Post by: funkyducky on November 05, 2011, 09:51:28 pm
Find
and:
Find the possible values of
Have fun :)
Here's another:
If the position vectors of A, B and C are,
and
respectively, determine whether the lines
and
intersect each other.
Post by: jane1234 on November 05, 2011, 10:19:30 pm
is defined by
. Find
Is the answer (x+1)^2 + y^2 = 25 for y E [-5,0)??
EDIT: On second thoughts, I don't think that's it...
EDIT2: Seems to be (x+1)^2 + y^2 = 25 for y E (0,5] instead... I'm probably completely wrong though :P
Post by: dc302 on November 05, 2011, 10:32:54 pm
Is the answer (x+1)^2 + y^2 = 25 for y E [-5,0)??
EDIT: On second thoughts, I don't think that's it...
Could be wrong but it might be (x-1)^2 + y^2 = 25?
Post by: syn14 on November 05, 2011, 10:39:24 pm
Yeh nvm it's just the top half.
Post by: dc302 on November 05, 2011, 10:41:13 pm
Post by: jane1234 on November 05, 2011, 10:41:47 pm
I also got (x+1)^2 + y^2 = 25. Where does the restriction on y come from?
Doesn't work for the bottom values if you substitute. tan^-1(y/x-4) - tan^-1(y/x+6) will give you pi/2, but these are not the proper arguments. I think the proper arguments only work for the top half of the circle...
Post by: Greatness on November 05, 2011, 10:46:02 pm
you guys are too good!
Post by: dc302 on November 05, 2011, 10:48:25 pm
:o anyone care to explain how to do this? hahaha i hate complex numbers...
you guys are too good!
Arg of a complex number is simply the inverse tan (arctan) of the real part/imaginary part.
So the equation actually looks like:
arctan( x-4 / y ) - arctan( x+6 / y) = pi/2
Then take the cosine of both sides, the RHS will become 0 and you use the double angle formula for the LHS.
Also, would the hyperbola (x+1)^2 - y^2 = 25 not also work? Do you have the solutions?
edit: uhhhhh how did i get that wrong. it's y/x-4 and y/x+6 :o
Post by: vea on November 05, 2011, 10:59:21 pm
Much thanks!
Post by: funkyducky on November 05, 2011, 10:59:45 pm
I've come across similar before where I've simply tan'd both sides, but obviously that doesn't work here, and I was stuck with a mess with this one.
Post by: dc302 on November 05, 2011, 11:01:48 pm
Can someone tell me how to simplify something of the form cos(arctanx) or sin(arctanx) ? x__x
Much thanks!
It might help to draw triangles.
If you let arctanx = P, where P is the angle of a triangle, and since tan = O/A, you know that tanP = x/1 so the opposite side has length x, and the adjacent side has length 1. So the hypotenuse has length, sqrt(1+x^2). Which means that sin(arctanx) = sinP = O/H = x/sqrt(x^2+1).
Hope this helps!
Post by: vea on November 05, 2011, 11:03:01 pm
Can someone tell me how to simplify something of the form cos(arctanx) or sin(arctanx) ? x__x
Much thanks!
It might help to draw triangles.
If you let arctanx = P, where P is the angle of a triangle, and since tan = O/A, you know that tanP = x/1 so the opposite side has length x, and the adjacent side has length 1. So the hypotenuse has length, sqrt(1+x^2). Which means that sin(arctanx) = sinP = O/H = x/sqrt*x^2+1).
Hope this helps!
Thanks for that, I was wondering if there was an algebraic way though?
Post by: dc302 on November 05, 2011, 11:05:09 pm
Can someone tell me how to simplify something of the form cos(arctanx) or sin(arctanx) ? x__x
Much thanks!
It might help to draw triangles.
If you let arctanx = P, where P is the angle of a triangle, and since tan = O/A, you know that tanP = x/1 so the opposite side has length x, and the adjacent side has length 1. So the hypotenuse has length, sqrt(1+x^2). Which means that sin(arctanx) = sinP = O/H = x/sqrt*x^2+1).
Hope this helps!
Thanks for that, I was wondering if there was an algebraic way though?
Sure is.
artanx = P
so tanp = x, sinp/cosp = x
sinp = xcosp
sinp ^2 = x^2 cosp ^2
sinp ^2 = x^2 (1-sinp ^2)
rearrange to get sinp = x/ sqrt(1+x^2)
edit: keep in mind that when you take the sqrt, you have to be wary of the signs, which depend on which quadrant your triangle is in.
Post by: vea on November 05, 2011, 11:09:18 pm
Can someone tell me how to simplify something of the form cos(arctanx) or sin(arctanx) ? x__x
Much thanks!
It might help to draw triangles.
If you let arctanx = P, where P is the angle of a triangle, and since tan = O/A, you know that tanP = x/1 so the opposite side has length x, and the adjacent side has length 1. So the hypotenuse has length, sqrt(1+x^2). Which means that sin(arctanx) = sinP = O/H = x/sqrt*x^2+1).
Hope this helps!
Thanks for that, I was wondering if there was an algebraic way though?
Sure is.
artanx = P
so tanp = x, sinp/cosp = x
sinp = xcosp
sinp ^2 = x^2 cosp ^2
sinp ^2 = x^2 (1-sinp ^2)
rearrange to get sinp = x/ sqrt(1+x^2)
edit: keep in mind that when you take the sqrt, you have to be wary of the signs, which depend on which quadrant your triangle is in.
nice!
thanks :)
Post by: funkyducky on November 05, 2011, 11:24:08 pm
Find
and:
Find the possible values of
Have fun :)
EDIT: I typed a mistake, was meant to be find
EDIT 2: EEP another typo! Find b for the second part, not a lol.
Post by: jane1234 on November 05, 2011, 11:49:25 pm
There's more to this question, if you want:is defined by
Findand
such that
is a subset of
and:and
has only one solution.
Find the possible values of
Have fun :)
EDIT: I typed a mistake, was meant to be find)
EDIT 2: EEP another typo! Find b for the second part, not a lol.
Just curious, where are these questions from?
Post by: funkyducky on November 05, 2011, 11:55:16 pm
Post by: luffy on November 06, 2011, 01:33:00 am
Let z = x + yi
Tan both sides.
There's more to this question, if you want:
Find
Let z = x +yi. You could simply do this question by recognition as you know that complex numbers of this form have the cartesian equation of a circle with centre z0.
Let z = x + yi and
Find the possible values of
For one solution, y = x - b is a tangent to the circle.
Hence, gradient of circle is 1.
When dy/dx = 1,
Sub this back into the circle equation.
Therefore, co-ordinates are P
and Q
Reject P as y > 0.
Sub Q into y = x - b
I get the feeling I made a fundamental error in step 1 of one of those problems. So, let me know where all my mistakes are :P
Post by: Mao on November 06, 2011, 04:50:11 am
Interesting. After some thought, I can arrive at the result via geometry. It is the top half of the circle centered about (-1,0) with a radius of 5, and excludes the points (4,0) and (-6,0).
I shall draw up a picture later.
Post by: funkyducky on November 06, 2011, 11:13:32 am
EDIT: Hang on,
What happened to compound angle formula?
EDIT: wait I get it, cos it =0...nevermind
EDIT: also, b can only equal
Post by: funkyducky on November 06, 2011, 12:30:19 pm
If the position vectors of A, B and C are
Post by: luffy on November 06, 2011, 12:37:44 pm
EDIT: also, b can only equalsince y>0 at the point of intersection
Sorry, this is probably a really noob question. But, why does y have to be greater than 0?
EDIT: Nevermind, I get it. I'll edit the post. Brilliant questions by the way.
Also, sorry for skipping a few steps in my working. In my defense, it was really late at night :P
Interesting. After some thought, I can arrive at the result via geometry. It is the top half of the circle centered about (-1,0) with a radius of 5, and excludes the points (4,0) and (-6,0).
I shall draw up a picture later.
Pleasseeeee post up the picture as soon as you get timee :D I really suck at recognising complex numbers by geometry and by basic inspection - my greatest weakness in specialist I think. Hence, it would really come in handy to learn it before the specialist exam.
EDIT: Nevermind, I just figured out the geometry of the relation.
Sorry, How exactly do you now that a=5 rather than a=5 or a=-5.
Edit: @below. I think you're right. I'll edit my post.
Post by: Lord of Tzeentch on November 06, 2011, 12:42:40 pm
Sorry, How exactly do you now that a=5 rather than a=5 or a=-5.
Post by: funkyducky on November 06, 2011, 12:59:23 pm
Post by: Lord of Tzeentch on November 06, 2011, 02:41:02 pm
Could someone tell me if there is an easier way to do these questions? Below is how I did them. (Sorry, in advance, for the huge post and for any errors I may have made )Is that first step, the fundamental error, 'Let z = x + yi'?
Let z = x + yi
I get the feeling I made a fundamental error in step 1 of one of those problems. So, let me know where all my mistakes are :P
Don't be so harsh on yourself, you did an amazing job. :)
Post by: Mao on November 06, 2011, 03:36:12 pm
Interesting. After some thought, I can arrive at the result via geometry. It is the top half of the circle centered about (-1,0) with a radius of 5, and excludes the points (4,0) and (-6,0).
I shall draw up a picture later.
Pleasseeeee post up the picture as soon as you get timee :D I really suck at recognising complex numbers by geometry and by basic inspection - my greatest weakness in specialist I think. Hence, it would really come in handy to learn it before the specialist exam.
EDIT: Nevermind, I just figured out the geometry of the relation.
Good to know you figured it out, but here's a picture anyways.
I have always been a fan of doing these questions geometrically. You can find a few more geometric descriptions in my bound notes (it's floating around in this board somewhere).
EDIT: I remember in my bound notes, there is a question similar to this, and I tackled it with the cartesian description. It was difficult and confusing. On top of that, you must consider the different quadrants and whether or not the locus will exist there.
Post by: funkyducky on November 06, 2011, 03:47:38 pm
Post by: Mao on November 06, 2011, 03:50:40 pm
Could someone tell me if there is an easier way to do these questions? Below is how I did them. (Sorry, in advance, for the huge post and for any errors I may have made )Is that first step, the fundamental error, 'Let z = x + yi'?
Let z = x + yi
I get the feeling I made a fundamental error in step 1 of one of those problems. So, let me know where all my mistakes are :P
Don't be so harsh on yourself, you did an amazing job. :)
The fundamental error is when you went from
Since inverse tan has a range of
Thus, you need to solve your cartesian equation using this hybrid function. When you have two Arg terms, you will have two Arg terms with which you need to combine into a hybrid function with more pieces in order to preserve all the conditions. After all that, you can obtain the bounds for your final solution.
That process is very complex, confusing, and I do not recommend it. (but you don't really have to worry about it. you won't have to deal with this kind of questions in an exam anyways.)
Post by: luffy on November 06, 2011, 03:53:14 pm
Could someone tell me if there is an easier way to do these questions? Below is how I did them. (Sorry, in advance, for the huge post and for any errors I may have made )Is that first step, the fundamental error, 'Let z = x + yi'?
Let z = x + yi
I get the feeling I made a fundamental error in step 1 of one of those problems. So, let me know where all my mistakes are :P
Don't be so harsh on yourself, you did an amazing job. :)
The fundamental error is when you went from
Since inverse tan has a range of, it is a bit different to Arg. The full description is:
Thus, you need to solve your cartesian equation using this hybrid function. When you have two Arg terms, you will have two Arg terms with which you need to combine into a hybrid function with more pieces in order to preserve all the conditions. After all that, you can obtain the bounds for your final solution.
That process is very complex, confusing, and I do not recommend it. (but you don't really have to worry about it. you won't have to deal with this kind of questions in an exam anyways.)
Oh - I originally had a similar thing to that hybrid in my answer. However, I edited it out when I was told my domains were wrong. It'll take too long to edit it back in haha. But thanks a lot for the picture and explanation - very much appreciated.
Post by: Mao on November 06, 2011, 03:58:00 pm
Here's another:
If the position vectors of A, B and C are
The easiest thing is to construct vector paths along these two lines. (i.e. if a plane flies with some vector velocity from some initial position vector, blah blah)
If at any point OA(t) = BC(s) for some arbitrary values of t and s, then the two paths do cross. We thus have a system of 3 equations (equating x y and z) and 2 variables (s and t):
Solving the first two simultaneously, we get s=0.5, t=0
Substituting into the last eqn, we see that this does not satisfy the final relationship. Thus the two lines do not cross.
Post by: luffy on November 06, 2011, 04:10:46 pm
Here's another:
If the position vectors of A, B and C are
The easiest thing is to construct vector paths along these two lines. (i.e. if a plane flies with some vector velocity from some initial position vector, blah blah)
If at any point OA(t) = BC(s) for some arbitrary values of t and s, then the two paths do cross. We thus have a system of 3 equations (equating x y and z) and 2 variables (s and t):
Solving the first two simultaneously, we get s=0.5, t=0
Substituting into the last eqn, we see that this does not satisfy the final relationship. Thus the two lines do not cross.
For 3 dimensional vectors, how do we check if the vectors 'paths' cross? (i.e. they don't necessarily intersect)
Post by: Mao on November 06, 2011, 05:07:59 pm
Let
If such a point of intersection exists, let this point be
If this point of intersection exists, then there must be a set of
Post by: funkyducky on November 06, 2011, 05:14:25 pm
Post by: luffy on November 06, 2011, 06:26:17 pm
@luffy, exactly as I did.
Letand
be two vector paths. We seek the point of intersection, i.e. when the coordinates are the same, though the time doesn't have to be the same.
If such a point of intersection exists, let this point be, and
reaches
.
reaches
. Then, we're saying
.
If this point of intersection exists, then there must be a set of
Fair enough. That seems really long-winded.
Just out of curiosity, if I make two Cartesian equations out of the i and j components (excluding the k-direction for now). Then, find the intersection between them and sub it into the two k-values of both vectors to check if they are the same. Would this approach work too?
Post by: dc302 on November 06, 2011, 08:24:29 pm
@luffy, exactly as I did.
Let
If such a point of intersection exists, let this point be
If this point of intersection exists, then there must be a set of
Fair enough. That seems really long-winded.
Just out of curiosity, if I make two Cartesian equations out of the i and j components (excluding the k-direction for now). Then, find the intersection between them and sub it into the two k-values of both vectors to check if they are the same. Would this approach work too?
If you just want the intersection of the paths, regardless of time, then yes that should also work.
Post by: Mao on November 07, 2011, 01:54:48 am
@luffy, exactly as I did.
Let
If such a point of intersection exists, let this point be
If this point of intersection exists, then there must be a set of
Fair enough. That seems really long-winded.
Just out of curiosity, if I make two Cartesian equations out of the i and j components (excluding the k-direction for now). Then, find the intersection between them and sub it into the two k-values of both vectors to check if they are the same. Would this approach work too?
That is exactly what I'm saying. =S awks.
I guess I should explain myself clearer next time. :P
Post by: Mao on November 07, 2011, 01:57:49 am
Mao, would you say the second question is outside of the spesh syllabus? I know how to solve this stuff from what I've learnt in unimaths, but other spesh kids wouldn't have an easy time of it...
Which is the second question?
The first locus question is a bit too hard for specialist level imo. The second and third locus questions aren't too difficult (though it relies on the result of the first question).
The vector question isn't above the syllabus, I have encountered them in my time in specialist. Though this type of questions would be considered very difficult at this level.