VCE Specialist 3/4 Question Thread!

[ashoni]

Need help with a questions from the Essentials textbook.

27. (Ch. Review 4)

If Arg(z) = pi/4 and Arg(z-3) = pi/2 find Arg(z-6i).

28a.

If Arg(z+2)= pi/2 and Arg(z) = 2pi/3, find z.

Thanks

[lzxnl]

For 27:
Let z=x+yi
Then arg z=pi/4 =>y/x=1 and x>0
arg (z-3) = pi/2 => z-3 is purely imaginary => Re(z)=3 => x=3 => z = 3+3i

For 28a:
arg(z+2) = pi/2 => z+2 is imaginary => Re(z)=-2
arg(z)=2pi/3 => y/x=tan 2pi/3=-sqrt 3
x=-2
y=2 sqrt 3
z=-2+2 sqrt 3

Just write z in cartesian form and use as much information that you're given as possible. It should fit together in the end (:

[brightsky]

alternatively, sketch the graphs and find the intersection. basically in each question, you want to find the set of all complex numbers z such that the two given conditions are satisfied. for 27, sketch the two graphs in your head, and you would see that the two graphs intersect of (3,3). to find Arg(z-6i) just visualise the point (3,3), move it down 6 and consider the angle that this new point makes with the pos dir of the real axis. you should find that the answer is -pi/4. for 28a, again sketch the two graphs in your head. the intersection point is (-2, 2sin(pi/3)) = (-2,2tan(pi/3))=(-2,2sqrt(3)).

[Henreezy]

if c, d and f are complex numbers such that c = 1/4cis(0.2pi), d=8cis(-0.1pi) and f=c/conjugate of d, then f is equal to:

I thought it would be
1/32cis(0.1pi) but apparently it is 1/32cis(0.3pi)... can anybody confirm or deny this?

[Jayward]

if c, d and f are complex numbers such that c = 1/4cis(0.2pi), d=8cis(-0.1pi) and f=c/conjugate of d, then f is equal to:

I thought it would be
1/32cis(0.1pi) but apparently it is 1/32cis(0.3pi)... can anybody confirm or deny this?

yeah i think youre right. whoever wrote the answer probably forgot to multiply the argument of 'd '  by -1 to make 'd bar'.

[Homer]

i dunno how to? 

[brightsky]

a = -0.4 x
d/dx(1/2v^2) = -0.4x
solve the diff equation, noting the initial condition given, and then substitute x=100.

[Homer]

Aeroplane can travel at a speed of 240km/h when there is no wind. today there's a wind velocity of -3i-7j km/h. Find

a) the position vector required if the aeroplane is to end up due east of its starting point after 2 hours.

[ashoni]

need help with these questions...

11.If r = 3i+3j-6k, s=i-7j+6k and t=-2i-5j+2k, find the values of xand y such that the vector r +xs+yt is parallel to the x axis.


17.ABC is a right-angled triangle with the right angle at B. If vector AC= 2i+4j and vector AB is parallel to i+j, find vector AB.

Thanks

[b^3]

(Try attempting the problem with the help before looking in the spoilers)

11. Combine each component.

Then for the vector to be parallel to the axis, the and components must be zero. So you can equate these two components to zero, get two equations with two unknowns, and solve for and .

Spoiler

17. Draw out the triangle to help you see the situation, then put the arrows for the vectors in the right direction, i.e. for vector AB, the arrow from A to B. Since is parallel to , .
Then


Now for the angle B to be a right angle, the dot product of the two vectors that meet at B must be zero, that is the dot product of and is zero.

Spoiler

But

Hope that helps

EDIT: Fixed the mistake as pointed out by 507

[zvezda]

find the gradient of the curve with equation x^2 - [(y-2)^2]/4 = 1 at the point where x = 2 and y<0.

I got dy/dx and i subbed in x = 2 into the expression for dy/dx. But the thing is (I know the answer and i understand it, im just wanting to see what was wrong with my working), i then graphed it.
So what i mean is, when x = 2, dy/dx = 8/(y-2). i graphed this and then said that dy/dx = (-4, 0). Why is this wrong?

[507]

Shouldn't it be :O

i graphed this and then said that dy/dx = (-4, 0). Why is this wrong?

Not quite sure what you mean here. I would just sub your x value into the initial equation which will give you a positive and negative value for y, so take the positive as y>0. Then sub this co-ordinate into the derivative that you found of and you should get your gradient. (Sorry if you already did this lol.)

[b^3]

Shouldn't it be :O

Yeh it should be, will fix now.

[zvezda]

Shouldn't it be :O

Not quite sure what you mean here. I would just sub your x value into the initial equation which will give you a positive and negative value for y, so take the positive as y>0. Then sub this co-ordinate into the derivative that you found of and you should get your gradient. (Sorry if you already did this lol.)

Yeah i looked at the answers and it had that process, and i understand it. I just cant understand why this one is wrong. To be clearer, I had 8/(y-2) = dy/dx after subbing in x=2. I then graphed this hyperbola and took all of the dy/dx values for the domain of y<0, which is (-4,0). Dont know why this is wrong though

[b^3]

Because you have a relation, and are given a value of , there are only two values of (in this case) that will correspond to that . So you have to take the positive one.

What you're doing is you're finding a for any value of , but evaluating the derivative with . You won't have all those values of the derivative on that curve, just those that have the value that corresponds to .