VCE Specialist 3/4 Question Thread!

[saba.ay]

Are you sure its minus between them? The answer you gives results from a plus between them.





Otherwise for the actual question you gave you would get

I feel like such an idiot. The question did have a plus, not a minus. I kept getting the second answer. :/
Thank you so much!

[zvezda]

Well, the cartesian plane is 2D, and since this vector involves three dimensions, finding a cartesian equation would not be possible.
The question asks to 'describe the motion'. That does not necessarily mean that you need only one equation to do this.
In this case, you can make three parametric equations. Let the planes be x, y and z.
So: , and

As you can see, in all three planes, the particle is moving linearly, as all equations involve with a degree of 1.
Thus, the particle's overall motion is also simply linearly. You can see this by substituting values for .
When , the particle is at .
When , the particle is at .
When , the particle is at ... etc.

Hope that helps.

Yeah fair enough. I understood the whole describing motion, though I was just wondering whether it would be possible to obtain some sort of equation

[zvezda]

q17 b im also having some slight troubles with

[Ancora_Imparo]

Yeah fair enough. I understood the whole describing motion, though I was just wondering whether it would be possible to obtain some sort of equation

Sorry for insulting your intelligence... I thought your main concern was about describing the motion...

q17 b im also having some slight troubles with

(using cosine double angle formula)
Thus: , a quadratic!

[zvezda]

Sorry for insulting your intelligence... I thought your main concern was about describing the motion...



(using cosine double angle formula)
Thus: , a quadratic!

hahah dont worry about it. But I guess that question there reveals just how intelligent I am. Totally missed the double angle formula. Cheers for the help

[saba.ay]

Can someone please explain how implied domain and range are found for inverse circular functions? I'm not understanding how this work. Specifically, the following example:

Cos^-1 ((x^2)-1)

[sin0001]

Cos^-1(x^2) is just the Cos^-1 graph with the part from x=[0,1], reflected in the y-axis, sorta looks like a semi-circle. And then you translate this graph by 1 unit to the right, in the x-axis.
That should be it, I think.

[Ancora_Imparo]

For :
Implied domain is .
Implied range is .

Thus, you know that must be greater than or equal to -1 and less than or equal to 1. So, to work out the implied domain of , solve these equations for x.








The implied domain of is simply the intersection of the two regions found above, which in this case, is simply or .

You then use this implied domain to work out the implied range. You know that when or , , which is the lowest possible point, and when , , which is the highest possible point. So the implied range is .

[Professor Polonsky]

Two questions, whoever's kind enough to help - both questions in the interval of



Now logically, the solutions are going to be at

But doing it mathematically didn't work.













Which is obviously wrong. Because if , then the second equation is also true - silly me, just figured that out now. But then how would you do it differently?

Also



Can't seem to get my head around this one.

[b^3]

When you squared the , you've introduced a new solution, as cases that would not have worked previously as they were negative, are now positive and thus give an answer. That is we cannot have both and being negative and giving . We can have one positive and the other zero to give negative one though. This rules out the solution as in that case we have sin being negative and cos being zero. Same goes for the solution, one is negative and the other is zero, which we can't have.

It's just a matter of being careful when you square things, so that you realise if you've introduced in new solutions  by squaring negatives.

Hope that makes sense and hope that helps

[Professor Polonsky]

Yup, thanks - that's what I figured out. I guess the question now is whether there's a way to solve it without squaring, or should I plug and into the equation and show that they're not true?

[Planck's constant]

Two questions, whoever's kind enough to help - both questions in the interval of



Now logically, the solutions are going to be at

But doing it mathematically didn't work.













Which is obviously wrong. Because if , then the second equation is also true - silly me, just figured that out now. But then how would you do it differently?

Also



Can't seem to get my head around this one.

Re the first question, I normally try to get rid of the '1' on the RHS.
Therefore introduce the half angle x/2 which is in [0, pi]

Your equation now becomes

2sin(x/2)cos(x/2) + 1 - 2sin^2(x/2)  = 1

This quickly reduces to sin(x/2) = 0 or tan(x/2) = 1

Which provides the right answer, remembering that x/2 in [0, pi]

[Conic]

I posted this a few pages back you can collect the terms:



then you solve:




 
which gives you:

    or 

[Jeggz]

For , this is what I did but don't know if it's correct :\




Re-write the equation with sin(x) and cos(x)





And then just replace the with , let a=cos(x), solve as a quadratic and hopefully you just get
Again, don't know if this correct!

EDIT: I have skipped  a few steps in between (because latex is not my field!), so let me know if you don't know how I got something 

[Planck's constant]

I posted this a few pages back you can collect the terms:



then you solve:




 
which gives you:

    or 

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