VCE Specialist 3/4 Question Thread!

[e^1]

Thank you b^3 for the answer, just to begin

I got another question (refer to attachment, NEAP 2010). I have a similar diagram as the one in the answer, but I assumed instead that OB = 2 metres. Doesn't the question say "2 metres from the vertical through O"?


edit: Good clearup Alwin. You can see my english skills are crap, and so I'm glad I don't have to do it anymore ... Although I think they could have also that "2 metres in the horizontal direction from the vertical through O"?

[Alwin]

Thank you b^3 for the answer, just to begin

I got another question (refer to attachment, NEAP 2010). I have a similar diagram as the one in the answer, but I assumed instead that OB = 2 metres. Doesn't the question say "2 metres from the vertical through O"?

no sorry! when it says "x distance from the vertical through O", it means if I draw a vertical line down through O, the mass is "x distance" away from this line. That's the interpretation the question is looking for.

btw, if it helps, imagine OB is a wall. Then the wall is the "vertical through O" and clearly BM is the distance that is 2m from the wall (the vertical) to the thingymabobbby (the mass)
if the visualisation doesn't work for you, dw Hope you kinda get what I'm getting at, at least

[Jeggz]

Any help with his question (4c) would be greatly appreciated
I'm really struggling with Vector Calc and visualising what's going on :\
Thanks in advance!

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf

[Stevensmay]

Any help with his question (4c) would be greatly appreciated
I'm really struggling with Vector Calc and visualising what's going on :\
Thanks in advance!

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf

I got a bit confused with that question as well. Turns out it's asking for the angle between the ground and the plane.

https://www.dropbox.com/s/v7dwdy903wmph5h/2013-11-01%2019.14.07.jpg
Pretend you are an observer standing on the ground and looking straight forwards, this is what you will see.

I did this by finding the velocity vector through differentiation.



So now we want to find this vector forms between the vector and the ground. The angle is how much the plane is pointing downwards if that makes it easier.

Set up a triangle perpendicular to the ground. The straight upright will be 4 units. The bit along the ground we can find by getting the length of the i and j components.



And we want the angle, so we have an adjacent side and an opposite side. Time to use tan.



Normal stuff from here on and we get

[Alwin]

Any help with his question (4c) would be greatly appreciated
I'm really struggling with Vector Calc and visualising what's going on :\
Thanks in advance!

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths2.pdf

Oh this question
Think UMEP unimaths, angles between lines and planes.

Now forget unimaths coz this is vce we're talking about

You just find the angle by going
It's the k component over the "distance" of the i and j components. Although, another method does come to mind but I'll try it out later

Checked the vcaa answers, and they had this method:

[b^3]

You want to know the angle at which the aircraft is going to hit the runway at. So to do that we can form a triangle in 3D from the velcoity vector, as it is the velocity vector that will give us the direction the plane is moving in and so angle it will hit the ground at. So we have the traingle(s) below.

Now to we know the hypotenuse will be the magnitude of the vector, and the length opposite to the angle is , so using trig ratios


EDIT: Beaten, gotta stop spending time on images ...also should be cramming for thermo... why am I on AN atm.

[Alwin]

So we have the traingle(s) below.
(Image removed from quote.)
EDIT: Beaten, gotta stop spending time on images ...also should be cramming for thermo... why am I on AN atm.

well crap, that diagram pwns the bejeus out of my paint one

And maybe you're on AN coz you're such a helpful soul? just a thought

[Jeggz]

WOAH! SO MANY REPLIES
Thanks to all of you guys! That makes so much more sense now

[ahat]

We should so have an AN party.

[ahat]

Let's set a record this year and make spesh scale by 13

How is this achieved?

[Jeggz]

I hope all of my fellow spesh students go well in your other subjects

Let's set a record this year and make spesh scale by 13


Good luck on your exams

How is it upto us lol?

EDIT: Beaten

[zvezda]

Do you mind giving an example? For the exponential, you can always isolate and solve for t, then sub it into your quadratic equation for y. But the thing is, you then have a quadratic equation for y, in terms of a log function of x.

The only "easy" method I can think of is to have a log scale on your x axis, but that might not go down well in spesh so you'd have to sketch it, "unlogarithm" the x-axis then draw it properly.
These are just my general thoughts on the problem, if you have a question post it and I can be more specific

The one i came across is too complex to write here lol.
But say you had e^-3t for the i component and 6t-9t^2 for the j component. How would you go about this? The exam that i saw this in was a kilbaha one, and they had some image of the CAS but couldnt deduce how they managed to graph the particle's path.
Cheers

[b^3]

The one i came across is too complex to write here lol.
But say you had e^-3t for the i component and 6t-9t^2 for the j component. How would you go about this? The exam that i saw this in was a kilbaha one, and they had some image of the CAS but couldnt deduce how they managed to graph the particle's path.
Cheers

Change the graph options to parametric and where it says put what you have for the i component, for put the j component.

Not sure which calc you have, but on the Ti-nspire (the old version at least), you can change to this mode by having the cursor in the entry bar, then pressing [ctrl]+[menu], then [2: Graph Type] and finally [2: Parametric]. The way of changing the mode might have changed slightly with the new calc though.

[Alwin]

And for the CASIO:

[ahat]

Is there a way to express 3^1/4 x 2^1/2 as 12^1/4?
Thanks.