VCE Specialist 3/4 Question Thread!

[soccerboi]

Here's the diagram i worked from

[BubbleWrapMan]

If you use a vector method you need to use a vector sum. Define i and j components and go from there.

Though, it's easiest to use Lami's theorem.

[Niskii]

Hey guys, quick question regarding VCAA 2010 Exam 2 M/C Q.22

Just confused with how they got D. I understand the iTute worked solutions for the question for the most part, just not sure where the V₀² comes from in the 3rd step? I understand why the V₀ is there, just not why it's squared! I managed to find a previous thread on the same question but I'm still a bit confused.

Thanks in advance

[Special At Specialist]

Hey guys, quick question regarding VCAA 2010 Exam 2 M/C Q.22

Just confused with how they got D. I understand the iTute worked solutions for the question for the most part, just not sure where the V₀² comes from in the 3rd step? I understand why the V₀ is there, just not why it's squared! I managed to find a previous thread on the same question but I'm still a bit confused.

Thanks in advance

I have no idea how they did what they did, but I'll show you how I solved it:

F = ma where F = F(x)
F(x) = ma
a = F(x) / m
d(1/2 v^2) / dx = F(x) / m
1/2 v^2 = 1/m * Integral of F(x) dx
Let Integral of F(x) dx = G(x) + C
1/2 v^2 = 1/m * (G(x) + C)
When v = v0, x = x0
1/2 (v0)^2 = 1/m * (G(x0) + C)
m/2 (v0)^2 = G(x0) + C
C = m/2 (v0)^2 – G(x0)
Therefore 1/2 v^2 = 1/m * (G(x) + m/2 (v0)^2 – G(x0))
v^2 = 2/m * (G(x) + m/2 (v0)^2 – G(x0))
When x = x1, v = v1
(v1)^2 = 2/m * (G(x1) + m/2 (v0)^2 – G(x0))
(v1)^2 = 2/m * (G(x1) – G(x0)) + (v0)^2
Since G(x1) – G(x0) = definite integral from x0 to x1 of F(x) dx
Then (v1)^2 = 2/m * definite integral from x0 to x1 of F(x) dx + (v0)^2
v1 = square root of all of that
OPTION D

[soccerboi]

Given z=(root2 - 1) + i (root2)

Find both values of root(1+z) in polar form.

I dont understand how they got (root2)cis(-7pi/8) as the second value. Anyone know how?
Thanks

[nina_rox]

I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

It's actually the same form, m and n are just constants, so you can multiply and divide them by whatever constants and they'll still be constants.

I have a question regarding this. For this general formular how do you know which of the three equations to make substitute in to find m and n when it just asks if this are linearly dependent. Does it matter which to equations to would add? Thank you! Any help is really appreciated

[Phy124]

Given z=(root2 - 1) + i (root2)

Find both values of root(1+z) in polar form.

I dont understand how they got (root2)cis(-7pi/8) as the second value. Anyone know how?
Thanks

Look for which have an angle between and :



Note: For the above is an element of (the set of real integers)

[Special At Specialist]

I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

It's actually the same form, m and n are just constants, so you can multiply and divide them by whatever constants and they'll still be constants.

I have a question regarding this. For this general formular how do you know which of the three equations to make substitute in to find m and n when it just asks if this are linearly dependent. Does it matter which to equations to would add? Thank you! Any help is really appreciated

It makes no difference. If they ask whether a, b and c are linearly dependant, you can make any of the following tests:
ma + nb = c
a + mb = nc
ma + b = nc
You only need to test out one of them and it really doesn't make a difference which one you choose. If the equation is solvable, then they are linearly dependent, otherwise they are linearly independent.

[BubbleWrapMan]

Not sure if I should bring this up but the determinant of is zero if and only if , and are linearly dependent. It's a fairly quick calculator test for a multiple choice question but it's probably a bad idea to show it in your working since it's outside the course.

[nina_rox]

I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

It's actually the same form, m and n are just constants, so you can multiply and divide them by whatever constants and they'll still be constants.

I have a question regarding this. For this general formular how do you know which of the three equations to make substitute in to find m and n when it just asks if this are linearly dependent. Does it matter which to equations to would add? Thank you! Any help is really appreciated

It makes no difference. If they ask whether a, b and c are linearly dependant, you can make any of the following tests:
ma + nb = c
a + mb = nc
ma + b = nc
You only need to test out one of them and it really doesn't make a difference which one you choose. If the equation is solvable, then they are linearly dependent, otherwise they are linearly independent.

Thank you so much! Just another thing to clear up, if two of the three equations are parallel or collinear does that affect where the equations go?

also Is it different when it's just two equations? Do you only look for a scalar multiple I think I've read? Which means they are linearly dependent? Thanks, really appreciated!

[BubbleWrapMan]

Yeah for two vectors, it's whether they're parallel/antiparallel

[Niskii]

I have no idea how they did what they did, but I'll show you how I solved it:

F = ma where F = F(x)
F(x) = ma
a = F(x) / m
d(1/2 v^2) / dx = F(x) / m
1/2 v^2 = 1/m * Integral of F(x) dx
Let Integral of F(x) dx = G(x) + C
1/2 v^2 = 1/m * (G(x) + C)
When v = v0, x = x0
1/2 (v0)^2 = 1/m * (G(x0) + C)
m/2 (v0)^2 = G(x0) + C
C = m/2 (v0)^2 – G(x0)
Therefore 1/2 v^2 = 1/m * (G(x) + m/2 (v0)^2 – G(x0))
v^2 = 2/m * (G(x) + m/2 (v0)^2 – G(x0))
When x = x1, v = v1
(v1)^2 = 2/m * (G(x1) + m/2 (v0)^2 – G(x0))
(v1)^2 = 2/m * (G(x1) – G(x0)) + (v0)^2
Since G(x1) – G(x0) = definite integral from x0 to x1 of F(x) dx
Then (v1)^2 = 2/m * definite integral from x0 to x1 of F(x) dx + (v0)^2
v1 = square root of all of that
OPTION D

Perfect! Thanks

[BubbleWrapMan]

I find that questions involving straight line motion with initial conditions are often made simpler by applying the fundamental theorem of calculus and then the substitution rule



(*)

The key part here is understanding why I integrated with respect to x in particular. On the left hand side we had , and we know from the substitution rule that the dxs will 'cancel';

, where

So applying this to the left hand side of (*):

(make sure the terminals match up)







, assuming is positive

[d3stiny]

Do we have to explicitly use u-substitution for anti differentiating f'(x)/f(x) functions?
i.e. would be sufficient working?

[BubbleWrapMan]

Yeah, if you're capable of going straight there it's fine, unless they specifically ask you to use substitution.