VCE Specialist 3/4 Question Thread!

[Stevensmay]

You can effectively pretend that the plane is a box on an inclined plane if it makes it easier.

Resolve the weight force into perpendicular and parallel components.
Now it's just sum of forces, acceleration of the plane is 0.

Perpendicular to movement.

Parrellel to movement.


Pretend the D in the picture is actually an R.
Best paint skillz.

[Jaswinder]

Perfect Thanks people!
question 8 this paper - http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths1.pdf

how do we sketch slope fields? 

[sin0001]

Perfect Thanks people!
question 8 this paper - http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths1.pdf

how do we sketch slope fields? 

Sub. in y-values into the differential equation and you'll get the gradient of the curve along that particular y-value (gradient will be the same for each y-value). Draw a horizontal line when dy/dx=0, a vertical one when dy/dx = ∞, draw a line with positive gradient when dy/dx > 0, etc.

[Stevensmay]

So find your derivative of the equation given.
Then find the gradient at each of the points they give us, and draw a small line on that grid co-ordinate with the gradient you find.

Edit: You got me this time sin!

[Jaswinder]

ahh niceee thanks again!

[jono88]

How do i show that cis(theta-3pi/2)=-sin(theta)+icos(theta)?

[Phy124]

Edit: took out a bracket

[e^1]

Hey everyone

Are there some situations where the answer cannot be in terms of , even if the answer needs to be in exact form? I ask this because in some trial exams they leave it as such, but other times they substitute 9.8 instead.

[BubbleWrapMan]

Hey everyone

Are there some situations where the answer cannot be in terms of , even if the answer needs to be in exact form? I ask this because in some trial exams they leave it as such, but other times they substitute 9.8 instead.

VCAA tells you to take , so both are exact. I think for the most part you can use either one, but if they ask for a specific kind of answer - e.g. a particular form involving , or a numerical answer rounded to a certain number of decimal places - then you should follow that. I guess the best way to get a feel for it is to do past exams.

[Jaswinder]

The co-ordinates of three points are A(1,0,5), B(-1,2,4) and C(3.5.2)

find the co-ordinates of the point D such that ABCD is a parrallelogram. I just feel there is alot to do and too many variables to solve for, for just 2 marks. How would you guys do this question? The right answer is: (5,3,3)

[BubbleWrapMan]

[Jaswinder]

could you please explain this step? thanks

[BubbleWrapMan]

Added to both sides.

[ahat]

The co-ordinates of three points are A(1,0,5), B(-1,2,4) and C(3.5.2)

find the co-ordinates of the point D such that ABCD is a parrallelogram. I just feel there is alot to do and too many variables to solve for, for just 2 marks. How would you guys do this question? The right answer is: (5,3,3)

Where's this question from, if I may ask?

[Jaswinder]

i forgot haha ill let you know if I find that paper again

Also P is on any point on the hyperbola with equation if m is the gradient of the hyperbola at P, then m could be?

I thought any real number in the interval (-2,2)

but the answer is any real number in the interval R\[-2,2]