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Mid-term exams...AAAUUGGGHHH!!. Taking a break.


Jedi Luke

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God, I had my maths one yesterday and my english one this morning. I don't think I did too badly at all. I'm looking for a very high 93% + score (hopefully). Now it's studying for Specialist Maths, Vectors (God that gets confusing). Damn, I've got a whole heap of physics to do as well on sound. Oh well, thanks again ppl! :)

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Originally posted by ET Warrior

I've taken 3 exams in the last week, and one more on friday. So far the scores are...

 

70 on Calc 3

88 on Psychology

100 on Physics

 

 

Data structures in a couple of days. Meh.

 

Your Physics sure is good. I've had my Physics paper, I don't know almost all of the questions. x_x I'm sure to fail it.

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Hey ET, I've got a question regarding Vectors.....:)

 

Q.) Express each vector as a product of its length and direction.

 

(a) i + 2j - 2k

 

Ok, the answer to this question is: 3( (1/3)i + (2/3)j - (2/3)k )

 

I understand how you get the answer. Find the magnitude (which is of course a scalar quantity and does not affect the direction of vectors), of the vector quantities, then for each vector quantity, divide by the magnitude so expansion of the final expression is the original expression.

 

My problem basically is that I didn't understand what the question asked me to do, or what this answer means. Could you please explain it to me ET or anyone? Thanks :)

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Well since you know how to find the magnitude, and divide by it I won't go into detail on how to do that, but now that you've divided your original vector by the magnitude you're left with a unit Vector (vector of length 1) pointing in the direction of the original vector. Unit Vectors can always be refered to as the direction of the vector.

 

And then the length of the vector is just the magnitude.

 

So displaying a vector as a product of length and direction is really just asking for the product between the magnitude and the unit vector.

 

Hope that helps.

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Originally posted by ET Warrior

So displaying a vector as a product of length and direction is really just asking for the product between the magnitude and the unit vector.

 

Hope that helps.

 

Yes that definately makes much more sense how you've explained it. In multiplying the unit vector by its magnitude has to result in the original expression. Thanks heaps ET for you help. :thumbsup::)

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Well I had my specialist maths assesment big test thing today. Wasn't too bad for the first section, but we're gonna finish part B and part C on monday. 20 minutes to go till the end of the lesson, he gave us part B, I had a look at it and boy it's gonna take a while to finish and it looks hard :(. Our teacher said for us to start it and have a look at it for over the weekend. I saw one question that looked tricky so looks like I'm gonna need help from ET, Rhett or anyone else that knows. Here it is:

 

(1 + itanx)^5 = (cis5x)/(cos^5x)

 

i is imaginary, ie root -1 as you know

assume x for theta (angle)

For the RHS equation, the denominator, cos^5x, is cos^5 , not 5x, so it's cos^5(x), just making sure no-one gets confused.

 

Basically, we need to proove this identity. The RHS is in fractional form (as you probably have guessed) but I can't put it as a fraction on the PC.

 

I'm too tired to think now, I'm busy studying standing waves and 2 dimensional interference........soo tired.....zzzzzzzz

 

Any help?

 

Thanks heaps everyone! :)

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Originally posted by ET Warrior

Is that supposed to be Cos(5x) in the numerator?

 

No that is supposed to be cis(5x)

 

Cis is the Polar form expression of the cartesian form expression.

 

cis is (cos(x) + i sin(x))

 

cis is short hand for: cos theta plus i sin theta. (that's the full cis expression).

 

i (root -1) is multiplying with sin(x). (x) is of course the angle.

 

You probably new all that ET and was just double-checking to be sure but just in case and for others, that's what it is.

 

Thanks.

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