversed?

need to transform graph y= square root x according to the following transformations

[ 2y+1 ]
[ -x+3 ]

[b^3]

Do you have to do it by matrix methods? because there is a simpler way if you don't.

[Tobias Funke]

well, i need to solve it with that matrix, and i don't know how

[pi]

With matrices - what happens to the transformations if it's been inversed?

need to transform graph y= square root x according to the following transformations

[ 2y+1 ]
[ -x+3 ]

Is this the practice SAC MC Questions?

You have to solve for x and y, not x' and y'. Type the equation into your calc (with all the matrices), use solve (maybe let a=x' and b=y'). Then plug in those solutions into y=sqrt(x). Done. Answer is E (from memory).

[Tobias Funke]

I'm getting x=-(b-3) and y=a-1/2

not entirely sure where to go from here?