Bazza's 3/4 Question Thread

[BubbleWrapMan]

^ Actually this is the formula in Spesh course you can apply directly without doing antiderivative

huh

[Lasercookie]

^ Actually this is the formula in Spesh course you can apply directly without doing antiderivative

huh

I think Jenny is saying that is one of those "short-cut" formulas you can use to check over your work quickly etc.

[BubbleWrapMan]

Oh yeah I see what you mean.

[Jenny_2108]

^ Actually this is the formula in Spesh course you can apply directly without doing antiderivative

huh

I think Jenny is saying that is one of those "short-cut" formulas you can use to check over your work quickly etc.

Yeah, this is what I meant 

[WhoTookMyUsername]

If you have an arbitrary point z on the complex plane, with no coordinate labels, is it possible to qualitatively graph the multiplicative inverse of z? 1/z
If so how?
Thanks

[Hancock]

Let's take an arbitrary point z = 1 + i.

So z^-1 = 1 / (1+i) = (1-i) / (1-i)(1+i) = (1-i) / (2) = (1/2)(1-i)

So we can see that the magnitude of z^-1 is the dilated by the reciprocal of the magnitude of z, and the vector from the origin to the point z^-1 is the conjugate of z.

[oneoneoneone]

[BubbleWrapMan]

Polar form is easier and more qualitative:



You can see you just negate the angle and invert the magnitude. They usually tell you |z| > 1 or |z| < 1 so you know 1/|z| < 1 or 1/|z| > 1, respectively.

[WhoTookMyUsername]

thanks for the help

Just did VCAA 07 (2)
and the solutions to one part had me stumped
(exam and corresponding assessment report are http://www.vcaa.vic.edu.au/Pages/vce/studies/mathematics/specialist/exams.aspx;
not needed to answer my question i don't think)

1) One question, involving vectors (plane path etc.) asked "when is the aircraft closest to the base of the control tower?"
I did d=mag(r(position vector))
and then found dd/dt
then dd/dt = 0 etc. which gave me the correct answer
however the solutions stated that
r . v = 0
(position and velocity vectors are perpendicular when distance from origin is at a minimum ))

I thought about it for a while and couldn't see why. I drew a circle; but that didn't help lol,
not sure about the reasoning behind this, could someone please explain?

Question based on VCAA 2007 Exam 2 ER Q4d)

EDIT: I "think" i can picture it now... but if anyone has a more mathematically grounded explanation that would be great!

2) (Unrelated)
Is it acceptable to use the, slightly dodgier, but more aesthetically pleasing sin^-1(x) as inversesin(x) for VCAA exams? VCAA uses arcsin(x) etc.

thanks in advance!

[Conic]

]2) (Unrelated)
Is it acceptable to use the, slightly dodgier, but more aesthetically pleasing sin^-1(x) as inversesin(x) for VCAA exams? VCAA uses arcsin(x) etc.

thanks in advance!

I would use arcsin(x), as it's definitely acceptable, and shorter than inversesin(x), and it wouldn't be confused with csc(x) like sin^-1(x).

[pi]

sin^-1(x) and arcsin(x) are generally the normal ways of writing inverse sine

[Jenny_2108]

1) One question, involving vectors (plane path etc.) asked "when is the aircraft closest to the base of the control tower?"
I did d=mag(r(position vector))
and then found dd/dt
then dd/dt = 0 etc. which gave me the correct answer
however the solutions stated that
r . v = 0
(position and velocity vectors are perpendicular when distance from origin is at a minimum ))

I thought about it for a while and couldn't see why. I drew a circle; but that didn't help lol,
not sure about the reasoning behind this, could someone please explain?

In geometry, if you draw a circle, you can see that the closest distance is the from the line perpendicular to the tangent right?



In this situation, velocity vector is the derivative of position vector, so you can consider it as "tangent line" AC (of course they are not the same thing but we are making comparision by a simple case) and position vector is vector r from the origin to the point of particle's position. In here, we can consider position vector as OB
Thus, the closet distance when position vector is perpendicular to velocity vector

If you wanna see the reason behind without using geometry, we can prove by vector methods

Given postition vector

So we can find the velocity vector:



To find the closest distance, we find the derivative and let it equal 0 so

[WhoTookMyUsername]

awesome explanation! thanks

sin^-1(x) and arcsin(x) are generally the normal ways of writing inverse sine

Did you use sin^-1(x) in your exams?

[pi]

awesome explanation! thanks

sin^-1(x) and arcsin(x) are generally the normal ways of writing inverse sine

Did you use sin^-1(x) in your exams?

Yeah, I used that one

[WhoTookMyUsername]

Ahk thanks (did anyone else for double confirmation xD?)

Anyone know how to solve parametric equations for x and y (cartesian) on cas?
With previous os you went solve(2eq, t)
But that doesn't work anymore D: