Post by: WhoTookMyUsername on January 03, 2012, 11:44:30 am
2) Difference with polar?
3) When should we use and not use it?
4) What's the best way to approach questions such as
Is it to remember that i^4 = 1 and then cancel down and remember that
Post by: paulsterio on January 03, 2012, 12:52:41 pm
1) What is rectangular mode on the CAS?
2) Difference with polar?
3) When should we use and not use it?
4) What's the best way to approach questions such as?
Is it to remember that i^4 = 1 and then cancel down and remember that
etc? (best to learn these off by memory?)
1) - Rectangular is a+bi form
2) - Polar, in VCE, is rcis(theta) form, but can also be e^(i*theta)
3) - What do you mean when should we use and not use it? Depends on the question?
This is how I would do 4)
Post by: WhoTookMyUsername on January 03, 2012, 02:16:20 pm
Like what mental process did you follow specifically?
thanks
Post by: Lasercookie on January 03, 2012, 02:23:34 pm
thanks paul, just with the last one how did you get from1 multiplied by itself 503 times is just 1to 1 ?
Like what mental process did you follow specifically?
thanks
Post by: Phy124 on January 03, 2012, 02:24:20 pm
thanks paul, just with the last one how did you get from1 to the power of any number = 1?
Like what mental process did you follow specifically?
thanks
edit: never mind laseredd beat me to it
Post by: paulsterio on January 03, 2012, 02:32:43 pm
If
If
Post by: WhoTookMyUsername on January 06, 2012, 10:39:57 am
in some places it suggest cartesion form is better as
thanks
Post by: paulsterio on January 06, 2012, 10:48:19 am
All forms are ok unless a specific form is required, this is usually always true, for example, 2(i+j+k) is as acceptable as 2i+2j+2k
Post by: WhoTookMyUsername on January 06, 2012, 11:04:49 am
1) if it says "express in the form x + yi" then is 2+2i better?
2) In year 11 we were taught to the arg(z) before quadrant was taken into account as A (alpha). Is this notation OK in year 12?
Post by: paulsterio on January 06, 2012, 11:23:28 am
2) Hmmm, I don't know if the notation is ok or not, but i'd just say let "alpha" = arg(z) and then go on to do the working, if you define it, it's defs ok.
Post by: WhoTookMyUsername on January 06, 2012, 11:37:21 am
Post by: WhoTookMyUsername on January 07, 2012, 10:50:32 am
and there's a 0 or 0i , do i write that down?
(e.g. 0 + i, 1 + 0i)?
2) is sina okay ? or is sin (a) preferable (do they take off marks for the former?)
Post by: monkeywantsabanana on January 07, 2012, 11:18:38 am
another "form" type quetions. If i'm asked to express in the form x + yi / cartesian form
and there's a 0 or 0i , do i write that down?
(e.g. 0 + i, 1 + 0i)?
2) is sina okay ? or is sin (a) preferable (do they take off marks for the former?)
No you just write the imaginary or real where they exist.
--> 0+i = i
I've been writing my trigs with and without the brackets whenever i feel like it... I don't think it's a big problem as long as it's clear
Post by: paulsterio on January 07, 2012, 11:13:41 pm
Similarly with the trigonometric functions, why would you leave out the brackets? - Marks won't be taken off, I don't think, but it's not much of a hassle to put in brackets, so do that, plus it's a good habit, it makes things clearer
Post by: WhoTookMyUsername on January 14, 2012, 08:28:21 pm
Thanks
Post by: brightsky on January 14, 2012, 08:38:35 pm
Post by: pi on January 14, 2012, 08:43:28 pm
How did you simplify
Post by: WhoTookMyUsername on January 14, 2012, 09:21:10 pm
It's 2 separate examples
with both denominators in linear form (what do i call this, a hyperbolic part?)
whilst
I can't understand why having
Post by: pi on January 14, 2012, 09:27:18 pm
etc.
Just takes practise to get used to them.
I can't understand why havingas the original denominator produces a "hyperbolic" part and a "truncus" part (1/x^2) whilst having a denominator like (x+3)(x-1) produces two hyperbolic parts
That really goes back to each expression being different, its not something that has an "explanation", its just something that just is what it is.
Post by: brightsky on January 14, 2012, 11:41:41 pm
(2x-3)/(4x-7)^2 = A/(4x-7) + B/(4x-7) = (A+B)/(4x-7), which is impossible since (2x-3) isn't a factor of (4x-7)^2
Post by: WhoTookMyUsername on January 15, 2012, 09:47:30 am
uh, is it possible to use "inspection" when dealing with partial fractions that where the numerator has the same or higher degree than the denominator to reduce the numerator power? (i don't think it is, but enwiabe wasn't aware of a place where long division of polynomials was actually necessary, is this it xD)
Post by: WhoTookMyUsername on January 15, 2012, 11:14:35 am
how come when
the resultant graph CROSSES the asymptote? WTF!!! Isn't the whole point of an asymptote that it can't be crossed???
Please explain!!! (firstly, in terms of WHY it can cross the asymptote, and secondly, in what cases / where do you know it will cross (without solving all the time))
thanks
Post by: dc302 on January 15, 2012, 11:53:33 am
I've just been completely mindf*****...
how come when
the resultant graph CROSSES the asymptote? WTF!!! Isn't the whole point of an asymptote that it can't be crossed???
Please explain!!! (firstly, in terms of WHY it can cross the asymptote, and secondly, in what cases / where do you know it will cross (without solving all the time))
thanks
It crosses the asymptote of the individual functions a/f(x) and b/g(x), but the whole thing does not have the same asymptote, so there is no reason to assume it shouldn't cross this. It may be easier if you plug in actual numbers to see how it works.
Also, I assume you are not talking about the f(x) = 0 or g(x) = 0 asymptote?
Post by: enwiabe on January 15, 2012, 11:53:49 am
I've just been completely mindf*****...
how come when
the resultant graph CROSSES the asymptote? WTF!!! Isn't the whole point of an asymptote that it can't be crossed???
Please explain!!! (firstly, in terms of WHY it can cross the asymptote, and secondly, in what cases / where do you know it will cross (without solving all the time))
thanks
It CAN be crossed, but not in the infinite limit. The asymptote is what happens as SOMETHING goes to infinity. In the finite realm, an asymptote can certainly be crossed.
Post by: TrueTears on January 15, 2012, 12:11:15 pm
yup exactly what enwiabe said, just to reiterate further, it's always good to go back and check the formal definitions to clear up any misunderstandings. Most of the time, misunderstandings come from not knowing the rigorous definition and rather applying your own definition of it.I've just been completely mindf*****...
how come when
the resultant graph CROSSES the asymptote? WTF!!! Isn't the whole point of an asymptote that it can't be crossed???
Please explain!!! (firstly, in terms of WHY it can cross the asymptote, and secondly, in what cases / where do you know it will cross (without solving all the time))
thanks
It CAN be crossed, but not in the infinite limit. The asymptote is what happens as SOMETHING goes to infinity. In the finite realm, an asymptote can certainly be crossed.
http://en.wikipedia.org/wiki/Asymptote
Post by: WhoTookMyUsername on January 15, 2012, 12:23:28 pm
i think i understand where i went wrong now
Post by: WhoTookMyUsername on January 15, 2012, 12:55:39 pm
Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)
thanks again :)
Post by: pi on January 15, 2012, 01:12:07 pm
Post by: Special At Specialist on January 15, 2012, 04:25:51 pm
No matter how many times you use partial fractions or re-arrange the equation, it is still the same equation, will still look exactly the same and will still have the same asymptotes.
Post by: WhoTookMyUsername on January 15, 2012, 06:21:15 pm
I don't think they do actually.
m... But because you're essentially adding the y values still, it should never "touch" the original two functions ? Am i missing something here
Moderator action: removed real name, sorry for the inconvenience
Post by: Planck's constant on January 15, 2012, 06:51:52 pm
Post by: kamil9876 on January 15, 2012, 07:02:20 pm
I was also going to ask about the silly function:
does it have an assymptote at
Post by: Planck's constant on January 15, 2012, 07:17:13 pm
Post by: Incommensura on January 15, 2012, 07:18:10 pm
Actually feel free to provide an answer. I was just going to explain about how I don't really know the precise definition of assymptote for graphs ( I guess for functions it's easy to define). TT's link doesn't really provide a rigorous definition.
I was also going to ask about the silly function:
does it have an assymptote at?
Odd question! My money would be on a hollow dot at x=1 and obviously 0 elsewhere. An asymptote wouldn't make sense because it can't 'approach' anything given it is zero for all values.
Quote
For something like
y=\frac {a} {f(x)} + \frac {b} {g(x)} + c
Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)
I'm pretty sure they do also. Just by common sense - even if c=0, the first term can never be 0 (it's a fraction) so the function can never equal the second term, and vice versa.
For the record though, it's almost certainly not mentioned in the book because (somebody correct me if I'm wrong) you never have to sketch or mark in a non-linear asymptote. Though it will still be useful for getting the correct shape, I guess.
Post by: kamil9876 on January 15, 2012, 07:32:59 pm
Quote
For something like
Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)
Post by: dc302 on January 15, 2012, 07:40:45 pm
Actually feel free to provide an answer. I was just going to explain about how I don't really know the precise definition of assymptote for graphs ( I guess for functions it's easy to define). TT's link doesn't really provide a rigorous definition.
I was also going to ask about the silly function:
does it have an assymptote at
This is a 'removable singularity' if my memory serves me correctly.
edit: also known as a 'pole of order 0'. Point is, things don't get this complicated in spesh so I wouldn't worry lol
Post by: kamil9876 on January 15, 2012, 07:46:51 pm
Post by: dc302 on January 15, 2012, 07:49:51 pm
edit: btw what is Walter Rudin lol
Post by: kamil9876 on January 15, 2012, 07:58:30 pm
Quote
btw what is Walter Rudin lol
He is a guy who wrote what many people (not so much me) treat as the bible of undergraduate Analysis.
Post by: Planck's constant on January 15, 2012, 08:14:52 pm
For the record though, it's almost certainly not mentioned in the book because (somebody correct me if I'm wrong) you never have to sketch or mark in a non-linear asymptote. Though it will still be useful for getting the correct shape, I guess.
Unless I am totally missing your point, you are quite often required to sketch non-linear asymptotes in spesh, eg
f(x) = x^2 + 1/x
Post by: WhoTookMyUsername on January 15, 2012, 08:35:15 pm
For something like
Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)
thanks again :)
So lol... uh... what's the answer xD ?
yea i was thinking along incomm's line i think... but i'm not sure;does anyone know?; can pi explain your answer a bit more?
(and i figured out the "crossing the asymptote" one i think, the books like sketch the graphs of
so c is the asymptote for the second one but not the overall one... i think...
Moderator action: removed real name, sorry for the inconvenience
Post by: Incommensura on January 15, 2012, 08:39:34 pm
Unless I am totally missing your point, you are quite often required to sketch non-linear asymptotes in spesh, eg
f(x) = x^2 + 1/x
Right, but questions like that (at least on the exam, SACs can be nastier) won't require you to draw in the asymptote for any mark, and our teacher told us we'd never have to actually mark that y=x^2 asymptote on the graph. Like I said, useful for shape of that graph.
Post by: Incommensura on January 15, 2012, 08:49:22 pm
^ I know what the graph looks like, just asking for what the definition of asymptote is. You're contradicting yourself there by saying no to my example but yes to:QuoteFor something like
Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)
Okay well answering that question I kind of implicitly assumed that f(x) and g(x) weren't the same and that b wasn't -a...
Asymptotes are defined as a curve which the function approaches, so knowing what the graph looks like is basically enough to tell if something's an asymptote - if there's no approaching, then there's no asymptote.
Of course I'm not actually super-pro at maths and if anybody does want to pull out their Rudin and prove me wrong then I'll wear it :)
Post by: WhoTookMyUsername on January 15, 2012, 08:50:31 pm
For something like
Do f(x) and g(x) become asymptotes as well ? (i think they do but the book doesn't mention it anywhere)
thanks again :)
So lol... uh... what's the answer xD ?
yea i was thinking along incomm's line i think... but i'm not sure;does anyone know?; can pi explain your answer a bit more?
(and i figured out the "crossing the asymptote" one i think, the books like sketch the graphs ofand
so c is the asymptote for the second one but not the overall one... i think...
anyone xD
Moderator action: removed real name, sorry for the inconvenience
Post by: dc302 on January 15, 2012, 09:06:38 pm
edit: Maybe I should provide more reasoning. When f(x) = 0, g(x) = 0, you get vertical asymptotes (because diving by zero gets you infinity). These asymptotes are almost always preserved, and the only time it doesn't is when they cancel each other out (think about infinity minus infinity). This can result in the the 'asymptote' going to some constant number, or even to infinity again (think about if you get infinity(1) minus infinity(2) where infinity(1) is 'bigger', so the overal effect goes to 1).
example: 1/x^2 - 1/x
Post by: WhoTookMyUsername on May 17, 2012, 07:28:20 pm
Do we have to take out the common factor?
E.g. ( simplified example)
Is 2(x+1)^2 + (x+1) + c Ok?
Or do we have to make it
(x+1)(2(x+1)+1)+c
?
Thanks
Post by: Bhootnike on May 17, 2012, 07:34:00 pm
Post by: WhoTookMyUsername on May 17, 2012, 07:40:41 pm
Post by: Bhootnike on May 17, 2012, 07:48:12 pm
maybe if the question explicitly states, in the form ... (x+a)(f(x+b) + c or something.. , then definitely.
Post by: pi on May 17, 2012, 09:03:38 pm
Post by: WhoTookMyUsername on May 19, 2012, 11:56:29 am
Are you allowed to have an integral in terms of u but diffing with respect to dx at any point during your working out? I thought you werent allowed to, but ive ween things like S y^2 dx for solids of rev
Like
S 4u dx
?
( s is integral sign)
Post by: pi on May 19, 2012, 12:29:53 pm
Post by: kamil9876 on May 19, 2012, 12:36:47 pm
Quote
I thought you werent allowed to, but ive ween things like S y^2 dx for solids of rev
Just think of y as shorthand for something in terms of x (i.e it could y=sin(x) or whatever). You can have something not in terms of x inside, just make sure that when you integrate you're not mistakenly integrating with respect to y. So in most practical cases, you would eventually express everything in terms of x and then start integrating.
Post by: WhoTookMyUsername on May 19, 2012, 01:15:12 pm
No that's not allowed (by VCAA in any case). If you have "dx" at the end, the variable inside the integral should be "x" only and furthermore, the terminals of the integral must be related to "x". Once you change it to "u", "dx" becomes "du" (with appropriate substitution) and the terminals of the integrals must be changed too.
QuoteI thought you werent allowed to, but ive ween things like S y^2 dx for solids of rev
Just think of y as shorthand for something in terms of x (i.e it could y=sin(x) or whatever). You can have something not in terms of x inside, just make sure that when you integrate you're not mistakenly integrating with respect to y. So in most practical cases, you would eventually express everything in terms of x and then start integrating.
D: are these contradictory?
If you say y is something in terms of x,
After youve let u = f(x)
Then having a u in there would effectivelye Be the same thing?
Is VCAA contradicting themselves allowing y^2 dx as a line of working?
Thanks
Post by: Mao on May 19, 2012, 01:18:56 pm
Post by: pi on May 19, 2012, 01:38:53 pm
D: are these contradictory?
Nope, it's just what Mao pointed to, there is a difference between y and y(x).
But personally, I'd just keep it simple and use dx when there is "x" as the variable, and du when "u" is the variable, most people do this in spesh.
Post by: paulsterio on May 22, 2012, 07:57:57 pm
Cause if you get the right answer, but factorise it and get the factorising wrong, then you will lose the answer mark for it, however, if you had just left it alone, you would have gotten full marks, like I always say, unless you need to express something in a certain form as stated by the question, avoid manipulating it, especially expanding and factorising, unless it's obvious or required.
Post by: WhoTookMyUsername on June 23, 2012, 01:24:03 pm
?
Ive tried to look it up online but i dont understand the implentation of the variation of the chain rule
Thanks!
Post by: kamil9876 on June 23, 2012, 02:25:09 pm
Where I used the chain rule to get to the second line, then I used the product rule.
Post by: WhoTookMyUsername on June 23, 2012, 03:03:42 pm
Can use the inside outside thingo
is there anyway to do it a bit more extended?
Like say m=e^xy
If you were to lengthen out the chain rule how would you get
Dm/dx ?
( e.g. Let u = e^x)
Thanks!
Post by: WhoTookMyUsername on June 24, 2012, 05:11:17 pm
Its driving me insane
Ugh finally got it it D:
How can i see what to integrate quickly ? Like what i should try and convert to?
Post by: pi on June 24, 2012, 05:19:15 pm
How do i integrate 1/(1+sinx)
Its driving me insane
dw about it, you'll never get anything that complicated in any SAC/exam without prompting on what to do next.
Post by: Somye on June 24, 2012, 06:04:43 pm
Post by: WhoTookMyUsername on June 24, 2012, 06:17:01 pm
Uh "rationalise it" then split into 2 integrals then integrate
Post by: WhoTookMyUsername on June 25, 2012, 04:35:57 pm
i did e^y = 2x
and then integrated with respect to y 1/2S(2,0 terminals)e^y dy is what i put
is that correct?
(my teachers answers have something random)
Post by: b^3 on June 25, 2012, 04:42:28 pm
what is the area bound by y=ln2x , axes and y=2?Thats they way I'd look at it, unless I've missed something. Ended up with
i did e^y = 2x
and then integrated with respect to y 1/2S(2,0 terminals)e^y dy is what i put
is that correct?
(my teachers answers have something random)
What did your teacher do?
Post by: WhoTookMyUsername on June 25, 2012, 04:49:01 pm
The area bounded by the curve y = ln2x , the coordinate axes, and the line y =2 is given by the integral :
he put C
S((e^2)/2, 1 terminals) ln2x dx
he put worked solutions up which make some reference to the "inverse" having the same amount under the x axes but i'm not sure why he did that or whether i'm missing somethign
Post by: Phy124 on June 25, 2012, 04:53:29 pm
Where the
(same answer as b^3)
So yeah you guys are right (AFAIK)
edit: changed x to 2x
Post by: b^3 on June 25, 2012, 05:57:23 pm
So if its exam 1, then you would need to do the first one.
Post by: WhoTookMyUsername on June 25, 2012, 07:58:00 pm
i think the infallible Dr. G. has made 2 mistakes in 3 of the past SACs combined
one was having 2 correct options, another was asking to prove something incorrect :O
does anyone have a good 2 (double sided) page calculus cheat sheet by any chance :D
Post by: WhoTookMyUsername on June 25, 2012, 08:15:39 pm
it was exam 2, both were options on the SAC
i think the infallible Dr. G. has made 2 mistakes in 3 of the past SACs combined
one was having 2 correct options, another was asking to prove something incorrect :O
does anyone have a good 2 (double sided) page calculus cheat sheet by any chance :D
Also
When it says evaluate the integral, do you need to account for when the equation crosses the x axis?
(im pretty sure not, but i almost made this mistake today, want to be sure)
Post by: kensan on June 25, 2012, 08:19:37 pm
Post by: brightsky on June 25, 2012, 09:14:51 pm
When it says evaluate the integral, you just do it without worrying about the graph interesting the x-axis. But if it asks specifically for the area, then you do account for it :)
yeah, don't equate integration and area calculation. integration is but a technique that facilitates the calculation of areas under curves. so when asked to evaluate a definite integral, don't even bother looking at graphs or adding units^2. just solve it as per usual.
Post by: sahil26 on June 25, 2012, 09:44:56 pm
Help would be appreciated :)
Post by: brightsky on June 25, 2012, 09:59:49 pm
Post by: kamil9876 on June 25, 2012, 10:00:43 pm
I assume that what you're really asking for is "why does Antidifferentiation give me area"
(because more sophisticated definitions of the integral pretty much already give you integral=area just from the definition)
Suppose
Let
We want to show that
what we really want to study is:
Now
(where
Now as
Hence
Post by: paulsterio on June 26, 2012, 12:27:34 am
Post by: brightsky on June 26, 2012, 11:31:47 am
I think the easier way to think of Kamil's post is to imagine dividing the area up into rectangles, then having the width of the rectangles tend towards 0 and then summing them, it will give the integral :D
nah that doesn't actually prove that A'(x) = f(x). it simply gives you a way of evaluating A(x).
Post by: paulsterio on June 26, 2012, 11:46:32 am
My bad!
Post by: kamil9876 on June 26, 2012, 12:00:02 pm
Quote
I assume that what you're really asking for is "why does Antidifferentiation give me area"
(because more sophisticated definitions of the integral pretty much already give you integral=area just from the definition)
paulsterio had the better definition of integral in mind, but I assumed that sahil26 was interested in the "antiderivative definition"
Post by: paulsterio on June 26, 2012, 01:00:45 pm
Post by: WhoTookMyUsername on June 26, 2012, 04:57:34 pm
still not 100% clear on what this concept actually means
for the attached graph, is point C a SPOI?
( i say no, somye says yes, ), but i'm actually not 100% sure what a POI is
i thought it was the place where the gradient is largest, but that's not the case because y=x^3 has a POI at x = 0 D: and dy/dx = 0
it was MC (i'm not sure any answers are right)
the only semi viable ones are
A) B and D represent SPOI
B) A and E are non-stationary POI
C) C is a stationary point of inflection
Post by: Somye on June 26, 2012, 05:04:06 pm
Post by: WhoTookMyUsername on June 26, 2012, 05:05:51 pm
fml
anyway
what's the best clear definition (practical) of a POI?
Post by: Panicmode on June 26, 2012, 05:22:38 pm
oh for f**** sake
fml
anyway
what's the best clear definition (practical) of a POI?
A point of inflection occurs if f(x) is continuous at that point and there is a change of concavity at that point.
ie.
f(x) is continuous at point x = a
and f"(x) > 0 for x > a
and f"(x) < 0 for x < a
or
f(x) is continuous at point x = a
and f"(x) > 0 for x < a
and f"(x) < 0 for x > a
Obviously this is conditional upon not running into other turning points in the double derivative but you get the idea.
The basic definition is a point at which a change in concavity occurs
Post by: b^3 on June 26, 2012, 05:32:52 pm
oh for f**** sakeA point of infection (POI) is where the gradient of the graph changes from increasing (
fml
anyway
what's the best clear definition (practical) of a POI?
E.g. For
For
(https://s3.amazonaws.com/grapher/exports/igtcwu1kwo.png)
To the left of
For
(https://s3.amazonaws.com/grapher/exports/lk0zfebi9s.png)
Now looking at
EDIT: Beaten by Panicmode....but since I typed all this out and generated the graphs I'll post it anyway. Hope the examples help.
EDIT2: Called a POI a stationary point, fixed it up now
Post by: WhoTookMyUsername on June 26, 2012, 05:53:59 pm
Another q:
Is there any way of rotating something around y, (volume) but where you can't get f(y),
Like can you determine the region rotating around y by considering the region you get when you rotate around x then doing something with this?
Post by: Panicmode on June 26, 2012, 05:58:54 pm
Thanks a lot for those explanations guys :D
Another q:
Is there any way of rotating something around y, (volume) but where you can't get f(y),
Like can you determine the region rotating around y by considering the region you get when you rotate around x then doing something with this?
I think I remember reading somewhere that you could rotate around the x-axis using the inverse equation... just need to make sure you get your end points correct.
Post by: WhoTookMyUsername on June 26, 2012, 06:34:09 pm
Thanks a lot for those explanations guys :D
Another q:
Is there any way of rotating something around y, (volume) but where you can't get f(y),
Like can you determine the region rotating around y by considering the region you get when you rotate around x then doing something with this?
I think I remember reading somewhere that you could rotate around the x-axis using the inverse equation... just need to make sure you get your end points correct.
Mm... Intresting , thanks for that, anyone know the full process?
I think i know how the inverse applies to area,
If you have f(x) bound to y at say y = 1 and y = 2 then S[inverse f(x)]( terminals 1 and 2) dx is the area bound to y?
( can someone verify thhis
But im not sure how to translate this to volumes
Thanks for everyones help :)
Post by: Panicmode on June 26, 2012, 06:37:08 pm
Thanks a lot for those explanations guys :D
Another q:
Is there any way of rotating something around y, (volume) but where you can't get f(y),
Like can you determine the region rotating around y by considering the region you get when you rotate around x then doing something with this?
I think I remember reading somewhere that you could rotate around the x-axis using the inverse equation... just need to make sure you get your end points correct.
Mm... Intresting , thanks for that, anyone know the full process?
I think i know how the inverse applies to area,
If you have f(x) bound to y at say y = 1 and y = 2 then S[inverse f(x)]( terminals 1 and 2) dx is the area bound to y?
( can someone verify thhis
But im not sure how to translate this to volumes
Thanks for everyones help :)
Again, I could be completely wrong. This was me studying for my calculus exam at 4 in the morning, 5 hours before the actual exam took place.
Post by: brightsky on June 26, 2012, 06:43:48 pm
Post by: paulsterio on June 26, 2012, 06:49:46 pm
y = x + 3 (for easiness sake)
for the inverse, x = y+ 3 so y = x - 3
so for the volume of rotation around the y - axis, we can do int((x-3)^2 dx) but that's essentially the same thing as int((y-3)^2 dy), you guys get what I'm trying to say right?
Post by: b^3 on June 26, 2012, 06:51:52 pm
Yes, but the inverse thing doesn't solve the problem, if you know the inverse, then essentially you can rotate it around the y-axis as well. For example,Don't forget the
y = x + 3 (for easiness sake)
for the inverse, x = y+ 3 so y = x - 3
so for the volume of rotation around the y - axis, we can do int((x-3)^2 dx) but that's essentially the same thing as int((y-3)^2 dy), you guys get what I'm trying to say right?
Post by: paulsterio on June 26, 2012, 06:56:12 pm
Post by: WhoTookMyUsername on June 26, 2012, 06:56:23 pm
so does the inverse thing work then?
(yeah this is irrelevant for the exam, but just something like this could come up on sac)
paul, it's mainly for something along the general lines of
y=f(x) is rotated around x axis between x = 1 and x = 2 (at these points y = 4 and y = 8)
the volume is 2(pi) units cubed
calculate the volume if it was rotated around the y axis
?
Post by: paulsterio on June 26, 2012, 07:01:32 pm
Post by: paulsterio on June 26, 2012, 07:06:02 pm
So with your question, y = f(x) is rotated around the x-axis between x = 0 and x = 2 (at these points, y = 0 and y = 8 ), the volume is...
What is the volume when rotated around the y-axis, it will be different depending on whether f(x) is x^3 or 2x^2 or whatever else, if you get what I mean.
Post by: WhoTookMyUsername on June 26, 2012, 07:07:33 pm
(but if there is an obscure conceptually difficult but possible way for a slightly modified quesiton i wouldn't be suprised if Dr. G. gives it to us xD)
thanks :)
Post by: paulsterio on June 26, 2012, 07:15:10 pm
Post by: pi on June 26, 2012, 07:15:59 pm
(if so, good! :D )
Post by: brightsky on June 26, 2012, 07:41:21 pm
find the volume of the solid obtained when y = e^x between y = 1 and y = 4 is rotated around the y-axis.
Post by: WhoTookMyUsername on June 26, 2012, 07:57:51 pm
V
= piS(4,1)[lny]^2 dy
for diff equations
dy^2/d^2 x
does that become (when flipped)
dx^2/d^2y?
Post by: HenryP on June 26, 2012, 08:01:30 pm
have a go at this:Well we know the formula to obtain the volume of the solid rotated about the y-axis is
find the volume of the solid obtained when y = e^x between y = 1 and y = 4 is rotated around the y-axis.
So we know
So we have Integral from
Making the substitution
Finally we have
Then Integration by parts to finish, although I don't think thats its in the course unless its changed from last year.
Hopefully there aren't any mistakes!
Post by: pi on June 26, 2012, 08:01:44 pm
and for brightsky's question, I'd just cheat and use int by parts :D
Post by: WhoTookMyUsername on June 26, 2012, 08:05:01 pm
(except second order)
and what's int by parts?
Post by: pi on June 26, 2012, 08:07:37 pm
i know it's not a fraction. i mean for d/e where dy/dx = f(y)
(except second order)
and what's int by parts?
If it's not a fraction, you can't reciprocate it under fraction laws.
Integration by parts (NOT on spesh course and NOT allowed in SACs/exams), so don't worry haha
Post by: brightsky on June 26, 2012, 08:09:19 pm
and yeah, in regards to my question, point is, you can use integration by parts, but considering the inverse function and rotating that around the x-axis is much quicker and easier way to go about it.
Post by: WhoTookMyUsername on June 26, 2012, 08:17:13 pm
ahi know it's not a fraction. i mean for d/e where dy/dx = f(y)
(except second order)
and what's int by parts?
If it's not a fraction, you can't reciprocate it under fraction laws.
Integration by parts (NOT on spesh course and NOT allowed in SACs/exams), so don't worry haha
it doesn't behave like a fraction D:
Post by: WhoTookMyUsername on June 26, 2012, 08:18:57 pm
have a go at this:
find the volume of the solid obtained when y = e^x between y = 1 and y = 4 is rotated around the y-axis.
loge(y) = x
V
= piS(4,1)[lny]^2 dy
for diff equations
dy^2/d^2 x
does that become (when flipped)
dx^2/d^2y?
was my answer wrong D:?
Post by: pi on June 26, 2012, 08:22:05 pm
loge(y) = x
V
= piS(4,1)[lny]^2 dy
was my answer wrong D:?
You don't have an answer yet :P
Post by: WhoTookMyUsername on June 26, 2012, 08:25:23 pm
(unless provided with additional information)
edit: what to include on cheat sheet
?
i have nothing for diff at the moment D:
(got a few nice WE for integ/ d/e)
Post by: paulsterio on June 26, 2012, 08:28:37 pm
For example, we can use integration by parts in order to antidifferentiate
Remember our Product Rule:
If
Then:
So essentially:
Taking the integral of both sides, we will have
Ok, so how does this help us solve the problem of integrating
Well, what we can do is say let
So essentially, we can see that
So it's easy enough now for us to be able to see now that
Which we can easily use a substitution and figure out using basic integration
Finishing off:
You won't need to be able to do that in Specialist Maths, but anyways, that's just a basic example of how integration by parts works.
Post by: WhoTookMyUsername on June 26, 2012, 08:29:58 pm
take the product rule, integrate both sides with respect to x, and rearrange, and you get your 'formula' for integration by parts.
and yeah, in regards to my question, point is, you can use integration by parts, but considering the inverse function and rotating that around the x-axis is much quicker and easier way to go about it.
OH ! WOW
NOW I GET IT
Brilliant example of what i was trying to illustrate, so you effectively integrate as is? with terminals at 1 and 4?
(e^x)
piS(1,4) e^(2x) dx ?
wait... i'm missing something here, what have i done wrong?
oh wait, aren't you still left with the same problem? having to integrate ln(x)? SOO confuzzled
and
what to include on cheat sheet
?
i have nothing for diff at the moment D:
(got a few nice WE for integ/ d/e)
Post by: pi on June 26, 2012, 08:38:47 pm
take the product rule, integrate both sides with respect to x, and rearrange, and you get your 'formula' for integration by parts.
and yeah, in regards to my question, point is, you can use integration by parts, but considering the inverse function and rotating that around the x-axis is much quicker and easier way to go about it.
OH ! WOW
NOW I GET IT
Brilliant example of what i was trying to illustrate, so you effectively integrate as is? with terminals at 1 and 4?
(e^x)
piS(1,4) e^(2x) dx ?
I don't think that's right :S
Post by: WhoTookMyUsername on June 26, 2012, 08:40:17 pm
take the product rule, integrate both sides with respect to x, and rearrange, and you get your 'formula' for integration by parts.
and yeah, in regards to my question, point is, you can use integration by parts, but considering the inverse function and rotating that around the x-axis is much quicker and easier way to go about it.
OH ! WOW
NOW I GET IT
Brilliant example of what i was trying to illustrate, so you effectively integrate as is? with terminals at 1 and 4?
(e^x)
piS(1,4) e^(2x) dx ?
I don't think that's right :S
no it's not... i thought that's where brightsky was heading but it's not D:
but considering the inverse function and rotating that around the x-axis is much quicker and easier way to go about it.is anyone able to elaborate / clarify this?
thanks :)
Post by: paulsterio on June 26, 2012, 08:46:35 pm
though integrating (ln(x))^2 is a pain, but you can do it by the substitution u = ln(x)
Post by: BubbleWrapMan on June 26, 2012, 08:51:55 pm
If you rotate a curve about a different axis, then in general you get a different shape and a different volume.but considering the inverse function and rotating that around the x-axis is much quicker and easier way to go about it.is anyone able to elaborate / clarify this?
thanks :)
Post by: pi on June 26, 2012, 08:53:54 pm
He just means taking y = ln(x) and rotating that around x = 0 and x = ln(4)
though integrating (ln(x))^2 is a pain, but you can do it by the substitution u = ln(x)
I don't think that works either as the solution to the original question, does it?
edit: and even if it does, I can's see how that simplifies the question...
Post by: WhoTookMyUsername on July 03, 2012, 06:43:46 pm
Do you have to check for turning points?
Similarly if it asks for max velocity or max distnce do you have to check if its a maximum or minimum?
I reckon technically you should do both but the textbook WS never does either
ThanksA!
Post by: kamil9876 on July 03, 2012, 09:40:33 pm
If it asks for the distance between say t = 1 and t = 3
Do you have to check for turning points?
Similarly if it asks for max velocity or max distnce do you have to check if its a maximum or minimum?
I reckon technically you should do both but the textbook WS never does either
ThanksA!
I don't understand the context of your question, can you provide an example of what you mean, i.e a question etc.
Post by: WhoTookMyUsername on July 03, 2012, 10:01:39 pm
find the distance travelled from t=0 to t= 2
for something (or a bit more complex, with something that's tougher to differentiate, but the question gives no indication of TP) similar to this, where either a) it's obvious there's no TP within the specified time period, or b) where it's "implied" there's no TP (all the TP quesitons in the textbook lead you through them) do you have to verify it and say there is a turning point only at t = 2, therefore distance traveled is equal to abs(displacement)?
Or is this an extra unnecessary step?
2) given that displacement = -(t-2)^2 + 3
determine when displacement is at a maximum
do we need to do the gradient sign test to verify that it is a maximum? (not so much a problem with this question, but more tricky questions where gradient sign tests can take quite a while)
Post by: WhoTookMyUsername on July 04, 2012, 06:27:45 pm
given that x=(t-2)^2 (x = displacement)
find the distance travelled from t=0 to t= 2
for something (or a bit more complex, with something that's tougher to differentiate, but the question gives no indication of TP) similar to this, where either a) it's obvious there's no TP within the specified time period, or b) where it's "implied" there's no TP (all the TP quesitons in the textbook lead you through them) do you have to verify it and say there is a turning point only at t = 2, therefore distance traveled is equal to abs(displacement)?
Or is this an extra unnecessary step?
2) given that displacement = -(t-2)^2 + 3
determine when displacement is at a maximum
do we need to do the gradient sign test to verify the TP is a maximum (or velocity = 0) (after differentiation) ? (not so much a problem with this question, but more tricky questions where gradient sign tests can take quite a while)
anyone know :D ?
thanks!
Post by: Jenny_2108 on July 04, 2012, 06:32:44 pm
Post by: Somye on July 04, 2012, 06:47:28 pm
why does your class study so fast? Mine just finished Euler's rule
Yeah that's where ours is up to as well (Bazza's as well), I'm assuming he's going ahead himself though...
Post by: HenryP on July 04, 2012, 07:28:34 pm
given that x=(t-2)^2 (x = displacement)In my opinion, when finding distance travelled its always good to verify that there is no turning point within the selected time period so that you can justify that distance travelled=abs(displacement). This may be just me being too careful but I guess its better to be safe than sorry.
find the distance travelled from t=0 to t= 2
for something (or a bit more complex, with something that's tougher to differentiate, but the question gives no indication of TP) similar to this, where either a) it's obvious there's no TP within the specified time period, or b) where it's "implied" there's no TP (all the TP quesitons in the textbook lead you through them) do you have to verify it and say there is a turning point only at t = 2, therefore distance traveled is equal to abs(displacement)?
Or is this an extra unnecessary step?
2) given that displacement = -(t-2)^2 + 3
determine when displacement is at a maximum
do we need to do the gradient sign test to verify that it is a maximum? (not so much a problem with this question, but more tricky questions where gradient sign tests can take quite a while)
As for your second question, I would definitely verify whether the turning point was a maximum or minimum.
If you're unsure whether to do these steps, take a look at how many marks its worth. If its only worth one, you wouldn't need to do any of these extra verification steps.
Post by: WhoTookMyUsername on July 06, 2012, 02:48:32 pm
I'm hugely struggling with vectors ATM (trying to go a bit ahead of class)
1) A Ball, just before it hits the ground, has the position vector
At what angle does the ball hit the ground?
2) What's the convention for x, y, z (axes) which 2 are the "ground"? If "z" represents upwards, but "y" represents "upwards" in a 2 D plane, how does this work :S? Does that mean 2D and 3D questions must be considered completely separately in terms of axes?
3) If somehow you use that dot product rule to find the angle, wouldn't letting the other vector = z, or 2z change the angle? (when it's the same angle created)
4) do you lose a mark if you forget the tilde?
I had absolutely no idea how to approach this and still don't know what's going on even after looking at WS
thanks
Post by: b^3 on July 06, 2012, 03:34:14 pm
Now consider the triangle this vector makes with the ground. Now to get the adjacent edge of this triangle, you will need to use pythagoras of the triangle the the
Then to find the angle we would have
2) For 2D planes you normally take y as upwards and x as across. For 3D, the x-y plane becomes the ground and the z axis becomes upwards. So it depends on the question really.
3) If the vector is parallel, then changing the vector from
4) We were told that you would lose if for forgetting it as if you don't then you are not denoting a vector. Having a look at past assesors reports they point out that it is a common problem.
Anyways hope that helps, and is right, I'm a little rusty on this.
Post by: WhoTookMyUsername on July 06, 2012, 03:50:36 pm
the textbook WS had invtan(3/sqrt(14)) (they found the 37 degree angle first)
is this coming from the dot product rule where they've taken the second vector as z?
Do they prefer if you use this method? (as it is more "vectorish")
Post by: b^3 on July 06, 2012, 03:56:02 pm
thanks a lot b^3 :)Oh right, yeh they used vectors, they used the velocity vector and the vector tht is straight up, that is they used
the textbook WS had invtan(3/sqrt(14)) (they found the 37 degree angle first)
is this coming from the dot product rule where they've taken the second vector as z?
Do they prefer if you use this method? (as it is more "vectorish")
EDIT: I checked back with a past exam (not VCAA) and they gave two methods, the original one I used and a vector method that used the dot product of the vel vector and the vector that was in the x-y plane that is in line with the vel vector (for this situation that would be
Post by: WhoTookMyUsername on August 18, 2012, 06:04:05 pm
Why is the tension in the string = mg
And not 2mg?
Struggling to understand this concept (what truly "is" tension)
Thanks!
Post by: nina_rox on August 18, 2012, 06:52:30 pm
I'm getting confused with signed and unsigned graphs. So for a velocity time graph is the distance the whole area under the graph (disregarding the negative areas and calculating them as positive - so unsigned) while the displacement regards the negative area (so signed)?
Thank you very much :)
Post by: paulsterio on August 18, 2012, 07:13:51 pm
Sorry, I have a quick question.
I'm getting confused with signed and unsigned graphs. So for a velocity time graph is the distance the whole area under the graph (disregarding the negative areas and calculating them as positive - so unsigned) while the displacement regards the negative area (so signed)?
Thank you very much :)
That's correct
Post by: WhoTookMyUsername on August 18, 2012, 07:19:55 pm
If you have two masses either side of a frictionless pulley ( hanging by a rope) of equal mass
Why is the tension in the string = mg
And not 2mg?
Struggling to understand this concept (what truly "is" tension)
Thanks!
Post by: paulsterio on August 18, 2012, 07:26:21 pm
If you have two masses either side of a frictionless pulley ( hanging by a rope) of equal mass
Why is the tension in the string = mg
And not 2mg?
Struggling to understand this concept (what truly "is" tension)
Thanks!
Because each mass is only mg
Think about it this way. If you hold a rope, in your hand, and hang a block of m on it, the tension in the rope will be mg
Easiest way to think of tension in this regard is that it's sorta like the normal reaction force - it stops the blocks falling to the ground.
Post by: WhoTookMyUsername on August 18, 2012, 07:30:32 pm
Should you add these together of is tension a force / unit area?
Post by: paulsterio on August 18, 2012, 07:36:04 pm
but isn't there a force of mg (tension) acting on both sides of the rope?
Should you add these together of is tension a force / unit area?
Yes, there are, but look at it this way.
Let's do this situation instead:
(http://i49.tinypic.com/wtw3np.jpg)
You see that of course, in both situations, the force meter will show the same value of mg?
Post by: WhoTookMyUsername on August 18, 2012, 07:46:17 pm
Post by: nina_rox on August 18, 2012, 08:18:55 pm
Yeah I thought my logic was correct and then I went to do some checkpoint questions and there was a question that asked for the distance and the area below the x axis has been subtracted from the area above the x axis in the answers? Is this incorrect?Sorry, I have a quick question.
I'm getting confused with signed and unsigned graphs. So for a velocity time graph is the distance the whole area under the graph (disregarding the negative areas and calculating them as positive - so unsigned) while the displacement regards the negative area (so signed)?
Thank you very much :)
That's correct
Post by: paulsterio on August 18, 2012, 08:22:27 pm
Post by: nina_rox on August 18, 2012, 08:24:31 pm
Yes that is incorrect.
Thanks very much Paul
Post by: WhoTookMyUsername on August 26, 2012, 06:56:33 pm
And when x <a f''(x) >0 and when x>a
F''(x) is < 0
Is the point x=a a point of inflection?
Is there a point of inflection for the graph x^(1/3)? ( gradient is undefined at 0,0, as is second derivative)
Thanks
Post by: brightsky on August 26, 2012, 09:14:56 pm
2) no x^(1/3) has no point of inflection. you should be able to visualise the graph as the inverse of x^3. the tangent at x =0 is x = 0.
Post by: WhoTookMyUsername on August 26, 2012, 09:17:59 pm
but if the answer for 1) is yes answer for 2) has to be yes double deriving x^(1/3) on CAS gives -2/(9x^(5/3)) and subbing in x = 1 and x = -1 gives -2/9 and 2/9 respectively
Post by: Phy124 on August 26, 2012, 09:29:24 pm
Remember;
In this case, the point
Post by: brightsky on August 26, 2012, 09:41:24 pm
thanks for help :)
but if the answer for 1) is yes answer for 2) has to be yes double deriving x^(1/3) on CAS gives -2/(9x^(5/3)) and subbing in x = 1 and x = -1 gives -2/9 and 2/9 respectively
note that f''(a) = 0 is a NECESSARY condition for x = a to be a point of inflection.
so basically when verifying whether or not x = a is a point of inflection:
1) verify that f''(a) = 0 (NECESSARY condition, but not SUFFICIENT, example:y = x^4)
2) ensure that f''(a - some small number) and f''(a+some small number) have opposite signs
only when both conditions are satisfied can x = a have the privilege of being called a point of inflection.
Post by: WhoTookMyUsername on August 26, 2012, 09:43:23 pm
Post by: brightsky on August 26, 2012, 09:57:53 pm
I think what ranga said is correct, i don't think the first point needs to be satisfied as concavity would still change regardless?
http://www.encyclopediaofmath.org/index.php/Point_of_inflection
scroll down to where they talk about necessary and sufficient existence conditions.
Post by: WhoTookMyUsername on August 26, 2012, 10:09:37 pm
Control f points of vertical tangent
Post by: brightsky on August 26, 2012, 10:35:37 pm
http://math.uchicago.edu/~vipul/teaching-0910/152/infinitycuspsasymptotes.pdf
Control f points of vertical tangent
ahh i stand corrected. never knew that!
Post by: Phy124 on August 26, 2012, 10:59:00 pm
ahh i stand corrected. never knew that!Not surprising, I know of many people who have already studied this level of mathematics and either don't know or have forgotten this rule. It's great that you've learnt it before you need it and will definitely never forget it ;)
Post by: WhoTookMyUsername on August 27, 2012, 07:34:57 am
http://math.uchicago.edu/~vipul/teaching-1011/152/concaveinflectioncusptangentasymptote.pdf
Cntrl f points of vertical ( different link)
Same university.. Seems to imply points of vertical tangent are NOT looool
Post by: WhoTookMyUsername on September 06, 2012, 04:23:42 pm
thanks!
Post by: Conic on September 06, 2012, 04:30:38 pm
Any quick way to find angle between two vectors on CAS?solve( a.b = |a|*|b|*cos(x),x) or arccos((a.b)/(|a|*|b|))
thanks!
a.b = dot product of a and b
Post by: WhoTookMyUsername on September 08, 2012, 05:37:04 pm
A body moves in a straight line so that at time t its accel is a where a =f(v)
Given that v=v0 when t=t0 the time taken for the velocity to change from v0 to v1 is given by?
I think it is S(v1,v0)( terminals) 1/f(v) dv
But the solution has a + t0 at the end of the equation,
I thought that t0 would give you the time at t1 not the change in time?
Am i missing something?
Thanks!
Post by: kamil9876 on September 08, 2012, 07:02:58 pm
Post by: pi on September 08, 2012, 07:05:12 pm
Post by: WhoTookMyUsername on September 08, 2012, 07:50:21 pm
Attached that question
(1) attachment
i feel there shouldn't be a t0 at the end of option A because it's a change in time question, not a "time" question
and for the other attached q
(2) attachment
i have absolutely no idea, conceptually what is required for part c (4c) and i'm quite worried about it lol, does anyone know how to (baby steps preferably :D) explain it (referring to paths, NOT particles themselves)
thanks a lot!
x=11/sqrt(174) y= 2/sqrt(174) z = 7/sqrt(174) for part 4a)
for b) r(0) = i - 2j
s(0) = j
also
3) how can we determine what info we're allowed to use in vector proofs? For e.g. can we assume that isoceles has 2 same angles in 2 corners?
Can we use any reasonable assumption that is not the thing we have to prove?
4) Let's say you sketch a hyperbola over the domain [0,4] (like
5) if sketching a parametric equation should we indicate the normal things like asymptotes? (i thought we should, but i saw a hyperbola graph
thanks very much :)
Post by: Jenny_2108 on September 08, 2012, 08:48:27 pm
Q2) Solve equations (by hand or put in CAS)
You get
b) Substitute t=0 into the postition vectors
c) you calculate s(t)-r(t) then find their distance = magnitude of the resulted vector
Then find derivative and let derivative = 0, you get minimum value of t
Sub back, you get the distance
Is the answer
Post by: WhoTookMyUsername on September 08, 2012, 10:17:35 pm
and with c) no, it's asking for the minimum distance between the paths of the particles, not the particles themselves
Post by: WhoTookMyUsername on September 09, 2012, 10:28:52 pm
Thankyou :) !
Post by: BubbleWrapMan on September 10, 2012, 01:25:59 pm
Post by: WhoTookMyUsername on September 10, 2012, 07:37:19 pm
though because it says the "time taken for the velocity to change from" it's t1-t0 isn't it? which is just the integral without the t0 at the end?
I stil have no idea about 4c) second attachment :'( (5 posts up)
Post by: Lasercookie on September 10, 2012, 07:42:41 pm
though because it says the "time taken for the velocity to change from" it's t1-t0 isn't it? which is just the integral without the t0 at the end?t0 is not necessarily equal to zero. We're calculating an area under a curve. If t0 has a value, then that curve will be translated up and down, correct? That will mean the area would be altered. By the fact that we're using a definite integral, we've already accounted that there's an initial time and a final time (the interval we're integrating over).
Post by: BubbleWrapMan on September 10, 2012, 09:34:52 pm
Post by: Jenny_2108 on September 10, 2012, 09:44:08 pm
Post by: BubbleWrapMan on September 10, 2012, 11:13:16 pm
^ Actually this is the formula in Spesh course you can apply directly without doing antiderivativehuh
Post by: Lasercookie on September 10, 2012, 11:21:11 pm
^ Actually this is the formula in Spesh course you can apply directly without doing antiderivativehuh
I think Jenny is saying that
Post by: BubbleWrapMan on September 10, 2012, 11:31:08 pm
Post by: Jenny_2108 on September 11, 2012, 05:50:00 pm
^ Actually this is the formula in Spesh course you can apply directly without doing antiderivativehuh
I think Jenny is saying thatis one of those "short-cut" formulas you can use to check over your work quickly etc.
Yeah, this is what I meant :P
Post by: WhoTookMyUsername on September 22, 2012, 12:50:14 pm
If so how?
Thanks
Post by: Hancock on September 22, 2012, 01:04:51 pm
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.
Post by: oneoneoneone on September 22, 2012, 01:11:03 pm
Post by: BubbleWrapMan on September 22, 2012, 01:14:23 pm
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.
Post by: WhoTookMyUsername on September 29, 2012, 08:54:31 pm
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! :)
Post by: Conic on September 29, 2012, 10:27:16 pm
]2) (Unrelated)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).
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! :)
Post by: pi on September 29, 2012, 10:40:30 pm
Post by: Jenny_2108 on September 29, 2012, 11:37:01 pm
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?
(http://grotongeometry.wikispaces.com/file/view/Theorem_12-2.JPG/34173149/Theorem_12-2.JPG)
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
Post by: WhoTookMyUsername on September 30, 2012, 09:42:46 am
sin^-1(x) and arcsin(x) are generally the normal ways of writing inverse sineDid you use sin^-1(x) in your exams?
Post by: pi on September 30, 2012, 09:44:35 am
awesome explanation! :) thankssin^-1(x) and arcsin(x) are generally the normal ways of writing inverse sineDid you use sin^-1(x) in your exams?
Yeah, I used that one :)
Post by: WhoTookMyUsername on October 01, 2012, 08:30:04 pm
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:
Post by: BubbleWrapMan on October 01, 2012, 10:43:35 pm
Also, I used sin^-1 during the year but I use arcsin these days. VCAA would accept both of those.
Post by: WhoTookMyUsername on October 01, 2012, 11:25:39 pm
yeah solving parametric by CAS is not necessary,
but especially for MC it can help a little bit.
I think in previous OS it was a little "glitch" that solving for "t" actually eliminated t from the equation
but that was a helpful glitch!
and there seems to be no way of solving for parametric ATM :(
Post by: WhoTookMyUsername on October 02, 2012, 05:59:19 pm
thanks (also anyone know how to find parametric equation with CAS xD)
Post by: meemz on October 02, 2012, 06:57:33 pm
Should wind force be labelled as a "pushing force" or a "pulling force :S" ?depends
thanks (also anyone know how to find parametric equation with CAS xD)
lets say a plane is travelling east, and the wind is travelling in the east direction, then it'll be a pushing force
and if the plane is travelling west and the wind in the east, then it'll a pulling force
Post by: WhoTookMyUsername on October 02, 2012, 08:29:37 pm
For this exam
Q 4a)
i labelled wind as a "pushing force" (on the other side) but not sure if it's correct :S or whether they'd accept both, vcaa had it as a pulling force
Post by: BubbleWrapMan on October 02, 2012, 08:50:00 pm
Post by: WhoTookMyUsername on October 02, 2012, 08:55:51 pm
i drew the arrow pointing "towards" the center from the left hand side
whereas the solutions drew it going from the centre to the "right."
I think my placement may be wrong because i think force arrows are supposed to be drawn with their tail on the object being acted on?
not 100% sure someone please help :D
(http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006specmath1-w.pdf
For this exam
Q 4a)
i labelled wind as a "pushing force" (on the other side) but not sure if it's correct :S or whether they'd accept both, vcaa had it as a pulling force)
(i.e.should a "pushing force" be labelled in the same way as a "pulling force" with the tail at the center of the object? is it technically incorrect to have the tail of the arrow not touching the object but only the head of the arrow touching :S?)
2)http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006specmaths2-w.pdf
Q 4cii)
For verifying a stationary point of inflection; how would you go about doing that when you have the double derivative in terms of Y only?
Can you still just go above and below and show it changes sign? Or do you have to make some kind of qualitative statement linking y to x or to dy/dx as well D:
Not sure what's going on here lol
Thanks!
Post by: BubbleWrapMan on October 02, 2012, 09:07:19 pm
For the calculus question, you only need to show that the second derivative changes sign at that point.
Post by: WhoTookMyUsername on October 03, 2012, 02:22:49 pm
Are we allowed to resolve force on a diagram for the purposes of working out for a following question?
My 1/2 teacher said they don't like resolution of forces drawn on a force diagram but im not sure
Thanks!
Post by: BubbleWrapMan on October 03, 2012, 04:50:00 pm
Post by: WhoTookMyUsername on October 06, 2012, 09:38:32 pm
Last question spurt most likely... ever :D !
1) Do we need to label POI for inverse cos etc. on graphs?
2) Anyway to convert an equation to hyperbolic form on CAS?
3) If the answer doesn't specify to give a numerical / approximate answer do we always leave the answer in terms of g?
4) For something like a=t^2 + t - 10 where a can only be positive or negative, should we indicate a as a vector (with tilde) or without tilde?
5) for a point on an argand plane should we put a "dotted line" to the origin?
6) where should friction be drawn from? the point of contact or the "centre" of the object?
7) Can you have a conjugate pair when the coefficients ARE NOT all real by coincidence?
Or does a conjugate pair for complex numbers imply that coefficients are definitely all real?
8) Anyway to graph complex numbers on cas? (not cartesian)
9) For volume of revolution do we have to account for intercepts? (my gut instinct was yes, but after playing around with x^2 - 4
S(0,2) + S(2,4) = S(0,4) (terminals for piSy^2 dx where S is integral symbol) so i'm not sure
thanks very much! :) Sorry for some questions which i know are slightly tedious :\
Post by: BubbleWrapMan on October 07, 2012, 12:21:00 am
2) What's hyperbolic form? :P
3) They usually say to take g to be 9.8 m/s^2 in the exam instructions.
4) If quantities like that are in one dimension you don't need the tilde.
5) No need.
6) Doesn't matter.
7) Only if there are other complex roots. If they are all in pairs then you always get real coefficients.
8 ) Not that I know of.
9) I don't think so? Are you saying that the sum of the integrals is not equal to the integral over the whole interval? Because it should be.
Post by: Jenny_2108 on October 07, 2012, 12:27:01 am
2) What's hyperbolic form? :PHyperbolic forms like sinh, cosh, tanh
Sometimes the CAS gives ans like that and we need to know how to convert it
2) Anyway to convert an equation to hyperbolic form on CAS?
I don't think so. CAS just gives this form
Maybe you can derive from CAS as
Post by: WhoTookMyUsername on October 07, 2012, 09:42:04 am
nah lol by hyperbolic i meant the "form of a hyperbola," although hyperbolic is technically quite incorrect xD
(not sure if cas can get into into form of a hyperbola form)
With 3) Would we lose a mark if it's in terms of g?
(e.g. obviously for tech free if we leave in terms of g less chance of making a mistake)
With 5) Would we lose a mark if we put the tilde?
With 9) As in, when you're doing area under the graph (or between graph and x axis) over a given interval, you have to split it up when you have intercepts over that region, do you have to do the same for volumes of revolution?
additionally (another tedious question)
10) how would you refer to a "region" of the x axis? Like, over the ... period (2,10) there are no x intercepts?
Over the ... region (2,10) there are no x intercepts? (if the domain is greater than (2,10)
Thanks a lot ! :D
Post by: BubbleWrapMan on October 07, 2012, 02:58:47 pm
5) Maybe, just don't do it :P
9) No you don't, volumes don't come out as negative when you use that formula.
10) I'd call it an interval. "There are no x-intercepts in the interval (2,10)." 'Region' usually refers to an area, and 'period' refers to the period of a function like circular functions.
Post by: pi on October 07, 2012, 03:12:30 pm
2) Anyway to convert an equation to hyperbolic form on CAS?
I don't think so. CAS just gives this form
Maybe you can derive from CAS as
You can do this on the CAS, I don't have a CAS on me to remember though -_- Look under the trig submenu from memory though.
Post by: Jenny_2108 on October 07, 2012, 09:06:22 pm
2) Anyway to convert an equation to hyperbolic form on CAS?
I don't think so. CAS just gives this form
Maybe you can derive from CAS as
You can do this on the CAS, I don't have a CAS on me to remember though -_- Look under the trig submenu from memory though.
Ignore my post. It wasn't the question he asked though
Post by: WhoTookMyUsername on October 08, 2012, 10:25:48 pm
1) anyway to get the cas to give you an equation in the form of a hyperbola ?
2) do we need to put in a "scale" along x and y axis if there isnt one on the graph?
3) when they ask to sketch a cartesian equation do we need to indicate direction?
Thanks :)
Post by: pi on October 08, 2012, 10:30:01 pm
2) Scaling is never needed
3) Direction? As in arrows on either end of the graph (if that's the case)? If the arrows, yep I would (and did).
Post by: Somye on October 08, 2012, 10:57:52 pm
3) Direction? As in arrows on either end of the graph (if that's the case)? If the arrows, yep I would (and did).
He's referring more to parametric equations where you have to put them in cartesian form and graph them.
From what I've seen in trials though, you don't need to...
1) What do you mean?
Post by: pi on October 08, 2012, 11:05:07 pm
By hyperbolic I thought you meant getting rid of cosh -> exponential form, woops.
Ah, cheers Somye :)
Post by: oneoneoneone on October 08, 2012, 11:08:54 pm
completesquare(x^2+6x-y^2-8x=0,x,y)
into the calculator itll give it in 'hyperbolic' form.
Post by: pi on October 08, 2012, 11:19:41 pm
if you type
completesquare(x^2+6x-y^2-8x=0,x,y)
into the calculator itll give it in 'hyperbolic' form.
Great tip! I stand corrected :)
Post by: TheRajinator on October 10, 2012, 09:37:41 am
Post by: WhoTookMyUsername on October 13, 2012, 09:13:45 pm
I just did a question asking about the formation of vectors,
"Find the times, when the the point (0,5) (0,0) and a boat forms an isosceles triangle of equal sides of 5km"
I got the answers correct, however i only showed two sides WERE equal, and did not show that the remaining side was a DIFFERENT length.
Not sure if i would have lost a mark for this?
thanks :)
Post by: Jenny_2108 on October 13, 2012, 09:19:23 pm
is an equilateral triangle an isosceles triangle?
yep
If 2 sides are equal, its an isosceles triangle, no need to show the remaining one has different length
Post by: paulsterio on October 13, 2012, 09:36:26 pm
an equilateral triangles are a subset of isosceles triangles.
Post by: WhoTookMyUsername on October 13, 2012, 09:41:52 pm
thanks for clearing it up! :)
Post by: WhoTookMyUsername on October 14, 2012, 09:10:47 pm
Thanks!
Post by: pi on October 14, 2012, 09:12:37 pm
General q: should we label undefined points? E.g. If it's x^2 over domain (0,oo) should we label (0,0) ?
Thanks!
Yep
Post by: WhoTookMyUsername on October 18, 2012, 05:45:01 pm
Mathemtatically notationally acceptable?
2) if the previous part asks for 4 dp, but we have a more detailed answer to 5 or 6dp, should we use this for the next question?
Thanks!
Post by: BubbleWrapMan on October 18, 2012, 07:09:40 pm
2) They usually don't put you in that situation, but you should use the value from the previous question.
Post by: StumbleBum on October 18, 2012, 08:50:24 pm
1) in substitution, is the step du=dx or something similar, e.g. du=2dxMy teacher said that it is not correct mathematically, can't remember why... In an exam i would just stick with du/dx=2 and then transpose it in your head before using it when you substitute. Although this does allow a risk for a very silly error! Maybe just write it then rub it out, although i doubt they would take any working marks off for that if you got the question wrong.
Mathemtatically notationally acceptable?
Post by: WhoTookMyUsername on November 11, 2012, 09:01:18 am
Also what are some methods of sketching the graph defined by
Arg(iz)<pi/4 ?
I converted it to cartesian, but was unsure but which side to shade, and which half of the cartesian to graph, had to sort of use intuition and transformations to decide, any better way? (Or alternative ways about thinking of the question?), how would you sketch this graph?
Thanks
Post by: brightsky on November 11, 2012, 09:40:39 am
consider i*z. it's z but rotated 90 degrees anticlockwise.
Post by: WhoTookMyUsername on November 11, 2012, 09:53:24 am
Sometimes cannot draw a triangle though, like for this one, i can't draw the triangle because i can't draw the graph without that rule :/ (without intuition)
How do you intuitively know that though?
Like what process did you go through to graph arg(iz)<pi/4?
Or what is your thought process / is there any way to do it mathematically on the page and work it out rather than from intuition?
Like i'd prefer a mathmatical process that always works for this type of thing, less risky in the exam
I didsome dodgy thing like
Arg(iz) = pi/4
Then iz=cis(pi/4)
Z = cis(pi/4 - pi/2)
Arg(z) = -pi/4
I know that's mathematically all over the place though i *think* my reasoning is sound
Just looking for something similar to that where i can apply a certain technique to questions like this (or a way to tidy what i did up)
Post by: brightsky on November 11, 2012, 09:58:17 am
and for Arg(iz), why not write Arg(iz) = Arg(z) + pi/2 and go from there?
Post by: WhoTookMyUsername on November 11, 2012, 10:09:19 am
I'm 99% sure Arg(iz) does not = Arg(z) + pi/2
IIRC
Arg(i+z) = Arg(i)Arg(z)
arg(a+b)=Arg(a)Arg(b)
cis(a+b)=cis(a)cis(b)
Still not sure how to go about graphing Arg(iz) < pi/4 without 'intuition' xD
Arg(iz) = pi/4
Then iz=cis(pi/4)
Z = cis(pi/4 - pi/2)
Arg(z) = -pi/4
Is closeish? Anyway to tidy :X
Post by: rife168 on November 11, 2012, 11:02:00 am
No it's not, iz becomes xi - y
I'm 99% sure Arg(iz) does not = Arg(z) + pi/2
IIRC
Arg(i+z) = Arg(i)Arg(z)
arg(a+b)=Arg(a)Arg(b)
cis(a+b)=cis(a)cis(b)
Still not sure how to go about graphing Arg(iz) < pi/4 without 'intuition' xD
Arg(iz) = pi/4
Then iz=cis(pi/4)
Z = cis(pi/4 - pi/2)
Arg(z) = -pi/4
Is closeish? Anyway to tidy :X
I think you have it the wrong way around...
Arg(ab)=Arg(a)+Arg(b)
Post by: WhoTookMyUsername on November 11, 2012, 11:06:52 am
Let z1=cis(a)
Let z2=cis(b)
Z1z2=cis(a)cis(b)
=cis(a+b)
Arg(z1z2) = a+b = arg(z1) + arg (z2)
Fml... Gotta go back and look over this stuff
Post by: rife168 on November 11, 2012, 11:11:35 am
Yes i do.
Let z1=cis(a)
Let z2=cis(b)
Z1z2=cis(a)cis(b)
=cis(a+b)
Arg(z1z2) = a+b = arg(z1) + arg (z2)
Fml... Gotta go back and look over this stuff
Well, it's better that you found that out now rather than after the exam... ;)
Post by: d3stiny on November 11, 2012, 11:18:18 am
Arg(iz) = pi/4
arctan(Im(iz)/Re(iz)) = pi/4
Im(iz)/Re(iz) = 1
and iz = xi - y
x / -y = 1
y = -x for x > 0
Not sure if thats right though.
Post by: WhoTookMyUsername on November 11, 2012, 11:29:31 am
Why not..Yeah that's sorta what i did the first time, but there's two problems with that,
Arg(iz) = pi/4
arctan(Im(iz)/Re(iz)) = pi/4
Im(iz)/Re(iz) = 1
and iz = xi - y
x / -y = 1
y = -x for x > 0
Not sure if thats right though.
Firstly how do you know for sure that x>0 (not a rhetorical question, i'm asking lol)
And secondly, how do you know which region to shade in?
(Arg(iz)<pi/4)
Haha yeah it is, i must have overlooked something... This has never troubled me in any practice exams though D:Yes i do.
Let z1=cis(a)
Let z2=cis(b)
Z1z2=cis(a)cis(b)
=cis(a+b)
Arg(z1z2) = a+b = arg(z1) + arg (z2)
Fml... Gotta go back and look over this stuff
Well, it's better that you found that out now rather than after the exam... ;)
Thanks! And thanks brightsky
Post by: WhoTookMyUsername on November 11, 2012, 03:09:13 pm
Is it R or [-1,1]?
I thought it would be R, but DHA had the domain as -1,1 (because he sketched it on the same graph as it's inverse)
But doesn't domain refer to the variable it's in terms of?
Like if it was m=cos(p)
The domain would be R, isn't it the same thing?
2)
Is piSx^2 dy an acceptable line of working? (Notationally wise)
Thanks
Post by: Somye on November 11, 2012, 03:15:12 pm
whereas in the question given, you cant really have x as your vertical axis, and so you have to use inverse etc.
Post by: pi on November 11, 2012, 03:18:03 pm
2) As per other thread. I wouldn't do it. But if you had already define x=... it should be ok. Play it safe.
Post by: brightsky on November 11, 2012, 03:19:00 pm
2) don't see any problem with that. although i'd err on the side of caution and just write everything in terms of y, i.e. write x(y).