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July 05, 2025, 05:30:56 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2534541 times)  Share 

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jono88

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Re: Specialist 3/4 Question Thread!
« Reply #2385 on: October 15, 2013, 11:21:32 pm »
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Those questions are quite common.
The main point is to use cos(nt) = Re(cis(nt))=Re((cis(t))^n) and likewise for sines
Again, make sure the expansion is correct...don't forget that i^3 = -i, for instance.
Have you tried expanding it using your CAS and getting the right answer?

Stevensmay

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Re: Specialist 3/4 Question Thread!
« Reply #2386 on: October 16, 2013, 12:32:20 pm »
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Using a TI-nspire cx and entering


Which is the expansion I believe given in the answers.

ahat

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Re: Specialist 3/4 Question Thread!
« Reply #2387 on: October 16, 2013, 09:33:32 pm »
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What's the best method for doing these type of questions, by hand or CAS?
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lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #2388 on: October 16, 2013, 09:40:31 pm »
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If you were to choose the answer by hand:
Firstly, the gradient is negative. A, C, E only viable choices.
Gradient independent of t. Eliminate A. Decrease V, decrease magnitude of gradient.
C


I can't do maths anymore -.-
« Last Edit: October 16, 2013, 10:14:23 pm by nliu1995 »
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availn

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Re: Specialist 3/4 Question Thread!
« Reply #2389 on: October 16, 2013, 10:03:15 pm »
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If you were to choose the answer by hand:
Firstly, the gradient is negative. A, C, E only viable choices.
Gradient independent of t. Eliminate A. Decrease V, decrease gradient. Answer is E.

I thought it would be C? When V is 0, gradient is 0.
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Re: Specialist 3/4 Question Thread!
« Reply #2390 on: October 16, 2013, 10:03:42 pm »
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From MAV exam 2 2006 Q2

A stone is projected upwards from the top of a 50m building at an angle of to the horizontal with a velocity of
m/s at time t, where t is greater than or equal to 0.

When I integrate this equation to find the displacement, do I integrate with respect to t or ?

ahat

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Re: Specialist 3/4 Question Thread!
« Reply #2391 on: October 16, 2013, 10:05:29 pm »
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Answer is C, did choose E first though ;)
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Stevensmay

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Re: Specialist 3/4 Question Thread!
« Reply #2392 on: October 16, 2013, 10:06:25 pm »
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Theta is a constant but t is not, so you would integrate with respect to t.

ahat

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Re: Specialist 3/4 Question Thread!
« Reply #2393 on: October 16, 2013, 10:07:01 pm »
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From MAV exam 2 2006 Q2

A stone is projected upwards from the top of a 50m building at an angle of to the horizontal with a velocity of
m/s at time t, where t is greater than or equal to 0.

When I integrate this equation to find the displacement, do I integrate with respect to t or ?

so you would integrate with respect to t (because vector is in terms of t)
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lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #2394 on: October 16, 2013, 10:13:51 pm »
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I thought it would be C? When V is 0, gradient is 0.

Yep, typo there. Fixed.
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Re: Specialist 3/4 Question Thread!
« Reply #2395 on: October 16, 2013, 10:14:39 pm »
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lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #2396 on: October 16, 2013, 10:15:27 pm »
+1
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ahat

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Re: Specialist 3/4 Question Thread!
« Reply #2397 on: October 16, 2013, 10:16:08 pm »
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Gradient independent of t.

How do you know this?

Ahhhh, I went back and did it on CAS, don't know how I got it wrong, do you guys have the "Plot differential equation" program on your CAS? Super useful.

Sure...

Calm down there bro, Nilu is a flipping beast at maths ♥
« Last Edit: October 16, 2013, 10:18:02 pm by ahat »
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lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #2398 on: October 16, 2013, 10:21:38 pm »
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For the gradient, scan across the page. The gradient should appear to be the same regardless of what t value you are, if you move your eyes horizontally.
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Re: Specialist 3/4 Question Thread!
« Reply #2399 on: October 16, 2013, 10:59:53 pm »
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Suggestion for anyone worried about that complex expansion on the last page: Pascal's triangle.

Also if the sines and cosines make it difficult to keep track of everything, try expanding then let and . In fact, is something you should probably be able to recite from memory.

For powers of , remember that multiplication by is just rotation by 90 degrees anticlockwise on an Argand diagram. That gives you the first four powers immediately and it also shows that the first four powers are the only ones you need to know.