Login

Welcome, Guest. Please login or register.

October 06, 2025, 09:01:37 pm

Author Topic: TrueTears question thread  (Read 67310 times)  Share 

0 Members and 1 Guest are viewing this topic.

kamil9876

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1943
  • Respect: +109
Re: TrueTears question thread
« Reply #435 on: July 27, 2009, 09:08:04 pm »
0
"angle between two vectors" means the angle formed when the two vectors are placed tail to tail. If you place the tail of a at the tail of b then you can see that the angle between the two vectors is in fact 90 degrees, hence cannot be theta.

In fact, another way to explain is this: you know that a.b=0 while Since a triangle cannot have more than one right angle (it already has one at the point where the lines meet the circle). Therefore.



Hence
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: TrueTears question thread
« Reply #436 on: July 27, 2009, 09:21:09 pm »
0
omfg, I thought meant the angle b/w vector a and b. FFS time to go suicide now, I can't read english for shit
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: TrueTears question thread
« Reply #437 on: July 28, 2009, 04:49:43 pm »
0
Okay now a really really interesting question heh



Well first part is very easy to proof. Now second part.

Hence would imply using the previous part to solve the next part.

Now





[I'm gonna leave out my steps for solving for x cause it's trivial]

Solving for x yields for

Now let's use the "otherwise" method.









Solving for x yields

My question is why does one method give only 1 solution and another method gives 3? Which is the right method? When I type to TI-89 calc to just solve with restricted domain I get all 3 solutions. However the answer for this question is just

I know why the "hence" method only gives 1 solution is because when you divide both sides by and if then it is undefined, but it still does not explain why you can not have 0 and as solutions because if you just have then it clearly satisfies.

Ideas anyone?
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: TrueTears question thread
« Reply #438 on: July 28, 2009, 05:07:00 pm »
0
Alright and the very last very interesting question:



I've sketched everything etc, now from the information given

There is no way you can get in terms of and because there is no relationship that links the with , the only information you have is the relationship linking the magnitudes of each but not the actual vectors? How do you go about this....?
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

kamil9876

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1943
  • Respect: +109
Re: TrueTears question thread
« Reply #439 on: July 28, 2009, 06:18:19 pm »
0
First question:

It seems as though the otherwise method is correct, here is a simple example of how to miss a solution:

Solve this equation:


Dividing both sides by x:



Where's the x=0? It's because by dividing both sides by x comes with the hidden assumption that Hence all you have proven is that "if then is the only solution". In order to complete the solution you have to now investigate the subset of possibilities that you discarded, and in this case it is: what if x=0?

Hence in your situation you have found the solution when . And so now you have to consider what happens when cosx-1=0 and see if any of the solutions coincide with sinx=0.

Hint: Good way to check if you've missed a solution in this fashion is to sub in the missed solution and see if you get something of the form
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: TrueTears question thread
« Reply #440 on: July 28, 2009, 06:33:25 pm »
0
Ahh well explained, thanks kamil.

Any ideas on the vector Q?
« Last Edit: July 28, 2009, 06:38:27 pm by TrueTears »
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: TrueTears question thread
« Reply #441 on: July 28, 2009, 07:39:04 pm »
0
without calc.

Find b.
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

kamil9876

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1943
  • Respect: +109
Re: TrueTears question thread
« Reply #442 on: July 28, 2009, 07:47:51 pm »
0
I think vector question has a lack of info (which two sides are parralel). But assume OA and CB are parralel. Then assume the other two sides are parralel. If the two ans are the same then that is the ans, if they are not then there isn't enough info.


Hint for next question: Try to "equate coefficients".
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: TrueTears question thread
« Reply #443 on: July 28, 2009, 08:05:38 pm »
0
Yeah that is what I did but what do you treat as the variable? I mean I treated as the variable and obviously you probs did the same, but isn't \pi just another number? That means you're equating numbers to numbers?
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

kamil9876

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1943
  • Respect: +109
Re: TrueTears question thread
« Reply #444 on: July 28, 2009, 08:29:11 pm »
0
All I'm doing is trying to find a solution.

If a+b=c+d   and you can find an instance where a=b AND c=d then that is a solution.

You can prove that this is the only solution by drawing the graphs of and . We know that one intersection happens when the trig function is at a max (b=7/2). Because the line is an increasing one there is no other solution to b greater than 7/2. To prove that there is no solution less than 7/2 we just have to argue that the line has a steeper gradient than the sin grpah. This is because the sin graph only has the same gradient as the line at only one point between a zero and maximum. Hence the linear graph falls down quicker.

Meh that was ugly.   
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: TrueTears question thread
« Reply #445 on: July 28, 2009, 10:21:08 pm »
0
If for some constant k and the 2 vectors are parallel then does also stand true if and only if the 2 vectors are parallel AND in the same direction?
« Last Edit: July 28, 2009, 10:29:08 pm by TrueTears »
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

Mao

  • CH41RMN
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 9181
  • Respect: +390
  • School: Kambrya College
  • School Grad Year: 2008
Re: TrueTears question thread
« Reply #446 on: July 28, 2009, 10:38:13 pm »
0
A trapezium is defined as a quadrilateral with two parallel (but not necessarily equal in length) sides.

There is one assumption though: OA || CB, in which case if you get the same trapezium, flip it and attach it to the right, you'll see that it forms a parallelogram, hence PQ + QP = OA + CB , and the ratio is
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015

Mao

  • CH41RMN
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 9181
  • Respect: +390
  • School: Kambrya College
  • School Grad Year: 2008
Re: TrueTears question thread
« Reply #447 on: July 28, 2009, 10:45:02 pm »
0
If for some constant k and the 2 vectors are parallel then does also stand true if and only if the 2 vectors are parallel AND in the same direction?


if k can be negative, then they don't need to point in the same direction.

In general, for some constant k element of real, if , then a and b are parallel, not necessarily pointing in the same direction.
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: TrueTears question thread
« Reply #448 on: July 28, 2009, 11:02:39 pm »
0
Hmm what if you got: a and b pointing in opposite directions.



and

so    []

but , in this case, meaning they have to be in same direction for it to be true? Or am I completely wrong.
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

Mao

  • CH41RMN
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 9181
  • Respect: +390
  • School: Kambrya College
  • School Grad Year: 2008
Re: TrueTears question thread
« Reply #449 on: July 28, 2009, 11:05:11 pm »
0
oh, right, that's what you mean. if |a| = k|b|, then a = kb can only be satisfied when they are pointing in the same direction, since k is restricted to non-negative numbers.

but you generally wouldn't be dealing with that kind of condition.
Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015