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March 28, 2024, 11:22:58 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164198 times)  Share 

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soccerboi

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Jenny_2108

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Re: Specialist 3/4 Question Thread!
« Reply #841 on: October 27, 2012, 02:12:41 pm »
0
^ When you calculate magnitude of acceleration, you take square root of i,j, k components right?
Eg: a= xi+yj+zk
      |a| = sqrt(x^2+y^2+z^2)
They don't take square root of i,j,k components so they have to square both sides
      |a|^2= x^2+y^2+z^2

sr, I'm using mobile so its hard to type LaTex :(

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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #842 on: October 27, 2012, 03:56:17 pm »
+1
If the square of |a| is constant then so is |a|, so you don't have have to square root it, but of course it's not wrong to square root it.
Tim Koussas -- Co-author of ExamPro Mathematical Methods and Specialist Mathematics Study Guides, editor for the Further Mathematics Study Guide.

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kensan

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Re: Specialist 3/4 Question Thread!
« Reply #843 on: October 29, 2012, 07:36:19 pm »
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and y=2 when x=0, then the value of y correct to three decimal places when x=3.

Ok so I'm having trouble integrating this, CAS just gives me the exact same thing. So how would I go about diong this? thanks :)
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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #844 on: October 29, 2012, 07:41:42 pm »
+5




Tim Koussas -- Co-author of ExamPro Mathematical Methods and Specialist Mathematics Study Guides, editor for the Further Mathematics Study Guide.

Current PhD student at La Trobe University.

rife168

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Re: Specialist 3/4 Question Thread!
« Reply #845 on: October 29, 2012, 08:09:39 pm »
+2
and y=2 when x=0, then the value of y correct to three decimal places when x=3.

Ok so I'm having trouble integrating this, CAS just gives me the exact same thing. So how would I go about diong this? thanks :)

Note that there in fact isn't actually an analytical solution to so your calculator has to solve the original question using some kind of numerical method, like taking the sum of areas of very small rectangles or trapeziods, which will end up being a decent, but importantly not exact, approximation. It's like the difference between finding an exact solution to a differential equation or finding an approximation using Euler's rule, which we did as a part of DEs.
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generalkorn12

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Re: Specialist 3/4 Question Thread!
« Reply #846 on: October 31, 2012, 09:22:11 am »
0
For Vector Proofing (such as, parallelogram and rhombus), whats the difference between a=b and |a|=|b|, I seem to get these confused...

An example would be from the 2008 VCAA Exam 1:

A (1,0,5), B(-1,2,4), C(3,5,2), D(x,y,z). Determine the Points of D such that ABCD is a Parallelogram.

The worked solutions use DC=AB, but I figured, since its a parallelogram wouldn't the lengths be equal?

paulsterio

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Re: Specialist 3/4 Question Thread!
« Reply #847 on: October 31, 2012, 11:25:55 am »
+1
For Vector Proofing (such as, parallelogram and rhombus), whats the difference between a=b and |a|=|b|, I seem to get these confused...

An example would be from the 2008 VCAA Exam 1:

A (1,0,5), B(-1,2,4), C(3,5,2), D(x,y,z). Determine the Points of D such that ABCD is a Parallelogram.

The worked solutions use DC=AB, but I figured, since its a parallelogram wouldn't the lengths be equal?

Remember that a vector has both direction and magnitude. If you say a = b, that means that both the magnitude and the direction of vectors a and b are the same - it means that a and b are the same vector. An example of this would be if both a and b were 2i + 3j (for example).

If you say that |a| = |b|, it means that the magnitudes of a and b are the same, but their directions might not be, for example, if a = i + 3j and b = 3i + j, they have the same magnitudes, but different directions and thus are different vectors.

If you want to prove that something is a parallelogram, you would have to prove DC = AB because the opposite sides have to be both EQUAL IN LENGTH and PARALLEL (same direction).

soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #848 on: November 03, 2012, 12:15:33 pm »
0
When plotting roots of a complex number on an argand plane, why is it incorrect to connect the point with the origin?
e.g VCAA 2006 ex 2, last question.


Also, can someone please explain to me how to do the attached question from VCAA 2006 ex2.
Thanks
« Last Edit: November 03, 2012, 12:24:01 pm by soccerboi »
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polar

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Re: Specialist 3/4 Question Thread!
« Reply #849 on: November 03, 2012, 12:36:23 pm »
+2
the complex number is just a point at because it is a number and not a line. if you drew a line from to that would be the line which is not just the complex number being represented anymore

the line joining and is the line but it needs to be written in terms of



note that in this case, it was and which meant that the real and imaginary components were positive, thus, didn't require the modulus signs
« Last Edit: November 03, 2012, 01:04:25 pm by polar »

soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #850 on: November 05, 2012, 12:07:40 pm »
0
A set of vectors are linearly dependent if one of them can be expressed as a sum of non zero multiples of another two vectors.

In VCAA 2011 exam 1, Q 9 c)ii), They have it as 3c-a=nb. This subtraction is not a"sum," so why is it linearly dependent?
I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

Thanks in advance  :)
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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #851 on: November 05, 2012, 12:28:26 pm »
+2
If one can be expressed as a linear combination of the other two, then they're linearly dependent. Doesn't matter if the coefficients are negative.

The geometric interpretation for dependence of three vectors in three dimensions is that they lie in the same plane.
Tim Koussas -- Co-author of ExamPro Mathematical Methods and Specialist Mathematics Study Guides, editor for the Further Mathematics Study Guide.

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paulsterio

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Re: Specialist 3/4 Question Thread!
« Reply #852 on: November 05, 2012, 12:31:22 pm »
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I know it can be expressed as 3c=a+nb; is this why its linearly dependent? I'm just a bit confused because it doesn't look like the general form that i'm familiar with: a=mb+nc.

It's actually the same form, m and n are just constants, so you can multiply and divide them by whatever constants and they'll still be constants.

soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #853 on: November 05, 2012, 12:33:28 pm »
0
ok thanks :)

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath1-w.pdf

For Q 7b),
I did mg+T2+T1=0
        mg+98+(98/root3)=0
and solved for m, getting a negative answer...

Whats wrong with me setting in out like this?
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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #854 on: November 05, 2012, 12:38:05 pm »
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You need to take into account the directions
Tim Koussas -- Co-author of ExamPro Mathematical Methods and Specialist Mathematics Study Guides, editor for the Further Mathematics Study Guide.

Current PhD student at La Trobe University.