Author Topic: VCE Specialist 3/4 Question Thread! (Read 2579389 times) Tweet Share

lzxnl · « **Reply #1455 on:** March 17, 2013, 08:46:48 pm »

Euler's method is really just applying linear approximation to differential equations
We have change in y = roughly dy/dx*change in x
so y-y0=dy/dx*(x-x0)
You're given the function dy/dx in terms of y and x and you're given x and y coordinates to plug into dy/dx; dy/dx=y/x=3/2 here
Also, the coordinates that are given are x0 and y0; in your case x0=2 and y0=3
This leaves y and x. However, they do ask for a specific x value; x=2.4
Therefore plug all the numbers in: y-3=3/2*(2.4-2)
y=3+1.5*0.4=3.6

Don't bother remembering all of these formulas until you know precisely where they come from. Euler's method is just linear approximation. If you can remember dy/dx is roughly change in y/change in x, you can derive the Euler method equation (:

dim_sim · « **Reply #1456 on:** March 17, 2013, 10:10:37 pm »

Hey I have a vector question.

I figured out 4a so I only need help with 4b.
(It's my first time attaching a picture so I hope it works.. :S)

Lasercookie · « **Reply #1457 on:** March 17, 2013, 11:02:33 pm »

I'm going to be lazy with my vector notation, but yeah a, b, a-hat, b-hat, i, j, k are vectors:

From the question:
$a = i - 2j + 2k$

b is the vector such that the vector resolute of a in the direction of b is equal to the unit vector of b
$(a.\hat{b})\hat{b} = \frac{a.b}{|b|^2}b = \hat{b}$

and the vector resolute of b in the direction of a is 2 times the unit vector of a
$(b.\hat{a})\hat{a} = \frac{b.a}{|a|^2}a = 2\hat{a}$

The dot product is commutative, $a.b = b.a$ , so this leaves us with enough information to solve this question

$a.b = \frac{\hat{b}|b|^2}{b} = \frac{2\hat{a}|a|^2}{a} \implies |b|^2 = \frac{2\hat{a}|a|^2b}{a\hat{b}} = 2|a| \times \frac{\hat{a}}{a} \times \frac{b}{\hat{b}}$

We'll need to simplify this out before we can solve it

$\hat{a} = \frac{a}{|a|}$ Which means that $\frac{\hat{a}}{a} = \frac{\frac{a}{|a|}}{a} = \frac{1}{|a|}$ and for $\frac{b}{\hat{b}} = \frac{b}{\frac{b}{|b|}} = |b|$

Which gives us
$|b|^2 = 2|a| \times \frac{1}{|a|} \times |b| = 2|b|$

and dividing by |b|
$|b| = 2$ which I hope is correct

dim_sim · « **Reply #1458 on:** March 18, 2013, 06:19:42 pm »

Quote from: laserblued on March 17, 2013, 11:02:33 pm

I'm going to be lazy with my vector notation, but yeah a, b, a-hat, b-hat, i, j, k are vectors:

From the question:
$a = i - 2j + 2k$

b is the vector such that the vector resolute of a in the direction of b is equal to the unit vector of b
$(a.\hat{b})\hat{b} = \frac{a.b}{|b|^2}b = \hat{b}$

and the vector resolute of b in the direction of a is 2 times the unit vector of a
$(b.\hat{a})\hat{a} = \frac{b.a}{|a|^2}a = 2\hat{a}$

The dot product is commutative, $a.b = b.a$ , so this leaves us with enough information to solve this question

$a.b = \frac{\hat{b}|b|^2}{b} = \frac{2\hat{a}|a|^2}{a} \implies |b|^2 = \frac{2\hat{a}|a|^2b}{a\hat{b}} = 2|a| \times \frac{\hat{a}}{a} \times \frac{b}{\hat{b}}$

We'll need to simplify this out before we can solve it

$\hat{a} = \frac{a}{|a|}$ Which means that $\frac{\hat{a}}{a} = \frac{\frac{a}{|a|}}{a} = \frac{1}{|a|}$ and for $\frac{b}{\hat{b}} = \frac{b}{\frac{b}{|b|}} = |b|$

Which gives us
$|b|^2 = 2|a| \times \frac{1}{|a|} \times |b| = 2|b|$

and dividing by |b|
$|b| = 2$ which I hope is correct

Hmm, the answer in the book says that |b| = 6 $:-\$

b^3 · « **Reply #1459 on:** March 18, 2013, 06:31:05 pm »

Quote from: laserblued on March 17, 2013, 11:02:33 pm

$a.b = \frac{\hat{b}|b|^2}{b} = \frac{2\hat{a}|a|^2}{a} \implies |b|^2 = \frac{2\hat{a}|a|^2b}{a\hat{b}} = 2|a| \times \frac{\hat{a}}{a} \times \frac{b}{\hat{b}}$

Laser lost the power on the magnitude of a here, so it should be |a|^{2}, which means at the end you get |b|=2|a|, evaluating the magnitude of |a| gives 3, making |b|=6. (also I got lazy with notation as well

)

dim_sim · « **Reply #1460 on:** March 18, 2013, 06:49:38 pm »

Quote from: b^3 on March 18, 2013, 06:31:05 pm

Laser lost the power on the magnitude of a here, so it should be |a|^{2}, which means at the end you get |b|=2|a|, evaluating the magnitude of |a| gives 3, making |b|=6. (also I got lazy with notation as well )

Ahh, I see. Thanks both of you.

Jeggz · « **Reply #1461 on:** March 19, 2013, 07:47:24 pm »

Guys, I have a Spesh SAC tomorrow. It's an Analysis task over 2 days, and its focuse on Partial Fractions, Trig, Co-ord Geom and Rational Functions. The thing is apparently the SAC is like a reasoning task, where you'll be given a series of multiple choice questions and you'll not only have to answer the question, but also explain why you picked that or why the others are wrong $:-\$ For instance they could give you a rational function and tell you to pick how many asymptotes the graph had and then the following question could be why can't it have "3" asymptotes ..? Has anyone else had this type of SAC? Or have any tips as to how I could prepare for this? Confused!

Daenerys Targaryen · « **Reply #1462 on:** March 19, 2013, 07:54:32 pm »

I have this kind of SAC.

It's more about giving some working to indicate how you got your answer. You can do working out and then circle the answer you got.
Or if the question doesn't have any working out, then you might have to disprove all but one option, but you must write it all down.

In terms of preparation, I'd just do exam style questions, especially the MC. And do them under the same conditions. You will begin to find it easier to know what you have to do for each kind of question. (prove one, or disprove all but one)

Jeggz · « **Reply #1463 on:** March 19, 2013, 08:37:14 pm »

Ohh thanks for that! I've been doing MC questions so I'll just finish of the rest, and hopefully that'll be enough

abcdqd · « **Reply #1464 on:** March 19, 2013, 08:45:37 pm »

should the answer be B?

Jaswinder · « **Reply #1465 on:** March 19, 2013, 09:00:16 pm »

i would say c

Holmes · « **Reply #1466 on:** March 19, 2013, 09:33:22 pm »

Quote from: abcdqd on March 19, 2013, 08:45:37 pm

(Image removed from quote.)
should the answer be B?

abcdqd · « **Reply #1467 on:** March 19, 2013, 09:35:01 pm »

but that says one of the roots is pi/8, when in the question its 7pi/8 ?

lzxnl · « **Reply #1468 on:** March 19, 2013, 09:48:33 pm »

Myurra, I have the exact same kind of SAC on Thursday and Friday. Prepare for this by doing chapter 5 of the Essentials book; it has 74 multiple choice questions and for each one, think about why the others are wrong.

Jeggz · « **Reply #1469 on:** March 19, 2013, 09:50:20 pm »

Quote from: nliu1995 on March 19, 2013, 09:48:33 pm

Myurra, I have the exact same kind of SAC on Thursday and Friday. Prepare for this by doing chapter 5 of the Essentials book; it has 74 multiple choice questions and for each one, think about why the others are wrong.

Thankyouuu! It's reassuring that my school's not the only weird one $:-\$
I'm onto it

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