My Specialist Maths Mega-thread :D

[funkyducky]

m=-1, since (equating co-efficients) 2my=-2y

[monkeywantsabanana]

yeah, i got that too... but then how do i find the cartesian equation from that?

I subbed m in as -1 so i had


...
and got 0=0, obviously but there can't be an equation?

[TrueTears]

m^2x^2 - x^2 +m^2y^2-y^2+2my+2y = x^2(m^2-1) +y^2(m^2-1)+2y(m+1) = 0

hint: complete the ...

[b^3]

let z=x+i*y
then we have
|m(x+i*y)+i|=|x+iy-i|
Factorise
|m*x+i(y+1)|=|x+i(y-1)|
Now remember |x+y|(mod)=






Hopefully there are no mistakes.

[monkeywantsabanana]

let z=x+i*y
then we have
|m(x+i*y)+i|=|x+iy-i|
Factorise
|m*x+i(y+1)|=|x+i(y-1)|
Now remember |x+y|(mod)=






Hopefully there are no mistakes.

The answers says

So you're right I think...

though I don't understand how you got from:

|m(x+yi)+i| to |mx+(y+1)i|

and TT .. I can see where you're going with it but I'm stuck



I tried letting a=(m+1) :| that didn't turn out so well...

OMG I'M HOPELESS

[TrueTears]

welll first note m^2-1 = (m+1)(m-1)

so we have x^2(m-1)+y^2(m-1)+2y=0

now look at y^2(m-1)+2y

complete the square on that expression is very simple, then just put it all back together

[monkeywantsabanana]

Thanks guys! I think i understand now

[dc302]

let z=x+i*y
then we have
|m(x+i*y)+i|=|x+iy-i|
Factorise
|m*x+i(y+1)|=|x+i(y-1)|
Now remember |x+y|(mod)=






Hopefully there are no mistakes.

You forgot the m in the 5th line so I don't think your final answer is correct.


edit: by my working, the answer should be a circle:

(y - 1/(1-m) )^2 + x^2 = 1/(1-m)^2

I don't know how the answer is supposed to be y= 1/4 x^2, where did the m even go?


edit 2: noting here that the above solution is for m not equal to 1 or -1.

[b^3]

let z=x+i*y
then we have
|m(x+i*y)+i|=|x+iy-i|
Factorise
|m*x+i(y+1)|=|x+i(y-1)|
Now remember |x+y|(mod)=






Hopefully there are no mistakes.

You forgot the m in the 5th line so I don't think your final answer is correct.


edit: by my working, the answer should be a circle:

(y - 1/(1-m) )^2 + x^2 = 1/(1-m)^2

I don't know how the answer is supposed to be y= 1/4 x^2, where did the m even go?


edit 2: noting here that the above solution is for m not equal to 1 or -1.

Ah yes I did too, missed the m coefficient in front of the (y+1). The small thing we miss on a computer screen that will cost up in the end.

[monkeywantsabanana]

Hi,

I HAVE FORGOTTEN MECHANICS !

Ok so... if you have an object placed on a rough ramp... do you have a friction force, a force going down the ramp, a force going up the ramp (if there's a string attached trying to pull it up), and a normal force acting on this object as well as gravity?

I'm not sure if what I said above made any sense...

Also, is the friction contained in the force going downward the ramp or are they two separate forces?

[dc302]

When you are drawing the diagram, there is no force going down the ramp. Any resultant force going down the ramp is a result of the sum of all the other forces (namely, the resolution of the gravity force in the downward direction, minus the tension from the string and the resistance).


edit: and no, the friction force is not 'contained' in any other force. It is a force on its own and is given by the equation (for spesh's intents and purposes) Fr = u*N, where u is the coefficient of friction, N is the magnitude of the normal force and Fr is the magnitude of the friction force.

edit2: oops I should also clarify that the 'edit' scenario only applies if the object is moving down the ramp or on the verge of moving down the ramp. Otherwise, the frictional force is equal and opposite to the resultant downward force. Uhh hopefully this isn't too confusing...

[monkeywantsabanana]

4 loose leafs and 3 pencil leads... I'm still stuck... Can someone help me with this question?

Prove:

[pi]

4 loose leafs and 3 pencil leads... I'm still stuck... Can someone help me with this question?

Prove:

This might be a bit long-winded, but here it goes:

























edit: made a typo in one line, oops

[monkeywantsabanana]

How did you just know to multiply the first by and the second one by ? Does it become second nature to you?

[pi]

Well, I just wanted the same denominator, so that seemed logical