My Specialist Maths Mega-thread :D

[aznxD]

Though, it may be useful if typing into the calculator 'cus there's no

You can use the define function on the inspire by:
define

Credit: b^3 Guide to Using the Ti-nspire for Specialist.

[Random_Guy]

How do I sketch circles, ellipses and hyperbolas on a Ti-89?

[pi]

How do I sketch circles, ellipses and hyperbolas on a Ti-89?

I believe that on the 89, you'll have to put the relation into y=... and then you'll find there will be two of these equations, you graph each one individually to produce the whole relation

[monkeywantsabanana]

Let and

Show that lies on the ellipse with equation:

[nubs]

Let and

Show that lies on the ellipse with equation:

Sub in into for

expand the cis into cos and isin

then for the 1/ part (1/w), rationlise it by multiplying the bottom and top by (cos(theta) - isin(theta))

you'll then end up with cos(theta)/2 - isin(theta)/2
add that to the 2cos(theta) + 2isin(theta) you found earlier (the w)

cbf writing theta, so I'll let theta = a

so you now have z = 5/2cos(a) + 3/2isin(a)
As we know that z=x+iy, we can equate coefficients

so x= 5/2cos(a) and y=3/2sin(a)
from here on in it's just parametric equations
cos(a)= 2x/5 and sin(a)=2y/3
we know that cos(a)^2 + sin(a)^2 = 1
so (2x/5)^2 + (2y/3)^2 = 1
so 4x/25 + 4y/9 = 1
x/25 + 4y/9 = 1/4

and yeah that's pretty much the maths part of it I think, sorry if it's not pretty enough for you

[monkeywantsabanana]

Let and

Show that lies on the ellipse with equation:

Sub in into for

expand the cis into cos and isin

then for the 1/ part (1/w), rationlise it by multiplying the bottom and top by (cos(theta) - isin(theta))

you'll then end up with cos(theta)/2 - isin(theta)/2
add that to the 2cos(theta) + 2isin(theta) you found earlier (the w)

cbf writing theta, so I'll let theta = a

so you now have z = 5/2cos(a) + 3/2isin(a)
As we know that z=x+iy, we can equate coefficients

so x= 5/2cos(a) and y=3/2sin(a)
from here on in it's just parametric equations
cos(a)= 2x/5 and sin(a)=2y/3
we know that cos(a)^2 + sin(a)^2 = 1
so (2x/5)^2 + (2y/3)^2 = 1
so 4x/25 + 4y/9 = 1
x/25 + 4y/9 = 1/4

and yeah that's pretty much the maths part of it I think, sorry if it's not pretty enough for you

OMG YOU GENIUS! THANKS I got up to this line then died.

"so you now have z = 5/2cos(a) + 3/2isin(a)"

Thanks!

[Planck's constant]

Let and

Show that lies on the ellipse with equation:

Sub in into for

expand the cis into cos and isin

then for the 1/ part (1/w), rationlise it by multiplying the bottom and top by (cos(theta) - isin(theta))

you'll then end up with cos(theta)/2 - isin(theta)/2
add that to the 2cos(theta) + 2isin(theta) you found earlier (the w)

cbf writing theta, so I'll let theta = a

so you now have z = 5/2cos(a) + 3/2isin(a)
As we know that z=x+iy, we can equate coefficients

so x= 5/2cos(a) and y=3/2sin(a)
from here on in it's just parametric equations
cos(a)= 2x/5 and sin(a)=2y/3
we know that cos(a)^2 + sin(a)^2 = 1
so (2x/5)^2 + (2y/3)^2 = 1
so 4x/25 + 4y/9 = 1
x/25 + 4y/9 = 1/4

and yeah that's pretty much the maths part of it I think, sorry if it's not pretty enough for you

OMG YOU GENIUS! THANKS I got up to this line then died.

"so you now have z = 5/2cos(a) + 3/2isin(a)"

Thanks!

Good method.

... excpet for the bit

'then for the 1/ part (1/w), rationlise it by multiplying the bottom and top by (cos(theta) - isin(theta))'

... more elegant to state,

1/ = /2     .... by DeMoivre

and proceed as per your method

[monkeywantsabanana]

How would you go about doing this question?:

Use the double angle formula for and the fact that to find the exact value of

[pi]

Hint: let x = pi/8 for the double angle tan formula and solve for tan(x)

[pi]

SPOILER:






(quadratic formula)

(due to quadrant)

[monkeywantsabanana]

AHHH I SEE, i kept on trying to get on the left hand side... silly me.

Thanks a lot !

[monkeywantsabanana]

Also, there's this question: , is an element of

Find









Then the answer says: is an element of and it takes the positive square root as the answer

What exactly does this mean

[pi]

It means that the angle x/2 is somewhere between and inclusive of pi/2 and 3pi/4 Sine values are positive in that domain (the domain is part of the second quadrant), hence the positive root

[monkeywantsabanana]

I don't understand the difference between scalar resolutes and perpendicular vector resolutes.



SCALAR RESOLUTE: OA = u^.v 

VECTOR RESOLUTE: v perpendicular = v - v|| (parallel)

THEN that means that v parallel = OA....

Why does v perpendicular, in the book equals v - u^(u^.v). I would expect it to equal v- (u^.v)

Why is there an extra u^?

Sorry, this might be hard to follow... but someone please enlighten me...

[TrueTears]

u^(u^.v) is the (parallel) VECTOR resolute (in other words it's a vector)

(u^.v) is just the SCALAR resolute (it's a magnitude)

obviously you can only "subtract" vector and vector