VCE Specialist 3/4 Question Thread!

[paulsterio]

what is the general approach to finding the complex roots of a polynomial if you are given a real root eg. z=a?
if you sub in that factor in the form (z-a) into the original polynomial, is that going to help in any way??

It will usually be a cubic, divide your cubic by z - a in order to get a quadratic, then just complete the square, or use the quadratic formula or factorise, whichever is appropriate.

[Bhootnike]

what is the general approach to finding the complex roots of a polynomial if you are given a real root eg. z=a?
if you sub in that factor in the form (z-a) into the original polynomial, is that going to help in any way??

It will usually be a cubic, divide your cubic by z - a in order to get a quadratic, then just complete the square, or use the quadratic formula or factorise, whichever is appropriate.

haha thanks a lot. forgot how to do it :S

[mxglasses]

can someone help me with this Q? i don't really know where to really start

A vessel, filled with liquid, is being emptied. The rate at which the liquid is flowing out at any instant is proportional to the volume remaining at that instant. If one quarter of the vessel is emptied in 5mins, what fraction remains after 10mins?

[b^3]

can someone help me with this Q? i don't really know where to really start

A vessel, filled with liquid, is being emptied. The rate at which the liquid is flowing out at any instant is proportional to the volume remaining at that instant. If one quarter of the vessel is emptied in 5mins, what fraction remains after 10mins?

The key here is to know that if the rate is proportional to the volume, then it is equal to constant K mutipled by the volume. Then to integrate it you need to flip it and you'll get t in terms of k, then rearrange.

Then from that you have two points and , from that you should be able to find and , then a fraction of the initial and sub in mins. Hope that helps

[mxglasses]

that's so smart! Man.. i should really get use to the wording of these problems.
anyways that's plenty of help! thankyou 

[mxglasses]

2 more Qs

2) If the water in a tank escapes through a hole in the bottom, the rate of flow is proportional to the square root of the depth of the water in the tank. Show that if the tank is cylindrical with its axis vertical, the time required for three-quarters of the water to flow out is equal to the time required for the remaining one-quarter to flow out.

3) Water is leaking from a cylindrical vessel at a rate that is proportional to the square root of the depth of water remaining. If the depth drops from 36cm to 25cm in one minute, how much longer will it take the vessel to empty?

is the starting point for both of them dv/dt= -k ? , i don't get what goes next after this

[b^3]

Basically yes it is the starting point, then you need to use the chain rule.

As it is a cylinder.

Now we don't know , but we know that as the water drains out that it isn't changing, i.e. it is a constant, so we can diff that equation like so
That will give
From there flip it and integrate it (remebering that is constant because its a cylinder (it won't be for things like a cone e.t.c). Then you will have in terms of and you have your two points. Then see if you can get it from there.

Just want to stress that will only be constant for cylinders standing upwards, for cones you would need to use ratios.

Anyway, hope that helps

EDIT: To make it a bit simpler, as is constant you could just represent it with (as it is the cross-sectional area after all). So you would have so

[soccerboi]

How do you construct a differential eqn for this:
The gradient of a normal to a curve at any point (x,y) is three times the gradient of the line joining the same point to the origin.

[brightsky]

-dx/dy=3y/x

[b^3]

The gradient will be given by the derivative, then remembering that tangents and normals are perpendicular, so
So the gradient of the normal is

The gradient of the line joining the point to the origin will just be rise over run, that is

So then the gradient of the normal is 3 times this
So we have


EDIT: Like brightsky has done, you probably wouldn't have to rearrange for dy/dx as it would still be a differential equation.

[tonyduong]

I have a problem finding the cartesian equation of |z+2i|=3^1/2*|z-3|. I tried to substitute z=x+yi except the equations were looking ugly. If you could help that would be great. Thanks.

[kamil9876]

I assume you mean:









Now complete the square to and and at the end of the day you will actually get some circle.

[mxglasses]

Ty b^3! i followed the steps and integrated it, but i don't know if the values i'm subbing in are the right ones. The ones i subbed in were t=1 when h= 25, and t=0 when h=36. i keep getting 5 mins as my answer but the solution says it's 6mins.

[b^3]

Integrating you should end up with
Did you add the ?
You have the two points, which give the correct answer of 6 mins.
Solving for you should get .
Then you should be able to solve for , getting .
Then substituting that those into the equation, you should get , then you want when the vessel is empty, that is when , putting that in you should get .

You might of forgot the C or swapped the h values around for the two coordinates.

EDIT: Just had another look, you probably did swap the h values when you worked it out, it should be as you said when and when . Using those you get the correct answer of 6 mins, swapping them around will give you the 5 mins that you got.

Hope that helps

[mxglasses]

hahaha nice observation  i did switch the h values LOL .. anyways ty again