VCE Methods Question Thread!

[deStudent]

Thx a lot!

[deStudent]

Kinda like my problem before, but I'm assuming solving literal equations where it asks for the answer to be given in terms of '[insert variable]' wouldn't be possible with the CAS?

For example:
s = 2 + ah + h^2 and h = a - 1/a? The question asked for s in terms of a.

[Jakeybaby]

Kinda like my problem before, but I'm assuming solving literal equations where it asks for the answer to be given in terms of '[insert variable]' wouldn't be possible with the CAS?

For example:
s = 2 + ah + h^2 and h = a - 1/a? The question asked for s in terms of a.

Casio have only just released a calculator which can solve systems of linear equations in terms of variables, the fx-CG20AU Colour. I'm unsure of the rules in VCE.
I don't think any TIs can do this, nor can any other Casio products.

[exit]

Casio have only just released a calculator which can solve systems of linear equations in terms of variables, the fx-CG20AU Colour. I'm unsure of the rules in VCE.
I don't think any TIs can do this, nor can any other Casio products.

CAS calculators are used in Methods so the CG20AU is out of the equation lol (no pun intended)

[deStudent]

Is it supposed to take forever when drawing trig function graphs with both x and y translations and dilations over a suitable domain?

It takes me a good 20 to 30 minutes? Sometimes even 40 minutes if the x axis translation throws me off. My method is: observe period and amplitude, find axis intercepts and then starting/ending points.

[Sine]

Is it supposed to take forever when drawing trig function graphs with both x and y translations and dilations over a suitable domain?

It takes me a good 20 to 30 minutes? Sometimes even 40 minutes if the x axis translation throws me off. My method is: observe period and amplitude, find axis intercepts and then starting/ending points.

yeah it takes a while with trig graphs imo for a 4 mark graph I would take up to 6 minutes (started at around 12 maybe)  but everything else would be below a mark a minute. I did my graphs through transformations i.e start with cos(x) then do a dilation and go to cos(2x) - translation - cos(2x)+1 this would be done neatly but in a rough sense on the side of the page and once a got to the last transformation i would do it on the grid lines.

Use properties of symmetry and patterns in the shape to cut down the time - personally i wouldn't find  x intercepts but can derive it by going through each transformation of the original function. I would use finding x-intercepts as a checking technique.

[deStudent]

yeah it takes a while with trig graphs imo for a 4 mark graph I would take up to 6 minutes (started at around 12 maybe)  but everything else would be below a mark a minute. I did my graphs through transformations i.e start with cos(x) then do a dilation and go to cos(2x) - translation - cos(2x)+1 this would be done neatly but in a rough sense on the side of the page and once a got to the last transformation i would do it on the grid lines.

Use properties of symmetry and patterns in the shape to cut down the time - personally i wouldn't find  x intercepts but can derive it by going through each transformation of the original function. I would use finding x-intercepts as a checking technique.

Ah right, thanks. But wouldn't you have to show x-intercepts on the graph?

[samuelbeattie76]

The volume (V) of a sphere varies directly with the cube of its radius (r). A sphere of radius 10 cm has a volume of 4188.8 cm^3. Find the radius of a sphere that has a volume of 1 cubic metre.

[wyzard]

The volume (V) of a sphere varies directly with the cube of its radius (r). A sphere of radius 10 cm has a volume of 4188.8 cm^3. Find the radius of a sphere that has a volume of 1 cubic metre.

Actually the volume of a sphere is directly proportional the the radius cubed.

Since you know the volume of a sphere with a radius of 10cm, you can use that fact to find the radius of a sphere of 1 cubic metre, which is 1,000,000 cubic centimetres.

[deStudent]

When solving something like: 2sin(2x + pi/3) = - sqrt(3), domain is 0 to 2pi.

Which method is it easier to solve it with? The book uses the method where you let theta = 2x + pi/3 and then adjust the domain of the question accordingly, however I just solve it directly. The method I use kinda takes some guess word to realise when you should stop adding 2pi.

Is one method better than the other?

[RuiAce]

When solving something like: 2sin(2x + pi/3) = - sqrt(3), domain is 0 to 2pi.

Which method is it easier to solve it with? The book uses the method where you let theta = 2x + pi/3 and then adjust the domain of the question accordingly, however I just solve it directly. The method I use kinda takes some guess word to realise when you should stop adding 2pi.

Is one method better than the other?

Here, both are fine.

"Better" is such a broad word. Because it depends on what "better" means.
By choosing to make a substitution, things are a bit more stepped out, and everything becomes clearer. However if your maths isn't wrong, the direct method is faster. Whatever you're aiming for between clearness and swiftness would be what decides the word "better".

[Sine]

Ah right, thanks. But wouldn't you have to show x-intercepts on the graph?

yeah show everything required and if your not sure just label more (correctly) until you are happy. Find x-intercepts through the period so you wouldn't actually need to manually find them.

[deStudent]

Thanks RuiAce and Sine.

[vcestressed]

can someone please factorise this?

x^4-5x^2-36
thanks ! 

[Jakeybaby]

can someone please factorise this?

x^4-5x^2-36
thanks ! 

(x^2 - 9)(x^2+4)
(x+3)(x-3)(x^2+4)