Advanced Higher Maths 2012-2013 : Discussion and Help Thread

Original post by Slumpy

It's been a while since I did mechanics, so I'm not sure, but I'd always err towards proving too much. I don't think the exam is particularly time pressured (is it?), so it shouldn't be a major issue if you understand those formulae. Have you had a look if they're on the learning outcomes?

They are in the learning outcomes and seemed to be in the marking scheme of one question I checked.
I'll check more

Reply 1182

Slumpy

16

Original post by I am Ace

They are in the learning outcomes and seemed to be in the marking scheme of one question I checked.
I'll check more

If they're in the learning outcomes, I'd have no hesitation about using them. I just tended to not bother learning formulae because deriving stuff saved effort in terms of learning.

Reply 1183

hollieeilloh

2

The fact the exam is so close is pretttty scary!, honestly focusing more on modern studies revision-wise, although I know I shouldn't be! :rolleyes:

Just aiming for a C I guess.. if I pass I'll be over the moon, ha.

Reply 1184

A matrix B is such that

[latex\]B^2=6B-9I

Where I is 2x2 identity matrix.

Find integers p and q such that

[latex\]B^3=pB++qI

Reply 1185

Original post by Slumpy

It's been a while since I did mechanics, so I'm not sure, but I'd always err towards proving too much. I don't think the exam is particularly time pressured (is it?), so it shouldn't be a major issue if you understand those formulae. Have you had a look if they're on the learning outcomes?

The Course spec (Applied Mathematics Mechanics) expects you to be able to derive your quoted formulae from first principles (i.e. calculus methods):
I would start with (to find total time of flight - your t formula) from:

\ddot {y} = -g

stating quite clearly that here g is constant of gravitation.

..you will end up with Newton's Laws of motion:

Unparseable latex formula:

y=usin(\theta )t -\fraction{1/2}gt^2

then obviously solve for

y=0

to get required t- formula!

Your range formula (R) is derived using:

\ddot {x} = 0

clearly no air friction assumption.

this will give

x=ucos(\theta )t

, substitute for 't' from earlier
giving you your range

R= x(t)

YES you will be expected to derive these as stated.

(edited 10 years ago)

Reply 1186

Original post by I am Ace

A matrix B is such that

[latex\]B^2=6B-9I

Where I is 2x2 identity matrix.

Find integers p and q such that

[latex\]B^3=pB+qI

Easy, but not obvious, just back substitute for [latex\]B^2=6B-9I such that;

[latex\]B^3=B^2.B=6B^2-9IB= 6(6B-9I)-9B etc!

(edited 10 years ago)

Reply 1187

Original post by tomctutor

The Course spec (Applied Mathematics Mechanics) expects you to be able to derive your quoted formulae from first principles (i.e. calculus methods):
I would start with (to find total time of flight - your t formula) from:

\ddot {y} = -g

stating quite clearly that here g is constant of gravitation.

..you will end up with Newton's Laws of motion:

Unparseable latex formula:

y=usin(\theta )t -\fraction{1/2}gt^2

then obviously solve for

y=0

to get required t- formula!

Your range formula (R) is derived using:

\ddot {x} = 0

clearly no air friction assumption.

this will give

x=ucos(\theta )t

, substitute for 't' from earlier
giving you your range

R= x(t)

YES you will be expected to derive these as stated.

Ah I didn't know you done applied mechanics.
Do you have videos on that website you suggested for applied mechanics (not just the basics of AH Physics)?

Reply 1188

Original post by tomctutor

Easy, but not obvious, just back substitute for [latex\]B^2=6B-9I

such that;

[latex\]B^3=B^2.B=6B^2-9IB= 6(6B-9I)-9B etc!

Do you have any resources for learning how to do the work done with kinetic and potential energy problems?
I don't understand them and I'm self-teaching, can't find any videos to make sense of it.

Reply 1189

Mathematical deitys of TSR, what the hell is a rotation matrix? I'm posting a question and my working so far, can anyone give me a nudge in the right direction?
It's part two onwords of the first question

Original post by Shengis14

Mathematical deitys of TSR, what the hell is a rotation matrix? I'm posting a question and my working so far, can anyone give me a nudge in the right direction?
It's part two onwords of the first question

Part (i) Reqd. ts det(

T_\alpha

)=1
Part(ii)
You will find that

T_\alpha T_\beta = T_{\alpha + \beta}

which us mathematicians like to call a linear transform!
Using the trig identities

sin(A+B)\equiv sin(A)cos(B)+cos(A)sin(B)

and

cos(A+B)\equiv cos(A)cos(B)-sin(A)sin(B)

So

T_\alpha

represents a rotation transform of unit vector in 2D space!

Part (iii)

T_\alpha T_\beta T_c= T_{\alpha + \beta +c}

Reply 1191

Hype en Ecosse

20

Original post by Shengis14

Mathematical deitys of TSR, what the hell is a rotation matrix? I'm posting a question and my working so far, can anyone give me a nudge in the right direction?
It's part two onwords of the first question

A rotation matrix is a matrix that...well, rotates something. Suppose we have a 2 dimensional vector

\begin{pmatrix} x \\ y \end{pmatrix}

. And I want to rotate it 90 degrees counter-clockwise. I just multiply x,y by my rotation matrix

\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

, with 90 degrees subbed in. That outputs

\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}

, so I multiply (x,y) by this, and the new co-ordinate that pops out is the new position in space that is 90 degrees counter-clockwise to x,y.

Do you recognise any trigonometric identities in your matrix there? :wink:

Original post by tomctutor

Part (i) Reqd. ts det(

T_\alpha

)=1
Part(ii)
You will find that

T_\alpha T_\beta = T_{\alpha + \beta}

which us mathematicians like to call a linear transform!
Using the trig identities

sin(A+B)\equiv sin(A)cos(B)+cos(A)sin(B)

and

cos(A+B)\equiv cos(A)cos(B)-sin(A)sin(B)

So

T_\alpha

represents a rotation transform of unit vector in 2D space!

Part (iii)

T_\alpha T_\beta T_c= T_{\alpha + \beta +c}

Boo. Too much mollycoddling. You've practically given him the answer. :tongue:

(edited 10 years ago)

Reply 1192

Original post by tomctutor

Part (i) Reqd. ts det(

T_\alpha

)=1
Part(ii)
You will find that

T_\alpha T_\beta = T_{\alpha + \beta}

which us mathematicians like to call a linear transform!
Using the trig identities

sin(A+B)\equiv sin(A)cos(B)+cos(A)sin(B)

and

cos(A+B)\equiv cos(A)cos(B)-sin(A)sin(B)

So

T_\alpha

represents a rotation transform of unit vector in 2D space!

Part (iii)

T_\alpha T_\beta T_c= T_{\alpha + \beta +c}

Det

T_\alpha

= -1 actually, and I've got

T_{\alpha - \beta}

For the answer is it enough to say that

T_\alpha T_\beta = T_{\alpha - \beta}

??

(edited 10 years ago)

Reply 1193

Hype en Ecosse

20

Original post by Shengis14

Det

T_\alpha

= -1 actually, and I've got

T_{\alpha - \beta}

For the answer is it enough to say that

T_\alpha T_\beta = T_{\alpha - \beta}

??

I'm definitely getting

T_{\alpha + \beta}

, double check your trig identities. I suspect you've made a mistake with the sine double angle.

Reply 1194

I've gone over it again and so has wolfram alpha, I still get

T_{\alpha - \beta}

, I'll upload a photo of my working so you can point out my error

Original post by Shengis14

I've gone over it again and so has wolfram alpha, I still get

T_{\alpha - \beta}

, I'll upload a photo of my working so you can point out my error

No, you're right. I got my "plus-minus" = "minus-plus" in the wrong formula, silly me. :tongue:

Reply 1196

Original post by I am Ace

Do you have any resources for learning how to do the work done with kinetic and potential energy problems?
I don't understand them and I'm self-teaching, can't find any videos to make sense of it.

A body in in central orbit (a planet, pendulum) or a body in free fall (projectile) can only have constant total energy (mechanical energy here) so ignore EM fields for instance.

E_{tot} = E_p + E_k

if we consider an object launched vertically upwards say

E_p = mgh

and

E_k = \frac{1}{2} mv^2

so

E_{tot} = mgy + \frac{1}{2} m\dot {y}^2= const.

we have a differential eqn. w.r.t time

t

with initial conditions,

y(t): y(0) = 0, y(T/2)=h

say and

\dot y(0) = u, \dot y(T/2)=0

where

T

is total time of flight!

particular solutions to this first ODE are suggested by

y(t) = ut - \frac{1}{2}gt^2, \dot y(t)= u - gt

which are simply Newtons eqn's of motion (aka SUVAT)
further however we also have

h=\frac{u^2}{2g}

and also for Time of flight T

T=\frac{2u}{g}

the point is that the sum of the potential + kinetic must be constant for conservative system!

(edited 10 years ago)

Reply 1197

TheFOMaster

Original post by Shengis14

Mathematical deitys of TSR, what the hell is a rotation matrix? I'm posting a question and my working so far, can anyone give me a nudge in the right direction?
It's part two onwords of the first question

Woah, I'm completely lost there... Whats T_b supposed to be? I feel like I've missed a massive part of the course...

Edit: Wait... if

T_a = \begin{pmatrix} \cos a & \sin a \\ \sin a & -\cos a \end{pmatrix}

would

T_b = \begin{pmatrix} \cos b & \sin b \\ \sin b & -\cos b \end{pmatrix}

?

(edited 10 years ago)

Reply 1198

Original post by TheFOMaster

Woah, I'm completely lost there... Whats T_b supposed to be? I feel like I've missed a massive part of the course...

Edit: Wait... if

T_a = \begin{pmatrix} \cos a & \sin a \\ \sin a & -\cos a \end{pmatrix}

would

T_b = \begin{pmatrix} \cos b & \sin b \\ \sin b & -\cos b \end{pmatrix}

?

Yes, that's what the questions getting at

Original post by Hype en Ecosse

No, you're right. I got my "plus-minus" = "minus-plus" in the wrong formula, silly me. :tongue:

Alright then, but the overall point is I can say that

T_\alpha T_\beta

is a rotation matrix because

T_{\alpha - \beta}

takes the same form as

\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

??

(edited 10 years ago)

Reply 1199