VCE Methods Question Thread!

[Sine]

This question has been asked numerous of times on this forum before.

Have a read of this document which VCAA released 7 years ago due to the confusion.
http://www.vcaa.vic.edu.au/Documents/bulletin/2011AprilSup2.pdf

[IonianDeo]

How do you address complex-to-complex transformations using any method (function, dash, mapping)? Using the attached question as an example.

[S200]

How do you address complex-to-complex transformations using any method (function, dash, mapping)? Using the attached question as an example.

You could try working out the initial transformations from
And then just add the transformations to each other...

[Lear]

How do you address complex-to-complex transformations using any method (function, dash, mapping)? Using the attached question as an example.

I would personally move all the y transformations to one the y side and all the x transformations to the other. Then you can individually solve the two functions to see what is being transformed and how.
I have an example attached where I found the transformation matrix to transform a complex cos function to another. See if you get the general idea and if not I’ll do it for your question too.

[PolySquared]

Hi guys.

How do you solve this question?

How do you know the restrictions on each of the equations because there aren't any open or closed circles on the ends of the lines.

https://imgur.com/a/WpLyFby

[WrongWong]

Hi guys.

How do you solve this question?

How do you know the restrictions on each of the equations because there aren't any open or closed circles on the ends of the lines.

https://imgur.com/a/WpLyFby

You'd assume the function is continuous, don't worry about it too much, a VCAA exam would specify.

[PolySquared]

You'd assume the function is continuous, don't worry about it too much, a VCAA exam would specify.

So how does that influence your answer? I have a SAC tomorrow and I dont understand hoe to find the restrictions on each of the graphs because there arent any open or closed circles on the ends.

[hawkrook]

Got a couple of questions. First, when i plug 0, 10 and 24 into the function for the t-value, to output the high tide of 8m, I get different values for the phase shift. When I plug in t = 0 i get pi/2 but then when i plug in t=10 i get phase shift = -3pi/2. Why is this happening and what would i say the value for the phase shift is?

Secondly, the textbook worked solutions always seem to start off with using a sine function for these worded questions but then they convert it to a cos function, once they find all the parameters for the function. Why do they do this and is it necessary for me to do this?

[A TART]

Got a couple of questions. First, when i plug 0, 10 and 24 into the function for the t-value, to output the high tide of 8m, I get different values for the phase shift. When I plug in t = 0 i get pi/2 but then when i plug in t=10 i get phase shift = -3pi/2. Why is this happening and what would i say the value for the phase shift is?

Secondly, the textbook worked solutions always seem to start off with using a sine function for these worded questions but then they convert it to a cos function, once they find all the parameters for the function. Why do they do this and is it necessary for me to do this?

I'm not sure where you got  0, 10 and 24 from, but you only need 1 piece of information for the phase shift. I assume you've figured out  A, n and b (Average value is 5, hence 5 up, b=5; Period is pi/6 because we know that 12=2pi/n; a=3 because we know that hightide (max) is 8m, 8-5=3m).

We also know that At noon (t=0) there is a high tide. Hence at need t=0, h(t)=8, we can solve for the final value e (the phase shift).

The textbook converts to cosine because sin (x+pi/2)= cos(x) (It's a bit easier to deal with when working)

[hawkrook]

I'm not sure where you got  0, 10 and 24 from, but you only need 1 piece of information for the phase shift. I assume you've figured out  A, n and b (Average value is 5, hence 5 up, b=5; Period is pi/6 because we know that 12=2pi/n; a=3 because we know that hightide (max) is 8m, 8-5=3m).

We also know that At noon (t=0) there is a high tide. Hence at need t=0, h(t)=8, we can solve for the final value e (the phase shift).

The textbook converts to cosine because sin (x+pi/2)= cos(x) (It's a bit easier to deal with when working)

Sorry, I meant 12 instead of 10. I understand that there will be a high tide at t = 0, and then you can use that to solve for n. But there will also be a high tide at t = 12 and t = 24 but when i use those values i get a different phase shift.

Will I lose marks if I don't convert back to cos?

[A TART]

Sorry, I meant 12 instead of 10. I understand that there will be a high tide at t = 0, and then you can use that to solve for n. But there will also be a high tide at t = 12 and t = 24 but when i use those values i get a different phase shift.

Will I lose marks if I don't convert back to cos?

I would just use the point (0,8) but technically you could also use (12,8) or (24,8)...etc. 

ie: 8=2sin(pi/2 x 12 +e)+5

1= sin (2pi+e)

(Be careful here, there is more than 1 solution)

3pi/2, pi/2, -pi/2, = 2pi + e

Rule of thumb: Pick the simplest term

So if we pick 3pi/2=2pi+e

e= pi/2

The answer is still the same, and will be for all high tides. If you graph it, both functions will look alike. I doubt you'd be penalised even if you used 12. pi/2 just looks a lot neater

[hawkrook]

How would i do question f here using the dash notation?

[heelloo112233]

How do you restrict the domain for the following equations and then write them in function notation.
y=60.5X10^-0.04x  +29
y= 0.25x^2  -6x +90
and i would like to restrict the domain so that x = 20
Thank you

[peachxmh]

How would you solve this question?

The graph of y=(x-2)/(x+m) has asymptotes with equations....

I'm not sure how to get it into a form where I can read it better (i.e. y= 1/(x-h) + k). Any help would be greatly appreciated!

[S200]

How would you solve this question?

The graph of y=(x-2)/(x+m) has asymptotes with equations....

I'm not sure how to get it into a form where I can read it better (i.e. y= 1/(x-h) + k). Any help would be greatly appreciated!

Well from what I can see, the graph would be a hyperbolic shape, with horizontal asymptote at 1 (I don't know why it's not 0, sorry... ), and the vertical asymptote at M.
Also, the graph has been translated in the positive direction of the X-axis.