jinny1's Specialist Questions Thread :D

[jinny1]

Thanks u heaps guys who needs a personal tutor these days when u hav atarnotes lol

[costa94]

(y-k)^2              (x-h)^2
----------       +    ----------      =    1
    b^2                    a^2

asymptotes: y - k = +- b/a (x - h)

[jinny1]

Does anyone know how to solve differential equation questions of vector forces??? Like a= sin(t)i + cos(t)j + sin(2t)k , fine velocity given v(0)=0. thank you!

by using a calculator

[dc302]

The relationship between acceleration a, velocity v and position r is:

a = dv/dt, d = dr/dt

So if you have a(t), you need to antidiff it to find v(t).

So if a = sint i + cost j + sin(2t) k,

then v = (-cost + c)i + (sint + d)j + (-0.5 cos(2t) + e)k, where c,d,e are constants. Now you just have to find the constants given that v(0)=0.

[jinny1]

The relationship between acceleration a, velocity v and position r is:

a = dv/dt, d = dr/dt

So if you have a(t), you need to antidiff it to find v(t).

So if a = sint i + cost j + sin(2t) k,

then v = (-cost + c)i + (sint + d)j + (-0.5 cos(2t) + e)k, where c,d,e are constants. Now you just have to find the constants given that v(0)=0.

oh shoot.. sorry!!!!! i meant on the calculator XD if i didn't know how to do them by hand by now then i wud be fked!

[Greatness]

I dont think its possible to do it on the cas? doesnt look like theres a  function for it in the menu :/

[jinny1]

So guys with complex numbers where z= x+yi

when we expand z * z(conjugate) we calculate : (x+yi)(x-yi)

but when we are trying to evaluate absolute function of z (abs(z)) why do we ignore the 'i' and just write: (x^2 + y^2)^1/2



shouldnt it be (x^2 + (yi)^2)^1/2 ?

[brightsky]

absolute value of z is effectively it's distance from the origin. it can be found simply by using pythagoras' theorem. and either way, they are equivalent:
zz* = (x+yi)(x-yi) = x^2 - y^2i^2 = x^2 - y^2(-1) = x^2 + y^2
square root that to give the absolute value.

[jinny1]

oh ok.. but isn't it technically wrong to just ignore the "i" when we are evaluating the absolute value of z

[dc302]

oh ok.. but isn't it technically wrong to just get rid of the "i" when we are evaluating the absolute value of z

Nope, it's not wrong. You can think of |z| as simply the square root of zz* (where z* is the conjugate).

[jinny1]

oh ok.. but isn't it technically wrong to just get rid of the "i" when we are evaluating the absolute value of z

Nope, it's not wrong. You can think of |z| as simply the square root of zz* (where z* is the conjugate).

hmmm.yea i understand but with any other absolution value functions: we always square the all the coeffcients of x and y. what makes 'i' so special that we can ignore it now??

[brightsky]

we did square it. i^2 = -1.

[dc302]

we did square it. i^2 = -1.

She is talking about |z|, because she is wondering why it's not sqrt( x^2 + (iy)^2 ), and I was saying that the i^2 is not supposed to be there.

oh ok.. but isn't it technically wrong to just get rid of the "i" when we are evaluating the absolute value of z

Nope, it's not wrong. You can think of |z| as simply the square root of zz* (where z* is the conjugate).

hmmm.yea i understand but with any other absolution value functions: we always square the all the coeffcients of x and y. what makes 'i' so special that we can ignore it now??

This may be hard to understand, but it is because i is linearly independent from the real numbers. That means no matter what real number you have, you cant make the number i. It is much easier if you think of the complex numbers as simply a means of writing coordinates in the plane (even though, of course, there is much more to complex numbers than just that). Sorry I can't really explain, maybe someone else can try.

[jinny1]

we did square it. i^2 = -1.

She is talking about |z|, because she is wondering why it's not sqrt( x^2 + (iy)^2 ), and I was saying that the i^2 is not supposed to be there.

oh ok.. but isn't it technically wrong to just get rid of the "i" when we are evaluating the absolute value of z

Nope, it's not wrong. You can think of |z| as simply the square root of zz* (where z* is the conjugate).

hmmm.yea i understand but with any other absolution value functions: we always square the all the coeffcients of x and y. what makes 'i' so special that we can ignore it now??

This may be hard to understand, but it is because i is linearly independent from the real numbers. That means no matter what real number you have, you cant make the number i. It is much easier if you think of the complex numbers as simply a means of writing coordinates in the plane (even though, of course, there is much more to complex numbers than just that). Sorry I can't really explain, maybe someone else can try.

yea thats exactly what i was trying to say i'm a guy btw..
If anyone has a definitive answer it would be great

it just seems like there are double standards for dealing with complex numbers...

[b^3]

we did square it. i^2 = -1.

She is talking about |z|, because she is wondering why it's not sqrt( x^2 + (iy)^2 ), and I was saying that the i^2 is not supposed to be there.

oh ok.. but isn't it technically wrong to just get rid of the "i" when we are evaluating the absolute value of z

Nope, it's not wrong. You can think of |z| as simply the square root of zz* (where z* is the conjugate).

hmmm.yea i understand but with any other absolution value functions: we always square the all the coeffcients of x and y. what makes 'i' so special that we can ignore it now??

This may be hard to understand, but it is because i is linearly independent from the real numbers. That means no matter what real number you have, you cant make the number i. It is much easier if you think of the complex numbers as simply a means of writing coordinates in the plane (even though, of course, there is much more to complex numbers than just that). Sorry I can't really explain, maybe someone else can try.

I'm not sure if this is entirely correct but here is how I look at it.
When we have |z|, it is similar to say taking the magnitude of a vector (because this is when the magnitude of z is say 2 e.t.c)
So when we do that with vectors, we take the coefficients in front of i, j, k and square them, then root them. Here we take the coffecients of the "real numbers" and "imaginary numbers" and square them then root the whole thing.