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April 27, 2024, 10:51:33 pm

Author Topic: Dekoyl's Questions  (Read 21575 times)  Share 

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dekoyl

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Dekoyl's Questions
« on: March 12, 2009, 10:56:45 pm »
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I think I should have a questions thread like my methods one from last year =]
Excuse me if some questions are obvious. Sometimes I need a little nudge for things to click.
Thanks in advance to everyone.

1. Let . A tangent to the graph of where makes an angle of with the positive direction of the x-axis. Find the possible values of .
Answer:


2. Find (to 2 d.p.), the angle between the line and the tangent to the graph of at the point of (2,2).
I got the tangent which is, I think,
Answer:
« Last Edit: March 12, 2009, 10:58:54 pm by dekoyl »

TrueTears

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Re: Dekoyl's Questions
« Reply #1 on: March 13, 2009, 11:33:15 am »
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Ok so we know that the angle the tangent makes with the positive direction of the x axis is (where m is the gradient)

but we also know that the derivative of tan(2x) also gives the gradient at any point on the graph.

so

so equate      (70 degerees in radians is )

so (just solve for a)

o crap lunch is over i g2g lol... anyways ill type up the rest when i get home, but that should help you if you see this :)

PhD @ MIT (Economics).

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

dekoyl

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Re: Dekoyl's Questions
« Reply #2 on: March 13, 2009, 02:51:39 pm »
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Ok so we know that the angle the tangent makes with the positive direction of the x axis is (where m is the gradient)

but we also know that the derivative of tan(2x) also gives the gradient at any point on the graph.

so

so equate      (70 degerees in radians is )

so (just solve for a)

o crap lunch is over i g2g lol... anyways ill type up the rest when i get home, but that should help you if you see this :)


Ah yeah =_= Forgot about trying to convert to radians.

Thanks TrueTears

TrueTears

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Re: Dekoyl's Questions
« Reply #3 on: March 13, 2009, 04:17:48 pm »
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k here it is



as you can see in the diagram, is the angle we are looking for and this is equal to

so (just form a triangle)



and

so

so
« Last Edit: March 13, 2009, 04:44:10 pm by TrueTears »
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dekoyl

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Re: Dekoyl's Questions
« Reply #4 on: March 14, 2009, 02:09:26 pm »
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Thanks for the diagram TrueTears =]

Another one.. I think I'm misinterpreting the question =\
3. Given that . The points are collinear. Which one of the following is not true?

is parallel to
is parallel to
is parallel to
is parallel to
is parallel to

Don't they all lie on the same, straight line?

TrueTears

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Re: Dekoyl's Questions
« Reply #5 on: March 14, 2009, 02:13:44 pm »
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Is it is parallel to ?
PhD @ MIT (Economics).

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dekoyl

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Re: Dekoyl's Questions
« Reply #6 on: March 14, 2009, 02:15:30 pm »
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Is it is parallel to ?
I'm not sure myself =P No answer given to this question.

TrueTears

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Re: Dekoyl's Questions
« Reply #7 on: March 14, 2009, 02:17:22 pm »
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wait nvm i think its the last one
« Last Edit: March 14, 2009, 02:49:35 pm by TrueTears »
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/0

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Re: Dekoyl's Questions
« Reply #8 on: March 14, 2009, 02:41:28 pm »
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There is no guarantee that and are parallel. It can be true under certain conditions however.

The first 4 will give vectors along the same line.
« Last Edit: March 14, 2009, 02:59:41 pm by /0 »

dekoyl

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Re: Dekoyl's Questions
« Reply #9 on: March 14, 2009, 05:35:52 pm »
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^Thanks TT and /0.

I can't visualise this very well =| (Unsure about where point D goes)
4. Let m > n >0 and a, b be the position vectors of two points A and B respectively. Point C, D divide the line segment AB in a ratio of m:n internally and externally respectively. i.e. C is between A, B and AC:CB = m:n. D is not between A,B and AD:DB = m:n.

I don't have a problem with what the question asks me to do just drawing the diagram/visualising part.

Thanks.

TrueTears

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Re: Dekoyl's Questions
« Reply #10 on: March 14, 2009, 05:50:11 pm »
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Okie so it means this, here is the diagram first



1. C is between A, B and AC:CB = m:n

so from this we can conclude

so cross multiply yields nAC = mCB

therefore

2. D is not between A,B and AD:DB = m:n.

doing the same as part 1. yields

But we know m > n > 0 so for example to get an idea of the diagram let m = 4 and n = 2

this means which is same as saying AC = 2CB, so this means AC is TWICE the length of CB, and in the diagram you can see this

the same goes for AD, so , so this means AD is TWICE the length of DB, again in the diagram you can see this.

Now that just gives an idea of the diagram, obviously we don't know the values of m and n, but at least now you can have a good idea of the diagram to do the question.
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Interested in asset pricing, econometrics, and social choice theory.

dekoyl

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Re: Dekoyl's Questions
« Reply #11 on: March 14, 2009, 05:54:59 pm »
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Ahh thanks TT. Didn't know it was a straight line.

[edit]

Yeah just that "internally,externally" made me visualise it as something that was enclosed =\
v
v
« Last Edit: March 14, 2009, 06:00:11 pm by dekoyl »

TrueTears

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Re: Dekoyl's Questions
« Reply #12 on: March 14, 2009, 05:57:12 pm »
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nps XD

everytime you see something like AD = kDB where (see how you have D at the 'end' of the LHS and D at the 'start' of the RHS), it means that they are collinear
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dekoyl

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Re: Dekoyl's Questions
« Reply #13 on: March 20, 2009, 09:35:53 pm »
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I just got one question that I'm not too sure about what it's asking.
This is the exact wording of the question.

Let and . If the vector is equal to the unit vector in the positive direction of the x-axis, find the constants of and .

I got     :-\

Thanks

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Re: Dekoyl's Questions
« Reply #14 on: March 20, 2009, 09:47:01 pm »
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I think this is what it is asking: