VCE Specialist 3/4 Question Thread!

[Homer]

sweet thanks alwin

[götze]

Everytime i sketch a circular function graph it sketches as a liner function eg a straight line though the y axis. Can anybody help ps i use a ti inspire calcualtor

[Alwin]

Everytime i sketch a circular function graph it sketches as a liner function eg a straight line though the y axis. Can anybody help ps i use a ti inspire calcualtor

I use Casio, but check if you're in degrees or radians. That's usually the problem. otherwise, check you scale eg if its y=1000cos(x) and you scale is y from -10 to 10 then there's your problem

[götze]

It says graphing angle radian so do i change it to degrees ?

[Alwin]

It says graphing angle radian so do i change it to degrees ?

well, it depends on the graph itself actually. For sin(x):

in degrees mode:


in radians mode:


On the casio, we have a quick trig initialise that scales the axis for us to show 1 period. If worst comes to worst. make you scale huge, and then zoom back in on the part you want.

Also, it might help if you post what graph you're trying to graph too

[saba.ay]

Thanks for the answer to the previous question guys- much appreciated!

Also, could someone please explain how I'd go about this? I think I'm off entirely in my approach.

[availn]

Thanks for the answer to the previous question guys- much appreciated!

Also, could someone please explain how I'd go about this? I think I'm off entirely in my approach.

I think you're meant to work this out with geometry. Draw 2 right angled triangles. Triangle 1 has an angle x, such that tan(x)=3/4, so it is a right angled triangle with sides 3, 4, and 5. Triangle 2 has an angle y, such that cos(y)=12/13, so it has sides 5, 12, and 13. If you multiply all of triangle one's sides by 3, you can fit it in triangle 2 (as they now both have a side of length 12), and find cos(x-y) using the cosine rule. I got 63/65, is this right?

[aestheticatar]

Vectors help please!

OABC is a quadrilateral
OA = a = 2i
OB = b = 3i + 2j
OC = c = 4i + j

D is a point on AC such that BD is perpendicular to AC.

What's the area of triangle ABC?

Thanks in advance

[Alwin]

Vectors help please!
Quadrilateral marked by OABC

OA = a = 2i
OB = b = 3i + 2j
OC = c = 4i + j

D lies on AC where BD is perpendicular to AC

What's the area of triangle ABC?

Thanks in advance

Hint: can you find the vector BD? think of vector addition and subtraction.
Now, since BD perpendicular to AC, can u see that BD represents the height of the triangle ABC.
So, you can easily find the base length which is |AC| and height which is |BD| 
I reckon you can do the rest from here

btw... vectors in methods?? did u post in the wrong thread by accident or am I forgetting something
edit: thanks for moving it b^3

Mod EDIT: Split, and merged topics into spesh thread

[aestheticatar]

LMFAOO shit my bad
But yeah I got it thanks heaps!

[availn]

Vectors help please!

OABC is a quadrilateral
OA = a = 2i
OB = b = 3i + 2j
OC = c = 4i + j

D is a point on AC such that BD is perpendicular to AC.

What's the area of triangle ABC?

Thanks in advance

http://en.wikipedia.org/wiki/Heron's_formula

Heron's formula makes it braindead, no thought required.

[brightsky]

http://en.wikipedia.org/wiki/Heron's_formula

Heron's formula makes it braindead, no thought required.

cross product also makes it a trivial exercise:

area  = 1/2*mod(AB x AC)

[random_person]

Question... How do you find out the volume of a torus (donut shape) by revolving a circle around the x axis?

[b^3]

I think I may have missed something, as the method that I've used isn't exactly in the course... using trigonometric substitution, but depending on your school you may cover it. Where exactly is the question from?

Anyways, lets start with a circle of radius shifted units up from the origin. Here's one I prepared earlier

Now that is a circle with equation .
Now for rotation around the axis we have:
.
So we need in terms of .


So now we integrate the square of the top curve minus the square of the bottom curve. Since the values of go from to , we have our terminals.


Now here comes the bit that isn't on the course.... A trigonometric substitution.


..Although there probably is a way to do it with things that are on the course...

EDIT: Fixed substitution mistake.

[barydos]

Good evening!

This question, seemed quite simple, but for some reason I'm not sure how to prove it properly haha. I must be looking at it from the wrong perspective, but here it is:


Answer to part a:

Spoiler

So it's part b), c) and d) that I don't know how to do.
If you could help out, I'd appreciate it