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March 29, 2024, 09:08:30 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164379 times)  Share 

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wyzard

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Re: Specialist 3/4 Question Thread!
« Reply #8655 on: April 27, 2017, 04:21:06 pm »
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^^^Thanks.

For the Ti Nspire is the easiest way to solve Euler Method type questions with the spreadsheets screen?

For specialist math, you're not required to use a spreadsheet to apply Euler's Method. I will not say it's the easiest way, but definitely the most efficient as repeated application of Euler's method becomes very tedious and repetitive. The main difficulty is setting up the correct formulas in a spreadsheet, which can be done in the CAS or Microsoft Excel, personally I prefer Excel :P
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deStudent

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Re: Specialist 3/4 Question Thread!
« Reply #8656 on: April 27, 2017, 04:41:06 pm »
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For specialist math, you're not required to use a spreadsheet to apply Euler's Method. I will not say it's the easiest way, but definitely the most efficient as repeated application of Euler's method becomes very tedious and repetitive. The main difficulty is setting up the correct formulas in a spreadsheet, which can be done in the CAS or Microsoft Excel, personally I prefer Excel :P
So in exam 2, how would I do these sorts of questions with the CAS?

wyzard

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Re: Specialist 3/4 Question Thread!
« Reply #8657 on: April 27, 2017, 04:48:03 pm »
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In exam 2 usually you'll be asked to find the new coordinates after at most 2 steps, which you can apply Euler's method easily without using a spreadsheet.
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lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #8658 on: April 27, 2017, 10:47:44 pm »
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Also Sine mentioned before sometimes they make it into Ae1/2x2 for example - this is often so they can get rid of the modulus sign, as A is ±ec
So if you had |y|=e1/2x2+c you could make it into y=Ae1/2x2, and solve for A without having to worry about whether it's positive or negative (finding A will determine that)

There's one more reason. This form allows A to be zero. If your initial condition is y(1) = 0, for instance, then your solution is y = 0.
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deStudent

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Re: Specialist 3/4 Question Thread!
« Reply #8659 on: May 06, 2017, 05:22:31 pm »
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http://m.imgur.com/a/E8tFg

For part d) I don't understand how the suggested answer's method actually gives us the area? They're just multiplying 2 lengths together?

Thanks

Quantum44

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Re: Specialist 3/4 Question Thread!
« Reply #8660 on: May 06, 2017, 05:40:52 pm »
+2
http://m.imgur.com/a/E8tFg

For part d) I don't understand how the suggested answer's method actually gives us the area? They're just multiplying 2 lengths together?

Thanks

Area of triangle = 0.5*base*height

I haven't actually done the question but since m is the midpoint, I'd assume that one of the lengths is the height of the triangle and the other length is half the base of the triangle.
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Syndicate

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Re: Specialist 3/4 Question Thread!
« Reply #8661 on: May 07, 2017, 11:42:11 am »
+2
http://m.imgur.com/a/E8tFg

For part d) I don't understand how the suggested answer's method actually gives us the area? They're just multiplying 2 lengths together?

Thanks

This is an isosceles triangle (or you can just say that two right angled triangles joined together).

Area of ABC = Area of ABM x 2
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peanut

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Re: Specialist 3/4 Question Thread!
« Reply #8662 on: May 07, 2017, 01:13:44 pm »
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Say we have a graph C(t) which has a horizontal asymptote at 1. How do we label this when sketched? y=1? C(t) = 1? C = 1?

Sine

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Re: Specialist 3/4 Question Thread!
« Reply #8663 on: May 07, 2017, 01:42:04 pm »
+1
Say we have a graph C(t) which has a horizontal asymptote at 1. How do we label this when sketched? y=1? C(t) = 1? C = 1?
If you are sketching it as y = C(t) it's fine to use y = 1
If you are sketching it as C as a function of t use C(t) = 1

deStudent

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Re: Specialist 3/4 Question Thread!
« Reply #8664 on: May 07, 2017, 05:46:40 pm »
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http://m.imgur.com/a/ClQyE
For Q30a) shouldn't cos be both positive in the working since angle 120 is in quadrant 4 when using true bearings?
Thanks

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8665 on: May 07, 2017, 09:01:23 pm »
+1
http://m.imgur.com/a/ClQyE
For Q30a) shouldn't cos be both positive in the working since angle 120 is in quadrant 4 when using true bearings?
Thanks

The unit vector is really
cos(-30º)i + sin(-30º)j
The component in the i (x) direction is cos(theta), the component in the j (y) direction is sin(theta), in this case theta = -30º. It also makes sense as the runner is travelling down, so the j component will be negative.
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deStudent

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Re: Specialist 3/4 Question Thread!
« Reply #8666 on: May 08, 2017, 12:13:36 am »
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The unit vector is really
cos(-30º)i + sin(-30º)j
The component in the i (x) direction is cos(theta), the component in the j (y) direction is sin(theta), in this case theta = -30º. It also makes sense as the runner is travelling down, so the j component will be negative.
Thanks. But how did you know you can use sin? The answer used cos for both, they used the formula costheta = a1/|vector|.

Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8667 on: May 08, 2017, 03:48:05 pm »
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Thanks. But how did you know you can use sin? The answer used cos for both, they used the formula costheta = a1/|vector|.
What I did was just using the unit circle. x=cos(theta) and y=sin(theta), and radius/modulus/length is 1.
For a question like this, it would be easier using sin and cos, but what I'm guessing they did was a.b=|a||b|costheta so v.i=1*1*costheta, x = cos(30) = (√3)/2 where x is the component of the unit vector in the i direction, and v is the unit vector
v.(-j) = -y=1*1*cos(60º)=1/2 so y= -1/2 where y is the component of the unit vector in the j direction (because it's multiplied with the unit vector in the negative j direction, it's -j). If it were multiplied by j (unit vector in the positive direction, up) it would be y = 1*1*cos(120º)= -1/2 (same answer), it's just the angle that would change.
Sorry for the horrible notation, I'm rusty but I hope you can follow my reasoning :)
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geminii

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Re: Specialist 3/4 Question Thread!
« Reply #8668 on: May 08, 2017, 05:56:55 pm »
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Hey everyone!
I was doing a question where it said convert the complex number -1 + i to polar form. I know r is |√((-1)^2+(1)^2) = √2, but this is what happens when I try to find theta.

theta = tan^-1(1/-1) = tan^-1(-1) = -pi/4.

My question is, when I draw -1 + i on an Argan diagram, it is in the second quadrant and the angle is 3pi/4. Why am I getting -pi/4??  ??? ??? ???

Thanks in advance!!
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Syndicate

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Re: Specialist 3/4 Question Thread!
« Reply #8669 on: May 08, 2017, 06:04:18 pm »
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Hey everyone!
I was doing a question where it said convert the complex number -1 + i to polar form. I know r is |√((-1)^2+(1)^2) = √2, but this is what happens when I try to find theta.

theta = tan^-1(1/-1) = tan^-1(-1) = -pi/4.

My question is, when I draw -1 + i on an Argan diagram, it is in the second quadrant and the angle is 3pi/4. Why am I getting -pi/4??  ??? ??? ???

Thanks in advance!!
are
You add -pi/4 to pi (and add whatever you get if the coordinates are in the third quadrant). Why? Because arctan only has a range of (-pi/2, pi/2).
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