VCE Specialist 3/4 Question Thread!

[zvezda]

let z=x+yi, all points that lie on z are in the form (3cos(t)+1,3sin(t)+2)

if w=2z, then w=2(x+yi) = 2x+2yi, thus, double both the real and imaginary components to get (6cos(t)+2,6sin(t)+4)
now, x=6cos(t)+2 and y=6sin(t)+4,

solve for cos(t) and sin(t)
cos(t) = (x-2)/6 and sin(t) = (y-4)/6

since cos^2(t) + sin^2(t) =1
((x-2)/6)^2 + ((y-4)/6)^2 = 1
(x-2)^2/36 + (y-4)^2/36 = 1
(x-2)^2 + (y-4)^2 = 36

hence, w is a circle with centre (2,4) and radius 6 units.

Are you familiar with the essentials textbook? If you are, is what you did from chapter 1H? parametric equations?

[Homer]

Yes, I think so

[Jaswinder]

how would i find the implied domain and range of tan(2cos-1(x)). I seem to have trouble understanding the concept behind implied domain and range questions :/

[Conic]

how would i find the implied domain and range of tan(2cos-1(x)). I seem to have trouble understanding the concept behind implied domain and range questions :/

It's a composite function, and the domain of arccos(x) is [-1,1] if that helps.

[polar]

how would i find the implied domain and range of tan(2cos-1(x)). I seem to have trouble understanding the concept behind implied domain and range questions :/

remember that where and needs to be satisfied for to be defined

[duhherro]

Could anyone help me with subtend angle arc stuff?

As seen in 1 D , Q2) b) of the essentials textbook. Not sure how it is 62 :/



And for Q10 b of 1C, how come we have to use the cosine rule but not the sin in working out the angle A ?

[b^3]

For the first one basically we have the situation below.

Now we know that TRW is subtended by the same arc as angle TSW, that is along the sector TW. That means we can say that the two angles are equal, so

For the sine rule to work, we need another angle, but we only know the three side lengths of the triangle. The cosine rule will work as we don't need to know the angle as long as we know the three side lengths.

EDIT: also LaTeX is still stuffing up

[duhherro]

For the first one basically we have the situation below.

Now we know that TRW is subtended by the same arc as angle TSW, that is along the sector TW. That means we can say that the two angles are equal, so

For the sine rule to work, we need another angle, but we only know the three side lengths of the triangle. The cosine rule will work as we don't need to know the angle as long as we know the three side lengths.

EDIT: also LaTeX is still stuffing up

Thanks b^3!

So subtend is pretty much two triangles with the same angles?

And also for the 2nd question, I actually did work out an angle in a). And I used that in the sin rule equation, but i think it has to do with the fact that it is NOT an non-included angle.

[b^3]

Subtend is more, well, the angle angles have the same cord (the lenght along the circle) in common, which means the two angles are the same.

As for the sine rule, you might have ended up with an ambiguous case of the sine rule, does minusing what you have from 180 give the correct answer?

[507]

A body is travelling at 20m/s when it passes point P and 40m/s when it passes point Q. Find its speed when it is halfway from P to Q, assuming uniform acceleration.

Also was looking for a bit of advice. I'm picking up spesh 3/4 without having done 1/2. Any advice on what should I do before school starts to prepare me for the 3/4 course? So far I've worked through the 1/2 vectors, polar coords/complex numbers, loci, kinematics and statics of a particle without any problems, about to to circular functions II, trig ratios and the Toolbox chapter from essentials. Thanks

[duhherro]

Subtend is more, well, the angle angles have the same cord (the lenght along the circle) in common, which means the two angles are the same.

As for the sine rule, you might have ended up with an ambiguous case of the sine rule, does minusing what you have from 180 give the correct answer?

so in our case, chord RT and SW the same??

And yeah haha , for the sine rule, minusing 180 gives us the answer :/ . Do you always use the cosine rule to be safe then?

[b^3]

What I mean is they both have the arc TW in common, and TW will be the same length as TW.... if that makes sense. So if they both 'subtend' this same arc then the angles are the same.

With the sine rule, you can get the ambiguous case sometimes. I would draw up a diagram, but essentials already has a nice on for example 12, and it goes through it a bit better too. In simple terms though, with the information given we could possibly have two triangles. So what you do is you work out firstly whether the information we have will lead us through the sine rule or the cos rule the quickest, and then if we have to use the sine rule, to check whether x or 180-x makes more sense to the situation.

Hope that helps

[polar]

A body is travelling at 20m/s when it passes point P and 40m/s when it passes point Q. Find its speed when it is halfway from P to Q, assuming uniform acceleration.

it's uniform acceleration so first find the acceleration:

let P be the initial position, Q be the final position:
v^2 = u^2 + 2as
(40)^2 = (20)^2 + 2as
a = ((40)^2-(20)^2)/(2s)
a = 600/s

now, let the distance be s/2 (half the original distance), take this point as the final position, use v^2 = u^2 + 2as again
v^2 = (20)^2 + 2a(s/2)
v^2 = (20)^2 + 2(600/s)(s/2)
v^2 = (20)^2 + 600
v^2 = 1000
v = sqrt(1000) - positive direction as P to Q
v= 10sqrt(10) ms^-1

[duhherro]

What I mean is they both have the arc TW in common, and TW will be the same length as TW.... if that makes sense. So if they both 'subtend' this same arc then the angles are the same.

With the sine rule, you can get the ambiguous case sometimes. I would draw up a diagram, but essentials already has a nice on for example 12, and it goes through it a bit better too. In simple terms though, with the information given we could possibly have two triangles. So what you do is you work out firstly whether the information we have will lead us through the sine rule or the cos rule the quickest, and then if we have to use the sine rule, to check whether x or 180-x makes more sense to the situation.

Hope that helps

Yeah thanks again b^3. I get what you mean by TW. as in if you make 2 triangles, the BW is common in both of them.

For ambiguous cases, how do you know whether to leave it at the base angle or do 180 minus?


And for 1D Q2) c)  I'm having trouble with working it out

Are these angle geometry questions important in the exam ? It is quite hard T_T

[Conic]

If the triangle has the following properties (copied from wikipedia) the angle will be 180-ø or ø:

- The only information known about the triangle is the angle A and the sides a and b
- The angle A is acute (i.e., A < 90°).
- The side a is shorter than the side b (i.e., a < b).
- The side a is longer than the altitude of a right angled triangle with angle A and hypotenuse b (i.e., a > b sin A).