VCE Specialist 3/4 Question Thread!

[Bhootnike]

consider , where m and n are non-zero positive integers

show that 


dafuk?

[brightsky]

let x = arctan(1/m) and y = arctan(1/n)
m = 1/tanx and n = 1/tany
tan(x+y) = (tanx +tany)/(1-tanx tany) = 1
tan x + tany = 1 - tanx tany

(m-1)(n-1)
=mn - m - n + 1
=1/(tanx tany) - 1/tanx - 1/tany + 1
= 1/(tanx tany) - (tanx + tany)/(tanx tany)+1
= (1 - tanx -tany + tanx tany)/(tanx tany)
= (1 - (1-tanx tany) + tanx tany)/(tanx tany)
= 2tanx tany/(tanx tany)
= 2

[Bhootnike]

wow.
haha thanks man!

your gonna kill spesh next year, sure you've already heard that ;p

[Hutchoo]

wow.
haha thanks man!

your gonna kill spesh next year, sure you've already heard that ;p

He was killing spesh in year 9 X_x.

[Homer]

Advanced Maths Question: A mass of 3 kg is suspended from a beam by two strings, 5 cm and 12 cm long. If the
strings are perpendicular to each other, find the tension in each string.

[BubbleWrapMan]

Here, hopefully that's clear enough. I would have typed it here but I'd rather have the diagram there and I'm not b^3 so yeah.

It says "13 cm" at the top, it was cut off slightly.

[Homer]

Thanks for your help! but we haven't learnt about "arctan" yet in our advanced class.  Is there possibly another way of solving such problems? Btw the answers given are 11.31 N; 27.14 N.

[BubbleWrapMan]

arctan is just the inverse tan function. I think you would know about it by now? And you could possibly do it without using arctan, by being careful with your geometry and finding values for the relevant sines of angles, but you'll have to learn about inverse trigonometric functions eventually so I guess you should start now.

[soccerboi]

I need help with this multiple choice please
Cheers

[b^3]

The area under a velocity time graph will give displacement. So integrating from 0 to 50 will give the displacement of particle A from the starting position, at t=50s.

Now if we do the same for , we will get the displacement of particle B. (Initially B is moving in the opposite direction until it turns around. When the integral is negative then it is to the left of O, i.e. it starts moving in the opposite direction at t=20 seconds but is still to the left of the starting position until the area under the graph and the area above the grap cancels out). <--- Probably didn't need that last bit but the explanation may help a little.

So basically if we know the displacement of each particle from the original position at t=50, take one away from the other and take the absolute value of that, we will get the distance between the particles. (since they are moving in a striaght line, drawing a number line may help visualise their movement and the distance between them). Hence E.

Hope that makes sense. I guess the key thing to remember here is

EDIT: Sorry if that didn't make sense, my explanations are all over the place tonight... not on enough sleep atm.

[soccerboi]

The area under a velocity time graph will give displacement. So integrating from 0 to 50 will give the displacement of particle A from the starting position, at t=50s.

Now if we do the same for , we will get the displacement of particle B. (Initially B is moving in the opposite direction until it turns around. When the integral is negative then it is to the left of O, i.e. it starts moving in the opposite direction at t=20 seconds but is still to the left of the starting position until the area under the graph and the area above the grap cancels out). <--- Probably didn't need that last bit but the explanation may help a little.

So basically if we know the displacement of each particle from the original position at t=50, take one away from the other and take the absolute value of that, we will get the distance between the particles. (since they are moving in a striaght line, drawing a number line may help visualise their movement and the distance between them). Hence E.

Hope that makes sense. I guess the key thing to remember here is

EDIT: Sorry if that didn't make sense, my explanations are all over the place tonight... not on enough sleep atm.

How would you know which to take away from? Why couldn't it have been f(t)-g(t) ?

[paulsterio]

How would you know which to take away from? Why couldn't it have been f(t)-g(t) ?

Given the modulus, it actually can be f(t)-g(t)

[generalkorn12]

I've just got two questions that are bugging me:

a) 'On a rough inclined plane, a mass of mg is being pushed off the top of the plane with initial velocity 1ms', sorry for butchering the question, but I was assuming that you would use the formula R=ma, and integrate that to get velocity, et cetera, but the question seems to assume we can use R=Ma, find the acceleration, and then use the four formulas, as it the solutions say because it has constant acceleration, is that even allowed on rough inclines?

b) For position vectors such as, r(t) = (4t-20)i + 2t j -+ (80 - 2t) k, if it asks for the angle when it hits the ground, wouldn't there be NO angle, because the vertical height would be zero (as it hits the ground?)...

Thanks!

[Jenny_2108]

I've just got two questions that are bugging me:

a) 'On a rough inclined plane, a mass of mg is being pushed off the top of the plane with initial velocity 1ms', sorry for butchering the question, but I was assuming that you would use the formula R=ma, and integrate that to get velocity, et cetera, but the question seems to assume we can use R=Ma, find the acceleration, and then use the four formulas, as it the solutions say because it has constant acceleration, is that even allowed on rough inclines?

b) For position vectors such as, r(t) = (4t-20)i + 2t j -+ (80 - 2t) k, if it asks for the angle when it hits the ground, wouldn't there be NO angle, because the vertical height would be zero (as it hits the ground?)...

Thanks!

a) yep
b) because it hits the ground so height = 0. You find angle based on i, j directions

[martin1106]

 I've got two questions

(1)  Is is true that a complex number is rotated by "θ" (anticlockwise) when it is multiplied by cis(θ)?

(2) If I use the vector formula a.b=a*b*cosθ to find the angle between two vectors,
 
 e.g OA.OB=OA.OB.cosθ (1)   

     OA.BO=OA.BO.cosθ (2) 
 
 The second one is incorrect, isn't it? I think the direction is restricted but some trial exam papers haven't specified that.

 Thanks