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Current Year 12 Thread Mark VI

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Reply 4120
Avatar for bananarama2
bananarama2
OP
I'm not looking at any unofficial mark schemes.
Original post by wcp100
That isn't as easy as the series log question 1987 paper.


I haven't looked at any papers that far back. :tongue:

Original post by wcp100
I'm not looking at any unofficial mark schemes.


That would be a wise idea - I have not done that and won't. How did FP1 go for you yesterday? :h:

Also... how has your day been?

I've spent much of my day out... socialising. :eek: I only just got back...
Reply 4122
Avatar for bananarama2
bananarama2
OP
Original post by Maths_Lover
I haven't looked at any papers that far back. :tongue:



That would be a wise idea - I have not done that and won't. How did FP1 go for you yesterday? :h:

Also... how has your day been?

I've spent much of my day out... socialising. :eek: I only just got back...


They're easier on the whole.

Alright I think. I wished it wasn't as long with short questions though.

Good. I went to see my brother. I'm just doing the rest of your questions.

How was your day?
Original post by wcp100
They're easier on the whole.

Alright I think. I wished it wasn't as long with short questions though.

Good. I went to see my brother. I'm just doing the rest of your questions.

How was your day?


I may have a look soon, then. :smile:

I do as well. :/

Sounds fun and I look forward to seeing your answers - I'll do your questions soon. :yep:

My day was really, really enjoyable (tanoshii desu!). I went to town with my family and then we went back to a family friend's house (round the corner from our house) and stayed there for hours. We chatted, ate, watched some TV and I played piano for them (I played hours of piano today! :eek:) and it was extremely fun and quite a refreshing change. :biggrin: However, I was fed too much though... "Have some more food!" "Nah, that's alright, I'm full..." "No really, eat some more." "Okay then..." *a few bites later* Gaah... my aching stomach........ :couchpotato:

I'm finishing off a STEP question which I was right in the middle of before we had to leave. :nooo:
A trophy fell on my head and now I have a lovely swollen cut by my eyebrow :colonhash: :colonhash:
Original post by Maths_Lover


I came across a very nice STEP I question this morning. No funny business involved, no trickery - it was just straight-forward and elegant. :love:

You guys can try it if you want... that is, if you haven't already but I suspect you have probably already done it in your sleep. :tongue:

2012-06-02 10.52.13.jpg


And with that, I am off again. :tongue:


I've probably done this question before but if I did it's been so long ago I can't remember. At least a year and a half...

Spoiler



Original post by wcp100
They're easier on the whole.

Alright I think. I wished it wasn't as long with short questions though.

Good. I went to see my brother. I'm just doing the rest of your questions.

How was your day?


Do you know on which page of this thread your questions are?

Original post by berryripple
A trophy fell on my head and now I have a lovely swollen cut by my eyebrow :colonhash: :colonhash:


Well, at least you softened the blow to the trophy before it hit the floor. :wink:
(edited 11 years ago)
CodeCogsEqn (2).pdf25.0KB
CodeCogsEqn (3).pdf27.4KB
Reply 4126
Avatar for bananarama2
bananarama2
OP
Original post by Blutooth

Do you know on which page of this thread your questions are?


ML is teaching me basic number theory. She PM'ed me the questions.
Original post by wcp100
ML is teaching me basic number theory. She PM'ed me the questions.


Oh, ok that's cool.
Reply 4128
Avatar for bananarama2
bananarama2
OP
Original post by Blutooth
Oh, ok that's cool.


I was teaching her a bit about partial differentiation if you want the questions I sent her?
Original post by wcp100
I was teaching her a bit about partial differentiation if you want the questions I sent her?


Sure. That'd be awesome. However, if you do I'm likely to return the favour. :wink:
Reply 4130
Avatar for bananarama2
bananarama2
OP
Original post by Blutooth
Sure. That'd be awesome. However, if you do I'm likely to return the favour. :wink:


Find all the first order partial derivatives of:

i) f(x,y)=arctan⁡(y/x) f(x,y)=\arctan (y/x)
ii) r(x,y,z)=x2+y2+z2 r(x,y,z)=\sqrt{x^2+y^2+z^2}

2) Show that the differential:

df=x2dy−(y2+xy)dx df = x^2 dy -(y^2+xy)dx

is not exact, but dg=(xy2)−1 dg=(xy^2)^{-1} is.

3)Show that

df=y(1+x−x2)dx+x(x+1)dydf=y(1+x-x^2)dx+x(x+1)dy

is not a an exact differential.
Find the differential equation that a function g(x) must satisfy if dϕ=g(x)df d \phi = g(x)df to be an exact differential. Verify that g(x)=e−x g(x)=e^{-x} is a solution of this equation and deduce that form on ϕ(x,y) \phi (x,y) .

4)Using the chain rule transform the coordinate system of the operator

∂2∂x2+∂2∂y2+∂∂z2 \displaystyle \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}+ \frac{\partial}{\partial z^2}

x=ρsin⁡θcos⁡ϕ x=\rho \sin \theta \cos \phi
y=ρsin⁡θsin⁡ϕ y= \rho \sin \theta \sin \phi
z=ρcos⁡θ z= \rho \cos \theta
ρ=x2+y2+z2 \rho = \sqrt{x^2+y^2+z^2}
Original post by wcp100


is not a an exact differential.
Find the differential equation that a function g(x) must satisfy if dϕ=g(x)df d \phi = g(x)df to be an exact differential. Verify that g(x)=e^{-x} is a solution of this equation and deduce that form on ϕ(x,y) \phi (x,y) .

4)Using the chain rule transform the coordinate system of the operator

∂2∂x2+∂2∂y2+∂∂z2 \displaystyle \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}+ \frac{\partial}{\partial z^2}

x=ρsin⁡θcos⁡ϕ x=\rho \sin \theta \cos \phi
y=ρsin⁡θsin⁡ϕ y= \rho \sin \theta \sin \phi
z=ρcos⁡θ z= \rho \cos \theta
ρ=x2+y2+z2 \rho = \sqrt{x^2+y^2+z^2}


Thanks for the questions. I haven't done much on DE but I'll give em a try. Here's the only interesting DE equation I know; it's an Oxford Maths DE interview question and actually quite beautiful.

There is a unit square with it's axes parallel to the x and y axis. The bottom left hand corner is centred on the origin. At the vertices on the square, there are 4 beatles. The beatles move in a special way: each beatle moves exactly towards the beatle on the next vertex going round clockwise.
i) sketch the paths of the beatles as they move across our coordinate system.
ii) Find an equation for the paths the beatles follow.
(edited 11 years ago)
Reply 4132
Avatar for bananarama2
bananarama2
OP
Original post by Blutooth
Thanks for the questions. I haven't done much on DE but I'll give em a try. Here's the only interesting DE equation I know; it's an Oxford Maths DE interview question and actually quite beautiful.

There is a unit square with it's axes parallel to the x and y axis. The bottom left hand corner is centred on the origin. At the vertices on the square, there are 4 beatles. The beatles move in a special way: each beatle moves exactly towards the beatle on the next vertex going round clockwise.
i) sketch the paths of the beatles as they move across our coordinate system.
ii) Find an equation for the paths the beatles follow.


I've ended up with simultaneous differential equations....
Original post by wcp100
I've ended up with simultaneous differential equations....


That was quick :smile: Just to find the right substitution now.
(edited 11 years ago)
Reply 4134
Avatar for bananarama2
bananarama2
OP
Original post by Blutooth
That was quick :smile: Just to find the right substitution now.
Edit: in one variable?


I don't know what to do with the pile of differential equations.

I'll tell you my thought process I feel like I've over complicated it.

I let each beatle be denoted a,b,c and d going clockwise of the origin. If they all move with speed u. Then:

Va⃗=ub⃗∣b∣ \vec{V_a}=u\frac{\vec{b}}{|b|}

I feel like this is completely wrong though. I did this for all of them.
Original post by wcp100
I don't know what to do with the pile of differential equations.

I'll tell you my thought process I feel like I've over complicated it.

I let each beatle be denoted a,b,c and d going clockwise of the origin. If they all move with speed u. Then:

Va⃗=ub⃗∣b∣ \vec{V_a}=u\frac{\vec{b}}{|b|}

I feel like this is completely wrong though. I did this for all of them.


Don't worry about the velocities. Just try and find an equation y in terms of x.

Look at the y and x coordinates of the points on a path, can you notice any symmetries... or anything to obtain a DE equation?

Hint:

Spoiler

(edited 11 years ago)
Reply 4136
Avatar for bananarama2
bananarama2
OP
Original post by Blutooth
Don't worry about the velocities. Just try and find an equation y in terms of x.

Look at the y and x coordinates of the points on a path, can you notice any symmetries... or anything to obtain a DE equation?

Hint:

Spoiler



Sorry I went to have my lunch.

dydx=1−x−yy−x \frac{dy}{dx}=\frac{1-x-y}{y-x}
(edited 11 years ago)
Original post by wcp100
Sorry I went to have my lunch.

dydx=1−x−yy−x \frac{dy}{dx}=\frac{1-x-y}{y-x}


Brill, now just to make the right sub...the same sub I used for ML's inequality question with the p-1/p+1, interestingly enough.
(edited 11 years ago)
Reply 4138
Avatar for bananarama2
bananarama2
OP
Original post by Blutooth
Brill, now just to make the right sub...the same sub I used for ML's inequality question, interestingly enough.


u=y-x ?
Original post by wcp100
u=y-x ?


That's what I said at interview but no, that's not it. See my previous edit.
(edited 11 years ago)

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